Linear Algebra and Representation Theory

Derives Proposition 5.2. By Equation (5.19), \(\det A\) is a sum of products of \(n\) entries, each entry from a different column. Differentiating a product of \(n\) differentiable functions — Proposition 7.30 extended to \(n\) factors by induction — replaces it by the sum of the \(n\) products in which one factor has been differentiated. Collecting, for each \(j\), the terms in which the differentiated factor came from column \(j\) gives exactly the Leibniz expansion of the determinant with that column replaced by its derivative, which is Equation (5.20).

Now let \(A\) be invertible, so that its columns are a basis of the coordinate space (Proposition 5.49), and expand each \(\dv{A_{j}}{t}\) in that basis. The array of coefficients is \(B=A^{-1}\dv{A}{t}\), since \(AB=\dv{A}{t}\) says column by column that \(\dv{A_{j}}{t}=\sum_{i}B_{ij}A_{i}\). Substituting this into the \(j\)-th term of Equation (5.20) and using linearity in the \(j\)-th column,

because the determinant on the right has the column \(A_{i}\) standing in the \(j\)-th place and therefore repeats a column, hence vanishes, unless \(i=j\). Summing over \(j\) produces \(\tr B\) and Equation (5.21).

Derivation. Derives Equation (5.22). Both are consequences of distributivity together with the existence of additive inverses in \((\mathbb{V},+)\); note that the two zeros involved are different objects, \(0\in\mathbb{K}\) and \(0_{\mathbb{V}}\in\mathbb{V}\), and that the content of the statements is exactly that they are compatible.

For Equation (5.22), apply distributivity h) with \(a=b=0\), using \(0+0=0\) in the field:

By axiom c) the vector \(0\cdot v\) possesses an additive inverse \(-(0\cdot v)\). Adding it on the left of both sides and using associativity a), the inverse axiom c) and the neutral element b),

For Equation (5.23) the argument is the same one run on the other distributive law. By axiom b), \(0_{\mathbb{V}}=0_{\mathbb{V}}+0_{\mathbb{V}}\), so distributivity g) gives

and adding \(-(a\cdot 0_{\mathbb{V}})\) on the left of both sides yields \(0_{\mathbb{V}}=a\cdot 0_{\mathbb{V}}\).

Derivation. Derives Definition 5.12. Write out what the set is:

the linear combinations being finite — \(N\) is arbitrary but each element of \(\gen{W}\) uses only finitely many vectors of \(W\). We verify the two conditions of Definition 5.7.

1. \(\gen{W}\neq\varnothing\). By hypothesis \(W\neq\varnothing\), so pick \(w\in W\); then \(w=1\cdot w\in\gen{W}\) by axiom f), the case \(N=1\), \(a^{1}=1\).

2. Closure under linear combination. Let \(u,v\in\gen{W}\) and \(c,d\in\mathbb{K}\). By definition there are finite families

   \[ u=\sum_{i=1}^{N}a^{i}w_{i}\ec\qquad v=\sum_{j=1}^{M}b^{j}w'_{j}\ec \]

   with all \(w_{i},w'_{j}\in W\). Then

   \begin{align*} cu+dv & = c\sum_{i=1}^{N}a^{i}w_{i}+d\sum_{j=1}^{M}b^{j}w'_{j}\\ & = \sum_{i=1}^{N}c\left(a^{i}w_{i}\right) +\sum_{j=1}^{M}d\left(b^{j}w'_{j}\right) && \text{by g)}\\ & = \sum_{i=1}^{N}\left(ca^{i}\right)w_{i} +\sum_{j=1}^{M}\left(db^{j}\right)w'_{j} && \text{by e)}\ec \end{align*}

   which is again a finite linear combination of elements of \(W\), with coefficients \(ca^{i},db^{j}\in\mathbb{K}\) by closure of the field. Hence \(cu+dv\in\gen{W}\).

So \(\gen{W}\) is a vector subspace. Note in passing that \(0_{\mathbb{V}}=0\cdot w\in\gen{W}\) by Equation (5.22), as it must be for a subspace.

Derives Proposition 5.20. If \(v=0_{\mathbb{V}}\) both sides vanish, by Equation (5.38) on the left and Equation (5.31) on the right, and \(u\) and \(0_{\mathbb{V}}\) are linearly dependent; so assume \(v\neq0_{\mathbb{V}}\), whence \(\braket{v}{v}>0\) by Equation (5.32) and we may divide by it. Put

that is, subtract from \(u\) its component along \(v\). Expanding \(\braket{w}{w}\) by distributivity Equations (5.35) and (5.37) and the two scalar rules Equations (5.34) and (5.36),

Now \(\braket{v}{u}=\braket{u}{v}^{\ast}\) by hermiticity Equation (5.33), and \(\braket{v}{v}\) is a positive real by Equation (5.39), so each of the last three terms is \(\abs{\braket{u}{v}}^{2}/\braket{v}{v}\), with signs \(-,-,+\):

By positive definiteness Equation (5.32) together with Equation (5.31) we have \(\braket{w}{w}\geq0\), and multiplying through by \(\braket{v}{v}>0\) gives Equation (5.43).

For the equality case: the chain above shows that equality holds if and only if \(\braket{w}{w}=0\), that is — by Equation (5.31) — if and only if \(u=\lambda v\), and a pair \(\set{u,v}\) with \(v\neq0_{\mathbb{V}}\) is linearly dependent precisely when \(u\) is a multiple of \(v\).

Derives Proposition 5.28. By induction on \(k\). For \(k=1\): \(v_{1}\neq0_{\mathbb{V}}\) by linear independence, so \(\norm{v_{1}}>0\) by Equations (5.32) and (5.44), and \(e_{1}=v_{1}/\norm{v_{1}}\) is a unit vector generating the same line, by Equation (5.41).

Suppose \(e_{1},\ldots,e_{k-1}\) have been constructed, orthonormal and satisfying Equation (5.50). Put

The subtraction is legitimate — \(\tilde{e}_{k}\neq0_{\mathbb{V}}\) — because otherwise \(v_{k}\) would lie in \(\gen{\set{e_{1},\ldots,e_{k-1}}}=\gen{\set{v_{1},\ldots,v_{k-1}}}\), contradicting linear independence. For \(j<k\),

so \(e_{k}\) is orthogonal to all its predecessors and, by construction, a unit vector. Equation (5.50) holds at step \(k\) because Equation (5.51) expresses \(e_{k}\) in terms of \(v_{k},e_{1},\ldots,e_{k-1}\) and, read backwards, expresses \(v_{k}\) in terms of \(e_{k},\ldots,e_{1}\); each generated subspace therefore contains the other.

Derives Corollary 5.29. \(\mathbb{S}^{\perp}\) is a subspace because \(\braket{s}{aw_{1}+bw_{2}}=a\braket{s}{w_{1}}+b\braket{s}{w_{2}}=0\) by Equations (5.34) and (5.35), and it is non-empty (\(0_{\mathbb{V}}\in\mathbb{S}^{\perp}\) by Equation (5.38)). Let \(\set{e_{1},\ldots,e_{m}}\) be an orthonormal basis of \(\mathbb{S}\), which exists by Proposition 5.28, and for \(v\in\mathbb{V}\) define

The computation just performed in Proposition 5.28 shows \(\braket{e_{j}}{v-Pv}=0\) for every \(j\), hence \(v-Pv\in\mathbb{S}^{\perp}\), so \(v=Pv+(v-Pv)\) exhibits \(\mathbb{V}\) as a sum. The sum is direct: if \(s\in\mathbb{S}\cap\mathbb{S}^{\perp}\) then \(\braket{s}{s}=0\), so \(s=0_{\mathbb{V}}\) by Equation (5.31).

Derives Proposition 5.31. Symmetry is Equation (5.33), the conjugation being the identity on \(\R\). For \(c=(c^{1},\ldots,c^{m})\in\R^{m}\) put \(w=\sum_{k=1}^{m}c^{k}v_{k}\); expanding by Equations (5.34), (5.35), (5.36) and (5.37),

by Equation (5.32), which is positive semi-definiteness; and by Equation (5.31) the value vanishes exactly when \(w=0_{\mathbb{V}}\).

If the family is independent, \(w=0_{\mathbb{V}}\) forces \(c=0\), so Equation (5.55) is strictly positive for every \(c\neq0\) and \(G\) is positive definite. If it is dependent, some \(c\neq0\) gives \(w=0_{\mathbb{V}}\) and the quadratic form vanishes there, so \(G\) is not positive definite. For the third condition: if \(G\) is positive definite and \(Gc=0\), then Equation (5.55) vanishes at \(c\), hence \(c=0\), so the array is injective and \(\det G\neq0\) by Proposition 5.49 and Equation (5.81). Conversely, if \(G\) is not positive definite there is a \(c\neq0\) with \(\braket{w}{w}=0\), hence \(w=0_{\mathbb{V}}\), hence \(\sum_{l}G_{kl}c^{l}=\braket{v_{k}}{w}=0\) for every \(k\); so \(Gc=0\) with \(c\neq0\) and \(\det G=0\).

Derives Lemma 5.33. Write \(L_{lmn}=\varepsilon_{ijk}a_{il}a_{jm}a_{kn}\). Exchanging two of \(l,m,n\) exchanges two of the three factors; relabelling the summed indices to put the factors back in order exchanges two indices of \(\varepsilon_{ijk}\) and so costs a sign. Hence \(L\) is totally antisymmetric in \((l,m,n)\): it vanishes whenever two of them are equal, and on the six remaining index triples it is \(L_{lmn}=\varepsilon_{lmn}L_{123}\). Finally

by Equation (5.19), the surviving terms of the sum over \(i,j,k\) being exactly those in which \((i,j,k)\) is a permutation of \((1,2,3)\), with \(\varepsilon_{ijk}\) its sign.

That \(\det a\transpose=\det a\) is Equation (5.19) reindexed by \(\sigma\longmapsto\sigma^{-1}\), a bijection of the permutations which preserves the sign and turns \(\prod_{j}a_{\sigma(j)j}\) into \(\prod_{j}a_{j\sigma(j)}\). The second equality of Equation (5.58) is therefore the first one applied to \(a\transpose\). Equation (5.59) is that second equality with \(a=R\), after renaming the summed indices \(i,j,k\) to \(a,b,c\) and the free indices \(l,m,n\) to \(i,j,k\); and \(\det R=\pm1\) because \(\left(\det R\right)^{2}=\det\left(R\transpose R\right)=1\).

Derives Lemma 5.34. If \(j=k\) both sides vanish: the left because \(\varepsilon_{ijk}=0\), the right because it becomes \(\delta_{jl}\delta_{jm}-\delta_{jm}\delta_{jl}\). The same argument disposes of \(l=m\). So let \(j\neq k\) and \(l\neq m\).

A term of the sum is non-zero only if \(i\notin\set{j,k}\) and \(i\notin\set{l,m}\). Since \(\set{j,k}\) has two elements and \(i\) ranges over three values, exactly one \(i\) meets the first condition, so the sum has at most one non-zero term. Both \(\set{j,k}\) and \(\set{l,m}\) are two-element subsets of \(\set{1,2,3}\); if they are different they share exactly one element, and the value missing from \(\set{j,k}\) then lies in \(\set{l,m}\), so the surviving term is killed by its second factor and the sum vanishes. The right-hand side vanishes too, since \((l,m)\) is then neither \((j,k)\) nor \((k,j)\) and neither product of deltas can be \(1\).

There remain the two cases \(\set{l,m}=\set{j,k}\). If \((l,m)=(j,k)\) the surviving term is \(\left(\varepsilon_{ijk}\right)^{2}=1\) and the right side is \(\delta_{jj}\delta_{kk}-\delta_{jk}\delta_{kj}=1-0=1\). If \((l,m)=(k,j)\) the term is \(\varepsilon_{ijk}\varepsilon_{ikj}=-1\) and the right side is \(\delta_{jk}\delta_{kj}-\delta_{jj}\delta_{kk}=-1\).

Derives Proposition 5.35. 1. \(\left(v\times u\right)_{i}=\varepsilon_{ijk}v_{j}u_{k} =\varepsilon_{ikj}v_{k}u_{j}=-\varepsilon_{ijk}u_{j}v_{k}\), the middle step being a relabelling of the two summed indices and the last the antisymmetry of Equation (5.56). For the orthogonality, \(\braket{u}{u\times v}=\varepsilon_{ijk}u_{i}u_{j}v_{k}\) vanishes because \(u_{i}u_{j}\) is unchanged and \(\varepsilon_{ijk}\) changes sign when \(i\) and \(j\) are exchanged, so the sum equals minus itself; and \(\braket{v}{u\times v}\) vanishes by the same argument on \(j\) and \(k\) after using part 1's antisymmetry.

2. The first equality is Equation (5.57) paired with \(u\). For the second, apply the second equality of Equation (5.58) at \((l,m,n)=(1,2,3)\) to the array \(a\) whose rows are \(u,v,w\): it gives \(\varepsilon_{ijk}a_{1i}a_{2j}a_{3k}=\varepsilon_{123}\det a=\det a\). The stated symmetries are those of \(\varepsilon_{ijk}\).

3. Using Equation (5.57) twice, then the cyclic symmetry \(\varepsilon_{ijk}=\varepsilon_{kij}\) and Equation (5.60),

which is Equation (5.62).

4. By Equation (5.57) and Equation (5.60),

which is Equation (5.63). Substituting Equation (5.18) turns the right-hand side into \(\norm{u}^{2}\norm{v}^{2}\left(1-\cos^{2}{\theta}\right)\), and \(\sin{\theta}\geq0\) on \([0,\pi]\); the parallelogram spanned by \(u\) and \(v\) has base \(\norm{v}\) and altitude \(\norm{u}\sin{\theta}\), the altitude computed in Section 5.1.5.

5. Multiply the first equality of Equation (5.58), taken with \(a=R\), by \(R_{dl}\) and sum over \(l\). On the left the two factors carrying \(l\) contract to \(\sum_{l}R_{il}R_{dl}=\delta_{id}\), which is \(R\transpose R=\identity\) read for \(RR\transpose\) — the two are the same statement in finite dimension, by Proposition 5.49. There results

Contracting this with \(u_{m}v_{n}\) and using Equation (5.57) twice gives \(\left[\left(Ru\right)\times\left(Rv\right)\right]_{d} =\left(\det R\right)R_{dl}\left(u\times v\right)_{l}\), which is Equation (5.64).

Derives Lemma 5.38. Exchange. We show by induction on \(j=0,1,\ldots,k\) that, after a renumbering of the \(w\)'s,

and in particular that \(j\leq m\). For \(j=0\) this is the hypothesis. Assume Equation (5.65) for \(j-1\), with \(j\leq k\). Then \(v_{j}\) is a finite linear combination

Not every \(b^{l}\) can vanish: otherwise \(v_{j}\) would be a linear combination of \(v_{1},\ldots,v_{j-1}\), which Definition 5.14 forbids. In particular the second sum is not empty, so \(j\leq m\). Renumber the remaining \(w\)'s so that \(b^{j}\neq0\) and solve:

which places \(w_{j}\) in \(\gen{\set{v_{1},\ldots,v_{j},w_{j+1},\ldots,w_{m}}}\). That set therefore generates everything the set of the previous step generated, namely \(\mathbb{V}\), which is Equation (5.65) at \(j\). Taking \(j=k\) gives \(k\leq m\).

Equal cardinality. If \(B\) and \(B'\) are two bases, each is independent and each generates, so the exchange applied twice gives \(\abs{B}\leq\abs{B'}\) and \(\abs{B'}\leq\abs{B}\).

Completion. Take \(j=k\) in Equation (5.65): the set \(G=\set{v_{1},\ldots,v_{k},w_{k+1},\ldots,w_{m}}\) generates \(\mathbb{V}\). If it is not linearly independent, some vanishing linear combination of its elements has a non-zero coefficient; that coefficient cannot belong only to the \(v\)'s, which are independent, so it may be taken to belong to some \(w_{l}\). Solving for that \(w_{l}\) expresses it through the others, so deleting it leaves a set that still generates \(\mathbb{V}\). Repeating terminates, \(G\) being finite, and leaves a generating, linearly independent set containing every \(v_{i}\) — a basis in the sense of Definition 5.15.

Derivation. Derives Definition 5.39. That the two sets are subspaces is the content asserted, and it is one line each. Both are non-empty, containing \(0_{\mathbb{V}}\) and \(0_{\mathbb{W}}\) respectively, since \(T(0_{\mathbb{V}})=T(0\cdot 0_{\mathbb{V}})=0\,T(0_{\mathbb{V}})=0_{\mathbb{W}}\) by Equation (5.22). If \(u,v\in\ker T\) and \(a,b\in\mathbb{K}\) then \(T(au+bv)=aT(u)+bT(v)=0_{\mathbb{W}}\); and if \(w=T(u)\), \(z=T(v)\) lie in \(\im T\) then \(aw+bz=T(au+bv)\in\im T\). Both verify Definition 5.7.

Derives Theorem 5.40. \(\ker T\) is a subspace of a space of dimension \(n\), so by Lemma 5.38 it has a finite basis \(\set{u_{1},\ldots,u_{k}}\) with \(k\leq n\), and that independent set completes to a basis \(\set{u_{1},\ldots,u_{k},x_{1},\ldots,x_{n-k}}\) of \(\mathbb{V}\). We claim that \(\set{T(x_{1}),\ldots,T(x_{n-k})}\) is a basis of \(\im T\).

It generates. Any element of \(\im T\) is \(T(v)\) for some \(v=\sum_{i}a^{i}u_{i}+\sum_{j}b^{j}x_{j}\), and linearity together with \(T(u_{i})=0_{\mathbb{W}}\) gives \(T(v)=\sum_{j}b^{j}T(x_{j})\).

It is independent. Suppose \(\sum_{j}c^{j}T(x_{j})=0_{\mathbb{W}}\). Then \(T\left(\sum_{j}c^{j}x_{j}\right)=0_{\mathbb{W}}\), so \(\sum_{j}c^{j}x_{j}\in\ker T\) and therefore \(\sum_{j}c^{j}x_{j}=\sum_{i}d^{i}u_{i}\) for some scalars \(d^{i}\). That is a vanishing linear combination of the completed basis, so every coefficient vanishes by Equation (5.29); in particular every \(c^{j}=0\).

Hence \(\dim{\im T}=n-k=n-\dim{\ker T}\), which is Equation (5.68).

Derives Proposition 5.42. Uniqueness. If \(S\) and \(S'\) both satisfy Equation (5.69), then \(\braket{Sw-S'w}{v}=0\) for every \(v\) by Equation (5.37); taking \(v=Sw-S'w\) and using Equation (5.31) gives \(Sw=S'w\) for every \(w\).

Existence. By Proposition 5.28 choose orthonormal bases \(\set{e_{i}}\) of \(\mathbb{V}\) and \(\set{f_{j}}\) of \(\mathbb{W}\), so that Equation (5.49) holds in each. Define \(T^{\dagger}\) on the basis of \(\mathbb{W}\) by

and extend by linearity, which is legitimate by the uniqueness of components used in Proposition 5.45. The map so defined is linear by construction, so both sides of Equation (5.69) are antilinear in \(w\) and linear in \(v\), and it suffices to check the identity on basis vectors:

so the two agree. This also establishes Equation (5.70).

Linearity is forced, not assumed. Whatever map satisfies Equation (5.69) is linear, so nothing was lost by building a linear one: for \(a,b\in\mathbb{K}\) and \(w,z\in\mathbb{W}\),

for every \(v\), and the uniqueness argument above forces the two arguments to coincide.

Involutivity. For all \(v,w\), \(\braket{\left(T^{\dagger}\right)^{\dagger}v}{w} =\braket{v}{T^{\dagger}w}=\braket{T^{\dagger}w}{v}^{\ast} =\braket{w}{Tv}^{\ast}=\braket{Tv}{w}\), using Equation (5.33) twice, so \(\left(T^{\dagger}\right)^{\dagger}=T\).

Derives Theorem 5.43. A preliminary: \(\left(\mathbb{S}^{\perp}\right)^{\perp}=\mathbb{S}\) for a subspace \(\mathbb{S}\) of a finite-dimensional inner-product space. The inclusion \(\mathbb{S}\subset\left(\mathbb{S}^{\perp}\right)^{\perp}\) is immediate from Equation (5.52) and the hermiticity Equation (5.33). Conversely let \(v\in\left(\mathbb{S}^{\perp}\right)^{\perp}\) and split \(v=s+w\) with \(s\in\mathbb{S}\), \(w\in\mathbb{S}^{\perp}\), which Corollary 5.29 permits. Then \(w=v-s\in\left(\mathbb{S}^{\perp}\right)^{\perp}\), being a difference of two of its elements, while also \(w\in\mathbb{S}^{\perp}\); hence \(\braket{w}{w}=0\) and \(w=0_{\mathbb{V}}\) by Equation (5.31), so \(v=s\in\mathbb{S}\).

Equation (5.71). By Equation (5.52), \(w\in\left(\im T\right)^{\perp}\) means \(\braket{Tv}{w}=0\) for every \(v\in\mathbb{V}\). Now

using Equation (5.33) and Equation (5.69), so the condition reads \(\braket{v}{T^{\dagger}w}=0\) for every \(v\); taking \(v=T^{\dagger}w\) and using Equation (5.31) this holds exactly when \(T^{\dagger}w=0_{\mathbb{V}}\), that is, when \(w\in\ker T^{\dagger}\).

Equation (5.72). Apply Equation (5.71) to \(T^{\dagger}\) in place of \(T\), which is legitimate because \(\left(T^{\dagger}\right)^{\dagger}=T\) by Proposition 5.42: it gives \(\left(\im T^{\dagger}\right)^{\perp}=\ker T\). Taking the orthogonal complement of both sides and using the preliminary yields \(\im T^{\dagger}=\left(\ker T\right)^{\perp}\).

The splittings and the ranks. Equation (5.73) is Corollary 5.29 applied to \(\im T\leq\mathbb{W}\) and to \(\ker T\leq\mathbb{V}\), rewritten through Equations (5.71) and (5.72). Finally, the second splitting gives \(\dim{\mathbb{V}}=\dim{\ker T}+\dim{\im T^{\dagger}}\), whereas Equation (5.68) gives \(\dim{\mathbb{V}}=\dim{\ker T}+\dim{\im T}\); subtracting, \(\dim{\im T}=\dim{\im T^{\dagger}}\).

Derivation. Derives Proposition 5.45. The claim has two halves, and it is worth separating them: the array \(\left[T\right]\) is well defined (its entries do not depend on how \(T(e_{i})\) happens to be written), and it determines \(T\) completely.

The entries are unique. Suppose the images of the basis vectors admitted two expansions,

Subtracting, and collecting terms by h),

But \(B_{2}=\set{f_{j}}\) is a basis and therefore linearly independent, so by Equation (5.29) every coefficient of such a vanishing combination is zero:

The array determines the transformation. Conversely, let \(v\) be any vector of \(\mathbb{V}\), with components \(v^{i}\) in \(B_{1}\) (Definition 5.17). Linearity gives

so the components of \(T(v)\) in \(B_{2}\) are \(\left(T(v)\right)^{j}=\sum_{i}T^{j}_{\ i}v^{i}\), the familiar matrix-times-column product. Two linear transformations with the same matrix therefore agree on every vector, and are the same transformation.

The two halves together say that, once the bases are fixed, \(T\mapsto \left[T\right]\) is a bijection between \(\mathcal{L}(\mathbb{V},\mathbb{W})\) and the \(\dim{\mathbb{W}}\times\dim{\mathbb{V}}\) arrays over \(\mathbb{K}\). What is not claimed is independence of the bases: change \(B_{1}\) or \(B_{2}\) and the same \(T\) acquires a different matrix, which is the remark made after Equation (5.74) and, later, the reason a representation is characterized by its character (Equation (5.181)) rather than by its matrices.

Derives Proposition 5.48. 1. If \(S\) satisfies Equation (5.77) then \(T\) is injective, since \(T(u)=T(v)\) gives \(u=S(T(u))=S(T(v))=v\), and surjective, since \(w=T(S(w))\) for every \(w\in\mathbb{W}\). Conversely, a bijection has a set-theoretic inverse, and that inverse satisfies Equation (5.77) by construction.

2. If \(S\) and \(S'\) are both inverses then

composition of maps being associative.

3. Let \(w,z\in\mathbb{W}\) and \(a,b\in\mathbb{K}\), and put \(v=T^{-1}(w)\), \(u=T^{-1}(z)\). Linearity of \(T\) gives \(T(av+bu)=aT(v)+bT(u)=aw+bz\), and applying \(T^{-1}\) to both sides,

So \(T^{-1}\) satisfies Definition 5.37. With part 1 this identifies the invertible linear transformations with the isomorphisms.

Derives Proposition 5.49. \(T\) is injective if and only if \(\ker T=\set{0_{\mathbb{V}}}\): one direction is immediate, and for the other \(T(u)=T(v)\) gives \(T(u-v)=0_{\mathbb{W}}\), hence \(u=v\). By Equation (5.68) this is equivalent to \(\dim{\im T}=n\), and a subspace of \(\mathbb{W}\) of dimension \(n=\dim{\mathbb{W}}\) is the whole of \(\mathbb{W}\) — a basis of it is an independent set of \(n\) vectors in \(\mathbb{W}\), which Lemma 5.38 completes to a basis of \(\mathbb{W}\) by adjoining \(n-n=0\) further vectors. So injectivity is equivalent to surjectivity, and each is therefore equivalent to bijectivity, which is invertibility by Proposition 5.48.

Derives Proposition 5.51. First, the matrix of a composition is the product of the matrices. Write \(B_{1}=\set{e_{i}}\), \(B_{2}=\set{f_{j}}\), and let \(S:\mathbb{W}\longrightarrow\mathbb{V}\) be any linear transformation, so that Equation (5.74) reads \(T(e_{i})=\sum_{j}T^{j}_{\ i}f_{j}\) and \(S(f_{j})=\sum_{k}S^{k}_{\ j}e_{k}\). Then

so that, by the uniqueness of Proposition 5.45, \(\left[S\circ T\right]^{B_{1}}_{B_{1}} =\left[S\right]^{B_{1}}_{B_{2}}\left[T\right]^{B_{2}}_{B_{1}}\); and the matrix of \(\id_{\mathbb{V}}\) in the basis \(B_{1}\) twice over is \(\identity\), since \(\id_{\mathbb{V}}(e_{i})=e_{i}\). Applying this to both halves of Equation (5.77),

which is Equation (5.78). For Equation (5.79), multiply out:

so \(B^{-1}A^{-1}\) is an inverse of \(AB\), and the uniqueness argument of Proposition 5.48 — which used nothing but associativity — makes it the inverse. The order reverses because the factor applied last must be undone first; the same reversal appears for the transpose in Equation (5.161) and for the pullback in Equation (5.98).

Derives Proposition 5.54. Let \(\mathbb{V}\) and \(\mathbb{W}\) be the two spaces, with bases \(B_{1}=\set{e_{i}}^{n}_{i=1}\) and \(B_{2}=\set{f_{i}}^{n}_{i=1}\); these have the same cardinality \(n\) by hypothesis, which is what allows the two to be listed with a common index. Define \(T:\mathbb{V}\longrightarrow \mathbb{W}\) on the basis by

and extend to all of \(\mathbb{V}\) by

Equation (5.80) is a legitimate definition precisely because the components \(v^{i}\) of a vector in a basis are unique — the argument just given for Proposition 5.45 — so no vector is assigned two images.

\(T\) is linear. If \(u=\sum_{i}u^{i}e_{i}\) and \(v=\sum_{i}v^{i}e_{i}\) then \(au+bv=\sum_{i}(au^{i}+bv^{i})e_{i}\), so

\(T\) is injective. Suppose \(T(v)=0_{\mathbb{W}}\). Then \(\sum_{i}v^{i}f_{i}=0_{\mathbb{W}}\), and \(B_{2}\) being linearly independent, Equation (5.29) forces \(v^{i}=0\) for every \(i\), whence \(v=0_{\mathbb{V}}\). Since \(T\) is linear, \(T(u)=T(v)\) implies \(T(u-v)=0_{\mathbb{W}}\) and hence \(u=v\).

\(T\) is surjective. Any \(w\in\mathbb{W}\) can be written \(w=\sum_{i}w^{i}f_{i}\), because \(B_{2}\) generates \(\mathbb{W}\); then \(w=T\left(\sum_{i}w^{i}e_{i}\right)\).

Thus \(T\) is a bijective linear transformation, that is, an isomorphism in the sense of Definition 5.53.

Derives Proposition 5.61. \(\mathbb{W}^{0}\) is defined by conditions linear in \(g\), so it is a subspace. Let \(\set{e_{1},\ldots,e_{m}}\) be a basis of \(\mathbb{W}\), \(m=\dim{\mathbb{W}}\), completed to a basis \(\set{e_{1},\ldots,e_{n}}\) of \(\mathbb{V}\) by Lemma 5.38, and let \(\set{f^{1},\ldots,f^{n}}\) be the dual basis Equation (5.86). By Equation (5.87) every \(g\in\mathbb{V}^{\ast}\) is \(g=\sum_{j}g_{j}f^{j}\) with \(g_{j}=g(e_{j})\). Since \(\mathbb{W}\) is generated by \(e_{1},\ldots,e_{m}\), the condition \(g(w)=0\) for all \(w\in\mathbb{W}\) is equivalent to \(g_{1}=\ldots=g_{m}=0\), that is, to \(g\in\gen{\set{f^{m+1},\ldots,f^{n}}}\). Those \(n-m\) functionals are independent, being part of a basis, so they are a basis of \(\mathbb{W}^{0}\) and Equation (5.89) follows.

Derives Corollary 5.62. Let \(A:\mathbb{V}\longrightarrow\mathbb{K}^{m}\) send \(v\) to \(\left(f^{1}(v),\ldots,f^{m}(v)\right)\); it is linear and its kernel is \(\mathbb{W}\). It is surjective: otherwise its image is a proper subspace \(\mathbb{S}\) of \(\mathbb{K}^{m}\), and Proposition 5.61 applied in \(\mathbb{K}^{m}\) gives \(\dim{\mathbb{S}^{0}}=m-\dim{\mathbb{S}}\geq1\), so some non-zero \(\left(c_{1},\ldots,c_{m}\right)\) satisfies \(\sum_{\alpha}c_{\alpha}f^{\alpha}(v)=0\) for every \(v\) — a vanishing non-trivial combination, contradicting independence. Hence Equation (5.68) gives \(\dim{\mathbb{W}}=n-m\).

By Equation (5.89), \(\dim{\mathbb{W}^{0}}=n-(n-m)=m\). Each \(f^{\alpha}\) vanishes on \(\mathbb{W}\) and so lies in \(\mathbb{W}^{0}\); the \(m\) of them are independent, hence a basis of \(\mathbb{W}^{0}\), which is the assertion, the coefficients being the unique components in that basis.

Derives Proposition 5.63. Part 1. \(\Phi(u)\) is by construction an element of \(\mathbb{V}^{\ast}=\mathcal{L}(\mathbb{V},\mathbb{K})\), that is, a linear functional on \(\mathbb{V}\). Hence, for \(a,b\in\mathbb{K}\) and \(v,w\in\mathbb{V}\),

which is Equation (5.34) (take \(b=0\)) and Equation (5.35) (take \(a=b=1\)). In the first slot the behaviour is that of \(\Phi\): by Equation (5.90),

which is Equation (5.36) and Equation (5.37). (For \(\mathbb{K}=\R\) the conjugation is the identity and antilinearity is linearity, so the same two lines cover both fields.)

For non-degeneracy: if \(\braket{u}{v}=0\) for every \(v\) then \(\Phi(u)\) is the zero functional, and \(\Phi\) being injective — with \(\Phi(0_{\mathbb{V}})=0\) by Equation (5.90) — gives \(u=0_{\mathbb{V}}\). If instead \(\braket{u}{v}=0\) for every \(u\), then \(f(v)=0\) for every \(f\in\mathbb{V}^{\ast}\), because \(\Phi\) is surjective; taking \(f\) to be the dual-basis functional \(f^{1}\) of Equation (5.86) built from a basis whose first element is \(v\) — possible if \(v\neq0_{\mathbb{V}}\) — would give \(f^{1}(v)=1\neq0\), a contradiction, so \(v=0_{\mathbb{V}}\).

Part 2. The three remaining axioms are, one by one, restatements of the two extra conditions.

• Hermiticity Equation (5.33) reads \(\braket{u}{v}^{\ast}=\braket{v}{u}\), which written out through Equation (5.91) is exactly Equation (5.92).

• Positive definiteness Equation (5.32) reads \(\braket{v}{v}=(\Phi(v))(v)>0\) for \(v\neq0_{\mathbb{V}}\), which is Equation (5.93).

• Identity of indiscernibles Equation (5.31): the direction (\(\Leftarrow\)) holds for any \(\Phi\), since \(\Phi(0_{\mathbb{V}})=0\) and the zero functional sends every vector, in particular \(0_{\mathbb{V}}\), to \(0\). The direction (\(\Rightarrow\)) is the contrapositive of Equation (5.93): if \(v\neq0_{\mathbb{V}}\) then \(\braket{v}{v}>0\neq0\).

Converse. Given an inner product on \(\mathbb{V}\), define \(\Phi(u)=\braket{u}{\ }\), the functional \(v\mapsto\braket{u}{v}\). It is a functional by Equations (5.34) and (5.35), it is antilinear in \(u\) by Equations (5.36) and (5.37), and it is injective, for \(\Phi(u)=0\) gives in particular \(\braket{u}{u}=0\) and hence \(u=0_{\mathbb{V}}\) by Equation (5.31). An injective map that is \(\R\)-linear between two real vector spaces of the same finite dimension — and \(\mathbb{V}\), \(\mathbb{V}^{\ast}\) have the same dimension over \(\mathbb{K}\), hence the same dimension over \(\R\) — is surjective, so \(\Phi\) is a bijection of the required kind, and it manifestly reproduces the given inner product through Equation (5.91). Uniqueness is immediate: two such maps agreeing in Equation (5.91) agree as functionals on every \(v\), hence are equal.

Derives Proposition 5.67. Linearity of \(T^{\ast}\). By definition \(T^{\ast}\omega=\omega\circ T\), which is a functional on \(\mathbb{V}\) as noted in Section 5.5.2. For \(a,b\in\mathbb{K}\) and \(\omega,\eta\in\mathbb{W}^{\ast}\), evaluation at an arbitrary \(v\in\mathbb{V}\) gives

the middle step being the pointwise definition of the operations on \(\mathbb{W}^{\ast}\). Since \(v\) was arbitrary the two functionals are equal.

Equation (5.96). Both sides are the same number, written twice:

It characterizes \(T^{\ast}\) because a functional is determined by its values: if \(\gen{\Phi\omega,v}=\gen{\omega,T_{\ast}v}\) for every \(v\) then \(\Phi\omega\) and \(T^{\ast}\omega\) agree at every point of \(\mathbb{V}\).

The two composition laws. Equation (5.97) is the definition of composition. For Equation (5.98), evaluate on \(\omega\in\mathbb{Z}^{\ast}\):

associativity of composition doing the whole of the work. The same conclusion is reached from Equation (5.96) alone: \(\gen{(S\circ T)^{\ast}\omega,v}=\gen{\omega,S_{\ast}T_{\ast}v} =\gen{S^{\ast}\omega,T_{\ast}v}=\gen{T^{\ast}S^{\ast}\omega,v}\) for all \(v\).

Equation (5.99). Expand \(T^{\ast}\phi^{a}\) in the dual basis of \(\mathbb{V}^{\ast}\) using Equation (5.87), whose coefficients are the values on the basis of \(\mathbb{V}\):

which is Equation (5.99). Note the index positions: the row index of \(\left[T\right]\) has become the label of the pulled-back functional and the column index the label of the dual basis vector, which is what “transpose” means here.

Derivation. Derives Definition 5.70. Two things have to be checked before Equation (5.103) defines anything: that it is a polynomial of degree \(n\), and that it does not depend on the basis in which \(\left[T\right]\) was written.

Degree. In the Leibniz sum Equation (5.19) for \(\det\left(\left[T\right]-\mu\identity\right)\) the identity permutation contributes \(\prod_{i=1}^{n}\left(T^{i}_{\ i}-\mu\right)\), a polynomial of degree \(n\) with leading term \((-1)^{n}\mu^{n}\). Any other permutation moves at least two indices, so at least two of its \(n\) factors are off-diagonal entries of \(\left[T\right]\), carrying no \(\mu\); such a term has degree at most \(n-2\) and cannot cancel the leading one. Hence \(\deg{p_{T}}=n\) exactly.

Basis-independence. A change of basis conjugates the array. Let \(e'_{i}=\sum_{j}A^{j}_{\ i}e_{j}\) with \(\left[A\right]\) invertible, and let \(\left[T'\right]\) be the matrix of \(T\) in the primed basis. Expanding \(T(e'_{i})\) in the two ways Equation (5.74) allows,

and equating coefficients, which Proposition 5.45 permits, gives \(\left[T\right]\left[A\right]=\left[A\right]\left[T'\right]\), that is \(\left[T'\right]=\left[A\right]^{-1}\left[T\right]\left[A\right]\) — the similarity law recorded again, in its algebra setting, in Remark 5.126. Since \(\left[A\right]^{-1}\identity\left[A\right]=\identity\),

the determinant of a product being the product of the determinants and \(\det\left(\left[A\right]^{-1}\right)\det\left[A\right]=\det\identity=1\) by Equation (5.78). So \(p_{T}\) is an attribute of the operator, not of the array.

Derives Theorem 5.71. The chain is one of equivalences, each already available. Equation (5.101) has a solution \(v\neq0_{\mathbb{V}}\) exactly when \(\ker\left(T-\lambda\identity\right)\neq \set{0_{\mathbb{V}}}\), that is, exactly when the endomorphism \(T-\lambda\identity\) fails to be injective. By Proposition 5.49 — and here the finite dimension is indispensable — that is exactly when \(T-\lambda\identity\) fails to be invertible, which by Equation (5.81) is exactly when \(\det\left(\left[T\right]-\lambda\identity\right)=0\). The last expression is \(p_{T}(\lambda)\).

Derives Corollary 5.72. By Definition 5.70 the polynomial \(p_{T}\) has degree \(n=\dim{\mathbb{V}}\geq1\), so by the fundamental theorem of algebra (Theorem 8.19) it has a root \(\lambda\in\C\), and Theorem 5.71 makes that root an eigenvalue. For the rotation, the matrix in the standard basis is \(\begin{pmatrix}\cos{\theta} & -\sin{\theta}\\ \sin{\theta} & \cos{\theta}\end{pmatrix}\), whose characteristic polynomial is \(\left(\cos{\theta}-\mu\right)^{2}+\sin^{2}{\theta} =\mu^{2}-2\mu\cos{\theta}+1\) with discriminant \(4\left(\cos^{2}{\theta}-1\right)=-4\sin^{2}{\theta}\), negative unless \(\sin{\theta}=0\).

Derives Proposition 5.74. The first inequality restates that an eigenvalue has an eigenvector, so \(\mathbb{E}_{\lambda}\neq\set{0_{\mathbb{V}}}\).

For the second, write \(g=\dim{\mathbb{E}_{\lambda}}\), take a basis \(\set{u_{1},\ldots,u_{g}}\) of \(\mathbb{E}_{\lambda}\) and complete it by Lemma 5.38 to a basis \(\set{u_{1},\ldots,u_{g},x_{1},\ldots,x_{n-g}}\) of \(\mathbb{V}\). Since \(T(u_{i})=\lambda u_{i}\), the first \(g\) columns of \(\left[T\right]\) in this basis carry \(\lambda\) on the diagonal and nothing below row \(g\):

A block array of this shape has \(\det=\det\left(\lambda\identity_{g} \right)\det{C}\), as the Leibniz sum Equation (5.19) shows: a permutation \(\sigma\) contributes a non-zero term only if \(\sigma(j)\leq g\) for every \(j\leq g\), because the entries below the first \(g\) rows in those columns vanish; being injective, such a \(\sigma\) maps \(\set{1,\ldots,g}\) onto itself and therefore maps the remaining indices onto themselves as well, so it splits as a pair \((\sigma_{1},\sigma_{2})\) with \(\sgn(\sigma) =\sgn(\sigma_{1})\sgn(\sigma_{2})\) and the sum factorizes into the two Leibniz sums. Applying this to \(\left[T\right]-\mu\identity\), which has the same shape,

so \(\left(\lambda-\mu\right)^{g}\) divides \(p_{T}\) and the algebraic multiplicity is at least \(g\).

That the inequality can be strict is shown by \(\left[T\right]=\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}\) on \(\C^{2}\), the shear that reappears in Remark 5.152 as a representation of \(\Z\): here \(p_{T}(\mu)=(1-\mu)^{2}\), so \(\lambda=1\) has algebraic multiplicity \(2\), while \(\left[T\right]-\identity=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}\) has a one-dimensional kernel, spanned by \(\bigl(1,0\bigr)\transpose\), so the geometric multiplicity is \(1\).

Derives Proposition 5.75. By induction on \(m\). For \(m=1\) the statement is that a single non-zero vector is independent, which Equation (5.29) gives at once, since \(a^{1}v_{1}=0_{\mathbb{V}}\) with \(v_{1}\neq0_{\mathbb{V}}\) forces \(a^{1}=0\) (the remark after Equation (5.23)).

Assume the claim for \(m-1\) vectors and let \(\sum_{i=1}^{m}a^{i}v_{i}=0_{\mathbb{V}}\). Apply the linear operator \(T-\lambda_{m}\identity\) to both sides. Each term gives \(a^{i}\left(T(v_{i})-\lambda_{m}v_{i}\right) =a^{i}\left(\lambda_{i}-\lambda_{m}\right)v_{i}\), and the last term vanishes, so

By the induction hypothesis every coefficient vanishes, and \(\lambda_{i}\neq\lambda_{m}\) for \(i<m\) then forces \(a^{i}=0\). The original relation reduces to \(a^{m}v_{m}=0_{\mathbb{V}}\) with \(v_{m}\neq0_{\mathbb{V}}\), whence \(a^{m}=0\) as well.

Derives Theorem 5.77. Read Equation (5.74) in a basis \(\set{e_{i}}\): the array \(\left[T\right]\) is diagonal, with entries \(\lambda_{i}\), precisely when \(T(e_{i})=\lambda_{i}e_{i}\) for every \(i\) — that is, precisely when every basis vector is an eigenvector. Both directions of the first claim are that one observation.

For the second, let \(p_{T}\) have \(n\) distinct roots \(\lambda_{1},\ldots,\lambda_{n}\) in \(\mathbb{K}\). Each is an eigenvalue by Theorem 5.71, so we may choose an eigenvector \(v_{i}\) for each; by Proposition 5.75 the set \(\set{v_{1},\ldots,v_{n}}\) is linearly independent, and \(n\) independent vectors in a space of dimension \(n\) are a basis — by Lemma 5.38 they complete to a basis by the adjunction of \(n-n=0\) further vectors.

Derives Theorem 5.79. 1. Let \(T(v)=\lambda v\) with \(v\neq0_{\mathbb{V}}\). Then

and \(\braket{v}{v}>0\) by Equation (5.32), so \(\lambda=\lambda^{\ast}\).

2. Let \(T(u)=\lambda u\) and \(T(v)=\mu v\) with \(\lambda\neq\mu\), both real by part 1. Then \(\braket{u}{Tv}=\mu\braket{u}{v}\), while \(\braket{u}{Tv}=\braket{T^{\dagger}u}{v}=\braket{Tu}{v} =\lambda^{\ast}\braket{u}{v}=\lambda\braket{u}{v}\). Subtracting, \(\left(\mu-\lambda\right)\braket{u}{v}=0\), and \(\mu\neq\lambda\) gives \(\braket{u}{v}=0\).

3. By induction on \(n\), the case \(n=0\) being vacuous. Let \(n\geq1\). By Corollary 5.72 — available because \(\mathbb{K}=\C\) and \(0<n<\infty\) — there is an eigenvalue \(\lambda_{1}\) with an eigenvector, which we normalize to a unit vector \(e_{1}\) using Equation (5.41). Put \(\mathbb{S}=\gen{\set{e_{1}}}\), so that Equation (5.52) gives \(\mathbb{S}^{\perp}=\set{w\in\mathbb{V}\mid\braket{e_{1}}{w}=0}\).

The complement is \(T\)-invariant. For \(w\in\mathbb{S}^{\perp}\),

so \(T(w)\in\mathbb{S}^{\perp}\).

Its dimension is \(n-1\). The map \(w\longmapsto\braket{e_{1}}{w}\) is a functional on \(\mathbb{V}\) by Equations (5.34) and (5.35), whose kernel is \(\mathbb{S}^{\perp}\) and whose image is the whole of \(\C\), since it sends \(e_{1}\) to \(1\). Equation (5.68) gives \(\dim{\mathbb{S}^{\perp}}=n-1\).

The restriction \(\left.T\right|_{\mathbb{S}^{\perp}}\) is an endomorphism of \(\mathbb{S}^{\perp}\), and it is self-adjoint for the restricted inner product, since Equation (5.69) holds in particular for vectors lying in the subspace. The induction hypothesis therefore supplies an orthonormal basis \(\set{e_{2},\ldots,e_{n}}\) of \(\mathbb{S}^{\perp}\) consisting of eigenvectors. Adjoining \(e_{1}\), which is a unit vector orthogonal to every element of \(\mathbb{S}^{\perp}\), gives an orthonormal set of \(n\) eigenvectors; it is a basis by Lemma 5.38, an orthonormal set being independent (a vanishing combination paired with \(e_{j}\) returns its \(j\)-th coefficient). Equation (5.105) is then Theorem 5.77, and the eigenvalues are real by part 1.

Finally, grouping the basis by eigenvalue writes \(\mathbb{V}\) as a sum of subspaces \(\mathbb{V}_{\lambda}\subseteq\mathbb{E}_{\lambda}\), mutually orthogonal by part 2. The inclusion is an equality: an eigenvector \(v\) with eigenvalue \(\lambda\) expands as \(v=\sum_{i}\braket{e_{i}}{v}e_{i}\), and pairing \(T(v)=\lambda v\) with \(e_{i}\) gives \(\lambda_{i}\braket{e_{i}}{v}=\lambda\braket{e_{i}}{v}\), so the coefficients with \(\lambda_{i}\neq\lambda\) vanish and \(v\in\mathbb{V}_{\lambda}\).

Derives Theorem 5.80. (\(\Rightarrow\)) The easy direction. Let \(\set{e_{i}}\) be a common eigenbasis, \(Ae_{i}=\alpha_{i}e_{i}\) and \(Be_{i}=\beta_{i}e_{i}\). Then

the scalars commuting in \(\C\); two linear operators agreeing on a basis agree everywhere, so \(\comm{A}{B}=\Theta\).

(\(\Leftarrow\)) Assume Equation (5.106). By Theorem 5.79 applied to \(A\), the space is the orthogonal direct sum of the eigenspaces \(\mathbb{E}_{\lambda}\) of \(A\), over the finitely many distinct eigenvalues \(\lambda\) of \(A\).

Each eigenspace of \(A\) is \(B\)-invariant. Let \(v\in\mathbb{E}_{\lambda}\). Then

the first equality being Equation (5.106), so \(Bv\in\mathbb{E}_{\lambda}\) by Equation (5.102). Note that this step, and only this step, uses the commutation.

The restriction \(\left.B\right|_{\mathbb{E}_{\lambda}}\) is therefore an endomorphism of \(\mathbb{E}_{\lambda}\), and it is self-adjoint for the restricted inner product. Theorem 5.79, applied to it on the space \(\mathbb{E}_{\lambda}\), supplies an orthonormal basis of \(\mathbb{E}_{\lambda}\) consisting of eigenvectors of \(B\); and every vector of \(\mathbb{E}_{\lambda}\) is already an eigenvector of \(A\) with eigenvalue \(\lambda\). Collecting these bases over all \(\lambda\) gives an orthonormal set — vectors from different eigenspaces are orthogonal by Theorem 5.79, part 2 — of \(\sum_{\lambda}\dim{\mathbb{E}_{\lambda}}=\dim{\mathbb{V}}\) vectors, each an eigenvector of both operators. It is a basis for the reason given at the end of the proof of Theorem 5.79.

Derives Corollary 5.82. Necessity. In an orthonormal eigenbasis \(\left[A\right]\) is diagonal, so \(\left[A^{\dagger}\right]=\left[A\right]^{\dagger}\) is the array of the conjugate eigenvalues by Equation (5.70), and two diagonal arrays commute.

Sufficiency. Put

so that \(A=H+\ii K\) and \(A^{\dagger}=H-\ii K\) by inspection. Both are self-adjoint: in an orthonormal basis Equation (5.70) makes \(\left[H\right]^{\dagger}=\left(\left[A\right]^{\dagger} +\left[A\right]\right)/2=\left[H\right]\), while conjugating the factor \(1/(2\ii)\) changes its sign and reversing the difference changes it back, so \(\left[K\right]^{\dagger}=\left[K\right]\) as well. Multiplying out,

so Equation (5.107) holds precisely when \(\comm{H}{K}=\Theta\). Theorem 5.80 then supplies an orthonormal basis \(\set{e_{i}}\) with \(He_{i}=h_{i}e_{i}\) and \(Ke_{i}=k_{i}e_{i}\), the eigenvalues real by part 1 of Theorem 5.79, whence \(Ae_{i}=\left(h_{i}+\ii k_{i}\right) e_{i}\): the basis diagonalizes \(A\), with eigenvalues sweeping the complex plane rather than the real line.

Unitary operators. \(U^{\dagger}U=UU^{\dagger}=\identity\) is Equation (5.107) outright. If \(Ue=\lambda e\) with \(e\neq0_{\mathbb{V}}\), then Equations (5.34), (5.36) and (5.69) give \(\braket{e}{e}=\braket{U^{\dagger}Ue}{e}=\braket{Ue}{Ue} =\abs{\lambda}^{2}\braket{e}{e}\), and \(\braket{e}{e}>0\) by Equation (5.32) leaves \(\abs{\lambda}=1\).

Derives Theorem 5.84. The two statements are one. Fix an orthonormal basis of \(\mathbb{V}\), which exists by Proposition 5.28, and let \(A\) be the matrix of \(S\) in it. Over \(\R\) the conjugation in Equation (5.70) is the identity, so self-adjointness reads \(A\transpose=A\); and a basis of eigenvectors written as the columns of \(R\) turns orthonormality into \(R\transpose R=\identity\) and the eigenvalue equations into Equation (5.108).

The roots are real. Read the same array \(A\) as the matrix of an operator on \(\C^{n}\) with the standard inner product \(\braket{u}{v}=\sum_{i}\left(u^{i}\right)^{\ast}v^{i}\). Its adjoint matrix is the conjugate transpose Equation (5.70), which for a real symmetric \(A\) is \(A\) itself, so the operator is self-adjoint and part 1 of Theorem 5.79 makes every one of its eigenvalues real. The characteristic polynomial is the same polynomial in both readings, being computed from the same array by Equation (5.19); over \(\C\) it has \(n\) roots with multiplicity by Theorem 8.19, and each of them is an eigenvalue by Theorem 5.71, hence real.

Each root has a real eigenvector. Let \(\lambda\) be one of them. It is a real scalar and a root of the characteristic polynomial of \(S\), so Theorem 5.71 — applied now over \(\mathbb{K}=\R\), where it holds verbatim — makes it an eigenvalue of \(S\): there is a \(v\in\mathbb{V}\setminus\set{0_{\mathbb{V}}}\) with \(Sv=\lambda v\). This is the step at which the complexification is discharged, and it is the only one that needed it.

Induction on \(n\). Normalize \(e_{1}=v/\norm{v}\) and let \(\mathbb{S}^{\perp}\) be the orthogonal complement Equation (5.52) of \(\gen{\set{e_{1}}}\). The computation made in part 3 of the proof of Theorem 5.79 used only Equation (5.69) and the inner-product axioms, never the complex field, so it holds here word for word: \(\mathbb{S}^{\perp}\) is \(S\)-invariant, \(\dim{\mathbb{S}^{\perp}}=n-1\) by Equation (5.68), and the restriction of \(S\) to it is self-adjoint for the restricted inner product. The induction hypothesis supplies an orthonormal basis \(\set{e_{2},\ldots,e_{n}}\) of \(\mathbb{S}^{\perp}\) made of eigenvectors, and adjoining \(e_{1}\) completes one for \(\mathbb{V}\).

Derives Theorem 5.86. \(M\) supplies an inner product. Put \(\braket{u}{v}_{M}=u\transpose Mv\) on \(\R^{n}\). It is linear in each argument by inspection, symmetric because \(M\transpose=M\), and positive definite by hypothesis, so it satisfies Equations (5.31), (5.32), (5.33), (5.34) and (5.35) and is an inner product in the sense of Definition 5.18, the conjugation being trivial over \(\R\).

A first change of basis makes \(M\) the identity. Proposition 5.28, applied to the standard basis of \(\R^{n}\) in this inner product, produces a basis \(\set{w_{1},\ldots,w_{n}}\) with \(\braket{w_{k}}{w_{l}}_{M}=\delta_{kl}\). Let \(W\) be the array whose columns are the \(w_{k}\): it is invertible, its columns being a basis, and the orthonormality relations are exactly \(W\transpose MW=\identity\).

A second, orthogonal, change of basis diagonalizes \(K\). The array \(\tilde{K}=W\transpose KW\) is real and symmetric, since \(\left(W\transpose KW\right)\transpose =W\transpose K\transpose W=\tilde{K}\). By Theorem 5.84 there is a real orthogonal \(R\) with \(R\transpose\tilde{K}R=\diag\left(\lambda_{r}\right)\) and every \(\lambda_{r}\) real. Put \(U=WR\), invertible as a product of invertible arrays. Then

which is Equation (5.109).

The second statement is the first one read column by column. Write \(u_{r}\) for the \(r\)-th column of \(U\). The first relation of Equation (5.109) says \(u_{k}\transpose Mu_{l}=\delta_{kl}\), the asserted \(M\)-orthonormality; and with \(\Lambda=\diag\left(\lambda_{r}\right)\) the two relations combine into \(U\transpose KU=\Lambda=U\transpose MU\Lambda\), so that \(U\transpose\left(KU-MU\Lambda\right)=0\) and, \(U\transpose\) being invertible, \(KU=MU\Lambda\) — whose \(r\)-th column is Equation (5.110) for \(u_{r}\) with \(\lambda=\lambda_{r}\). That these are all the roots follows from

a polynomial of degree \(n\) whose roots with multiplicity are the \(\lambda_{r}\).

Signature. Equation (5.109) exhibits \(\diag\left(\lambda_{r}\right)\) as congruent to \(K\) in the sense of Equation (5.139). Rescaling each column of \(U\) by \(\abs{\lambda_{r}}^{-1/2}\) wherever \(\lambda_{r}\neq0\) is a further congruence and brings the array to Equation (5.142) with \(p\), \(q\) and \(r\) the counts of positive, negative and vanishing \(\lambda_{r}\); Theorem 5.116 says those counts are the same for every basis reaching that form.

Derives Proposition 5.88. The denominator never vanishes, \(M\) being positive definite. Expand \(u\) in the \(M\)-orthonormal basis of Theorem 5.86, \(u=\sum_{r}c^{r}u_{r}\), not all \(c^{r}\) zero. Then \(u\transpose Mu=\sum_{r}\left(c^{r}\right)^{2}\) and \(u\transpose Ku=\sum_{r}\lambda_{r}\left(c^{r}\right)^{2}\), so

a weighted average of the roots with non-negative weights not all zero, which lies between the smallest and the largest of them — that is Equation (5.112). Subtracting,

a sum of non-negative terms, which vanishes exactly when \(c^{r}=0\) for every \(r\) with \(\lambda_{r}>\lambda_{1}\) — that is, exactly when \(u\) lies in the span of the \(u_{r}\) belonging to \(\lambda_{1}\), which by Equation (5.110) is exactly the solution set for \(\lambda=\lambda_{1}\). The maximum is the same computation with the signs reversed, and both are attained, at \(u_{1}\) and at \(u_{n}\).

Derives Corollary 5.89. The inequality \(\lambda_{1}\leq\mathcal{R}(u)\) is Equation (5.112). For the estimate, write \(w=\sum_{r\geq2}c^{r}u_{r}\) — legitimate because \(w\) is \(M\)-orthogonal to \(u_{1}\) — so that \(\varepsilon^{2}=\sum_{r\geq2}\left(c^{r}\right)^{2}\) and, by Equation (5.113) with \(c^{1}=1\),

each factor \(\lambda_{r}-\lambda_{1}\) being at most \(\lambda_{n}-\lambda_{1}\).

Derives Proposition 5.90. Existence. \(B=A\transpose A\) is symmetric, and for \(x\neq0\), \(x\transpose Bx=\left(Ax\right)\transpose\left(Ax\right)>0\), since \(A\) is invertible and therefore injective (Proposition 5.49), so \(Ax\neq0\). By Theorem 5.84, \(B=R\diag\left(\mu_{r}\right)R\transpose\) with \(R\) orthogonal, and each \(\mu_{r}>0\), being \(x\transpose Bx\) for the corresponding unit eigenvector. Put

symmetric by inspection, positive definite because its eigenvalues \(\sqrt{\mu_{r}}\) are positive, and satisfying \(P^{2}=B\) and hence invertible. Set \(O=AP^{-1}\); then, \(P^{-1}\) being symmetric,

Uniqueness. Suppose \(A=OP\) as in Equation (5.115). Then \(A\transpose A=P\,O\transpose O\,P=P^{2}\), so \(P\) is a positive definite symmetric square root of the array \(B\) fixed by \(A\). There is only one. Indeed such a \(P\) commutes with \(B=P^{2}\), so it maps each eigenspace \(\mathbb{E}_{\mu}\) of \(B\) into itself: \(Bx=\mu x\) gives \(B(Px)=P(Bx)=\mu\,Px\). The restriction of \(P\) to \(\mathbb{E}_{\mu}\) is symmetric, so by Theorem 5.84 that subspace has a basis of eigenvectors of \(P\), whose eigenvalues \(\nu\) satisfy \(\nu^{2}=\mu\) and \(\nu>0\), hence \(\nu=\sqrt{\mu}\): \(P\) acts on \(\mathbb{E}_{\mu}\) as multiplication by \(\sqrt{\mu}\). Since the eigenspaces of \(B\) span \(\R^{n}\) by Theorem 5.84, \(P\) is determined by \(B\), and then \(O=AP^{-1}\) is determined too.

Derives Corollary 5.91. \(\norm{Lx}^{2}=x\transpose\left(L\transpose L\right)x\), and \(L\transpose L\) is symmetric with \(x\transpose L\transpose Lx=\norm{Lx}^{2}\geq0\), hence positive semi-definite. Taking \(M=\identity\) in Proposition 5.88, which is legitimate because the identity is positive definite, the quotient \(\norm{Lx}^{2}/\norm{x}^{2}\) has maximum \(\lambda_{\max}\left(L\transpose L\right)\), attained at a corresponding eigenvector; restricting to \(\norm{x}=1\) gives the first equality of Equation (5.116) and the attainment.

For the second, let \(L\transpose Lv=\lambda v\) with \(v\neq0\) and \(\lambda\neq0\). Then \(Lv\neq0\), since \(L\transpose Lv\neq0\), and \(LL\transpose\left(Lv\right)=L\left(L\transpose Lv\right) =\lambda\,Lv\): every non-zero eigenvalue of \(L\transpose L\) is one of \(LL\transpose\), and by symmetry of the argument the converse holds too. If instead \(\lambda_{\max}\left(L\transpose L\right)=0\) then, all its eigenvalues being non-negative and at most \(0\), \(L\transpose L\) has every eigenvalue zero, so \(\norm{Lx}^{2}=0\) for every \(x\) by Equation (5.108), so \(L=0\) and both sides vanish.

Derives Proposition 5.92. 1. Normalize \(v\); then \(\abs{\lambda}=\norm{\lambda v}=\norm{Av}\leq\norm{A}\) by the definition of the operator norm.

2. For symmetric \(A\), Equation (5.108) gives \(A\transpose A=A^{2}=R\diag\left(\lambda_{r}^{2}\right)R\transpose\), so the eigenvalues of \(A\transpose A\) are the \(\lambda_{r}^{2}\) and Equation (5.116) gives \(\norm{A}^{2}=\max_{r}\lambda_{r}^{2}\).

3. The characteristic polynomial of Equation (5.117) is \(\left(\lambda-\mu\right)^{2}\), so \(\lambda\) is its only eigenvalue, whatever \(c\); the padding contributes only the eigenvalue \(0\). Applying \(A\) to the second standard basis vector gives the vector with components \((c,\lambda)\), of norm at least \(\abs{c}\), so \(\norm{A}\geq\abs{c}\).

Derives Proposition 5.94. Substituting \(x=Ry\) gives \(q=y\transpose\left(R\transpose AR\right)y\), and Equation (5.108) makes the array in the middle diagonal with the eigenvalues along it, which is Equation (5.118). The level set is then read off Equation (5.118) term by term, and the pattern of signs is the signature of Theorem 5.116 and Definition 5.117, hence independent of every choice made along the way.

Derives Proposition 5.95. The two descriptions agree. Multiplying Equation (5.119) out and using \(r\cos{\varphi}=x\) gives

which requires \(ex<p\) because \(r>0\); squaring it and substituting \(r^{2}=x^{2}+y^{2}\) produces Equation (5.120). Conversely let a point satisfy Equation (5.120) with \(ex<p\). It is not the origin, which would give \(-p^{2}=0\), so \(r>0\); and Equation (5.120) says \(r^{2}=\left(p-ex\right)^{2}\) with \(p-ex>0\), so Equation (5.122) holds and, divided by \(r\), returns Equation (5.119). For \(e>0\) the distance from the point to the line \(x=p/e\) is \(p/e-x\), positive by the same inequality, and Equation (5.122) says exactly \(r=e\left(p/e-x\right)\). Putting \(\varphi=\pi/2\) in Equation (5.119) gives \(r=p\).

\(0\leq e<1\). Completing the square in Equation (5.120),

and dividing by the right-hand side gives

with \(a\) and \(b\) as in Equation (5.121), since \(pe/(1-e^{2})=ae\). This is an ellipse centred at \((-ae,0)\), and

so the two foci, at distance \(\sqrt{a^{2}-b^{2}}=ae\) from the centre along the major axis, are the origin and \((-2ae,0)\). The side condition \(ex<p\) is automatic here: Equation (5.123) confines \(x\) to \(x\leq a(1-e)=p/(1+e)\), and \(ep/(1+e)<p\).

The area. For each \(u=x+ae\) in \([-a,a]\), Equation (5.123) gives a vertical segment of length \(2b\sqrt{1-u^{2}/a^{2}}\), so the area enclosed is, substituting first \(u=at\) and then \(t=\sin{\theta}\) by Corollary 7.44,

the integral of \(\cos{2\theta}\) vanishing over a full period of it.

\(e=1\) and \(e>1\). For \(e=1\) the quadratic part loses its \(x^{2}\) term and Equation (5.120) becomes \(y^{2}=p^{2}-2px\), a parabola with vertex \((p/2,0)\) opening towards decreasing \(x\); written as \(y^{2}=-2p\left(x-p/2\right)\) it has focal length \(p/2\), so its focus is at \((p/2-p/2,0)\), the origin. For \(e>1\) the eigenvalues \(1-e^{2}\) and \(1\) have opposite signs, so by Proposition 5.94 the quadric is a hyperbola; the side condition \(ex<p\) selects the branch containing the point \(\varphi=\pi/2\), since the other branch lies where \(p-ex<0\).

Derives Lemma 5.97. The kernels stabilize by step \(n\). The subspaces \(\ker\left(N^{k}\right)\) increase with \(k\). If two consecutive ones agree, \(\ker\left(N^{k}\right)=\ker\left(N^{k+1}\right)\), the chain is constant from there on: \(N^{k+2}v=0\) gives \(N^{k+1}\left(Nv\right)=0\), so \(Nv\in\ker\left(N^{k+1}\right)=\ker\left(N^{k}\right)\) and \(N^{k+1}v=0\). Every strict increase costs at least one dimension and the dimensions are bounded by \(n\), so a repetition has occurred by \(k=n\) and \(\ker\left(N^{n}\right)=\ker\left(N^{2n}\right)\).

The sum is direct and fills the space. Let \(v\) lie in both summands, \(v=N^{n}w\) with \(N^{n}v=0\). Then \(N^{2n}w=0\), so \(w\in\ker\left(N^{2n}\right)=\ker\left(N^{n}\right)\) and \(v=N^{n}w=0_{\mathbb{V}}\). The two dimensions add up to \(n\) by Equation (5.68), so Equation (5.124) holds.

Invariance and the two behaviours. \(T\) commutes with \(N\) and hence with \(N^{n}\), so it carries kernel and image of \(N^{n}\) into themselves. On \(\ker\left(N^{n}\right)\) the power \(N^{n}\) vanishes by definition, so \(N\) is nilpotent there. On \(\im\left(N^{n}\right)\) the map \(N\) is injective — if \(v\) lies there and \(Nv=0\), then \(v\) lies in \(\ker N\), a subspace of \(\ker\left(N^{n}\right)\), so \(v\) lies in both summands and vanishes — and therefore invertible on it, by Proposition 5.49.

Derives Theorem 5.98. By induction on \(n\). Corollary 5.72 supplies an eigenvalue \(\lambda_{1}\), and Lemma 5.97 applied to it splits \(\mathbb{V}=\mathbb{G}_{1}\oplus\mathbb{W}\) with \(\mathbb{W}=\im\left(N^{n}\right)\), \(N=T-\lambda_{1}\identity\). The first summand contains an eigenvector, so it is not trivial and \(\dim{\mathbb{W}}<n\). If \(\mathbb{W}\) is trivial, then \(k=1\) and there is nothing left to prove; otherwise \(T\) restricts to an endomorphism of \(\mathbb{W}\) and the induction hypothesis decomposes \(\mathbb{W}\) into the generalized eigenspaces, computed inside \(\mathbb{W}\), of the distinct eigenvalues of that restriction. Two identifications complete the argument, and both rest on one observation:

\(T-\mu\identity\) is invertible on \(\mathbb{G}_{1}\) for every \(\mu\neq\lambda_{1}\). On that subspace \(T-\mu\identity=N+\left(\lambda_{1}-\mu\right)\identity\) is a nilpotent operator plus a non-zero scalar, and such a sum is invertible: with \(c=\lambda_{1}-\mu\) and \(N^{n}=\Theta\) there, \(\left(N+c\identity\right)^{-1} =c^{-1}\sum_{j=0}^{n-1}\left(-N/c\right)^{j}\), a finite sum, as multiplying out and telescoping shows.

The eigenvalues of \(T\) are \(\lambda_{1}\) together with those of the restriction. An eigenvector of the restriction is one of \(T\). Conversely let \(Tv=\mu v\) with \(v\neq0_{\mathbb{V}}\) and \(\mu\neq\lambda_{1}\), and split \(v=g+w\) along Equation (5.124); both summands being \(T\)-invariant, \(Tg=\mu g\), so \(\left(T-\mu\identity\right)g=0_{\mathbb{V}}\) and the observation forces \(g=0_{\mathbb{V}}\): the eigenvector lies in \(\mathbb{W}\).

The generalized eigenspaces agree. Fix \(\mu\neq\lambda_{1}\) and let \(v\in\ker\left(\left(T-\mu\identity\right)^{n}\right)\). Split \(v=g+w\) along Equation (5.124); both summands are \(T\)-invariant, so \(\left(T-\mu\identity\right)^{n}g=0_{\mathbb{V}}\), and the observation — which makes that operator invertible on \(\mathbb{G}_{1}\), a power of an invertible operator being invertible — forces \(g=0_{\mathbb{V}}\). Hence \(\ker\left(\left(T-\mu\identity\right)^{n}\right)\) lies inside \(\mathbb{W}\). Computed inside \(\mathbb{W}\) it is the kernel of the \(\dim{\mathbb{W}}\)-th power, which by the first paragraph of the proof of Lemma 5.97 is the same subspace as the kernel of the \(n\)-th power. So Equation (5.126) is the induction hypothesis with \(\mathbb{G}_{1}\) adjoined.

Finally \(\mathbb{E}_{\lambda_{i}}\subseteq\mathbb{G}_{i}\) because a vector killed by \(T-\lambda_{i}\identity\) is killed by its powers (Equation (5.102)); invariance is Lemma 5.97; and the nilpotency is Equation (5.125) itself.

Derives Corollary 5.99. Any basis of \(\mathbb{G}_{i}\) consists of eigenvectors of \(S\) with eigenvalue \(\lambda_{i}\), so collecting such bases over \(i\) diagonalizes \(S\), by Equation (5.126). On \(\mathbb{G}_{i}\) the operator \(N\) is \(T-\lambda_{i}\identity\), nilpotent with \(N^{n}=\Theta\) there by Equation (5.125); since the subspaces exhaust \(\mathbb{V}\), \(N^{n}=\Theta\) throughout. And on \(\mathbb{G}_{i}\) the operator \(S\) is a scalar, which commutes with everything that preserves the subspace, so \(\comm{S}{N}=\Theta\) on each summand and hence everywhere.

Derives Proposition 5.102. Let \(\set{e_{i}}^{p}_{i=1}\) and \(\set{f_{j}}^{q}_{j=1}\) be bases of \(\mathbb{V}\) and \(\mathbb{W}\). The \(p+q\) vectors \((e_{i},0_{\mathbb{W}})\) and \((0_{\mathbb{V}},f_{j})\) generate: any \((v,w)\) equals \(\sum_{i}v^{i}(e_{i},0_{\mathbb{W}}) +\sum_{j}w^{j}(0_{\mathbb{V}},f_{j})\) by Equation (5.127). They are independent: a vanishing combination has first slot \(\sum_{i}a^{i}e_{i}=0_{\mathbb{V}}\) and second slot \(\sum_{j}b^{j}f_{j}=0_{\mathbb{W}}\), and each forces its own coefficients to vanish by Equation (5.29). So they form a basis, and Lemma 5.38 makes the count \(p+q\) the dimension.

Derives Proposition 5.104. \((1)\Rightarrow(2)\): existence of a decomposition for every \(v\) is \(\mathbb{U}+\mathbb{W}=\mathbb{V}\). If \(x\in\mathbb{U}\cap\mathbb{W}\) then \(x=x+0_{\mathbb{V}}=0_{\mathbb{V}}+x\) are two decompositions of the same vector, the first with \(\mathbb{U}\)-part \(x\) and the second with \(\mathbb{U}\)-part \(0_{\mathbb{V}}\); uniqueness forces \(x=0_{\mathbb{V}}\).

\((2)\Rightarrow(1)\): existence is again the first half. For uniqueness, let \(u+w=u'+w'\) with \(u,u'\in\mathbb{U}\) and \(w,w'\in\mathbb{W}\). Then \(u-u'=w'-w\) lies in \(\mathbb{U}\) and in \(\mathbb{W}\), hence is \(0_{\mathbb{V}}\), so \(u=u'\) and \(w=w'\).

The isomorphism. The map \(\Sigma:(u,w)\longmapsto u+w\) is linear by Equation (5.127), surjective by the first half of (2), and injective because \(\Sigma(u,w)=0_{\mathbb{V}}\) gives \(u=-w\in\mathbb{U}\cap\mathbb{W}=\set{0_{\mathbb{V}}}\). It is therefore an isomorphism (Definition 5.53), and isomorphic spaces have equal dimension (Remark 5.55), so Equation (5.130) follows from Equation (5.128).

Derives Proposition 5.105. \(\mathbb{U}\cap\mathbb{W}\) is a subspace, being defined by membership of two subspaces at once. Let \(\set{z_{1},\ldots,z_{r}}\) be a basis of it and complete it, by Lemma 5.38, to a basis \(\set{z_{1},\ldots,z_{r},u_{1},\ldots,u_{p}}\) of \(\mathbb{U}\) and to a basis \(\set{z_{1},\ldots,z_{r},w_{1},\ldots,w_{q}}\) of \(\mathbb{W}\). We claim that \(\set{z_{1},\ldots,z_{r},u_{1},\ldots,u_{p},w_{1},\ldots,w_{q}}\) is a basis of \(\mathbb{U}+\mathbb{W}\), which gives Equation (5.131) at once, the count being \(r+p+q=(r+p)+(r+q)-r\).

It generates. Every element of \(\mathbb{U}+\mathbb{W}\) is a sum of an element of \(\mathbb{U}\) and an element of \(\mathbb{W}\), each of which expands in its own basis.

It is independent. Suppose

Then \(x=\sum_{j}b^{j}w_{j}=-\sum_{k}c^{k}z_{k}-\sum_{i}a^{i}u_{i}\) lies in \(\mathbb{W}\), by the left-hand expression, and in \(\mathbb{U}\), by the right-hand one; so \(x\in\mathbb{U}\cap\mathbb{W}\) and \(x=\sum_{k}d^{k}z_{k}\) for some scalars. Comparing the two expressions for \(x\) in the basis of \(\mathbb{W}\), \(\sum_{j}b^{j}w_{j} -\sum_{k}d^{k}z_{k}=0_{\mathbb{V}}\) forces every \(b^{j}=0\) by Equation (5.29). The original relation then reduces to a vanishing combination of the basis of \(\mathbb{U}\), so the remaining coefficients vanish too.

Derives Proposition 5.107. (\(\Leftarrow\)) If one contains the other the union is the larger of the two, which is a subspace by hypothesis.

(\(\Rightarrow\)) Suppose neither containment holds, so that there are vectors \(u\in\mathbb{U}\setminus\mathbb{W}\) and \(w\in\mathbb{W}\setminus\mathbb{U}\). If \(\mathbb{U}\cup\mathbb{W}\) were a subspace it would contain \(u+w\), which must then lie in \(\mathbb{U}\) or in \(\mathbb{W}\). Both are impossible:

• if \(u+w\in\mathbb{U}\) then \(w=(u+w)-u\in\mathbb{U}\), since \(\mathbb{U}\) is closed under linear combinations, contradicting the choice of \(w\);

• if \(u+w\in\mathbb{W}\) then \(u=(u+w)-w\in\mathbb{W}\), contradicting the choice of \(u\).

Hence \(\mathbb{U}\cup\mathbb{W}\) is not a subspace.

Derives Corollary 5.108. \(\mathbb{U}+\mathbb{W}\) is a subspace containing \(\mathbb{U}\cup \mathbb{W}\) — take \(w=0_{\mathbb{V}}\) or \(u=0_{\mathbb{V}}\) in Equation (5.129) — so it contains the smallest such subspace, which is \(\gen{\mathbb{U}\cup\mathbb{W}}\) by Remark 5.13. Conversely every \(u+w\) is a linear combination of two elements of \(\mathbb{U}\cup\mathbb{W}\) and therefore lies in \(\gen{\mathbb{U}\cup\mathbb{W}}\) by Definition 5.12. The two inclusions give Equation (5.134).

Derivation. Derives Proposition 5.112. Existence. Both \(B_{\mathrm{s}}\) and \(B_{\mathrm{a}}\) are bilinear, being linear combinations of the bilinear maps \((u,v)\longmapsto B(u,v)\) and \((u,v)\longmapsto B(v,u)\); exchanging \(u\) and \(v\) in Equation (5.140) leaves the first unchanged and reverses the sign of the second, so they are respectively symmetric and antisymmetric in the sense of Definition 5.111; and adding them returns \(B\). The halving is where the characteristic hypothesis is spent here — \(\tfrac{1}{2}\) must exist in the field — and it is spent again, in the other direction, in the next paragraph.

Uniqueness. Suppose \(B=S+A=S'+A'\) with \(S,S'\) symmetric and \(A,A'\) antisymmetric, and put \(C=S-S'=A'-A\). Then \(C\) is symmetric, being a difference of symmetric forms, and antisymmetric, being a difference of antisymmetric ones, so for all \(u,v\)

Since \(2\neq0\) in the field, \(C=0\), that is \(S=S'\) and \(A=A'\). In matrix terms Equation (5.140) is the familiar \(\left[B\right]=\tfrac{1}{2}\left(\left[B\right] +\left[B\right]\transpose\right) +\tfrac{1}{2}\left(\left[B\right]-\left[B\right]\transpose\right)\), and the uniqueness argument says that the only array equal to both its own transpose and minus it is the zero array.

Derives Proposition 5.114. \(\flat_{B}(u)\) is a functional by Equation (5.136), and \(u\longmapsto\flat_{B}(u)\) is linear by Equation (5.135), so \(\flat_{B}\in\mathcal{L}(\mathbb{V}, \mathbb{V}^{\ast})\).

\((1)\Rightarrow(2)\): the first half of Definition 5.113 says exactly that \(\ker\flat_{B}=\set{0_{\mathbb{V}}}\), so \(\flat_{B}\) is injective; and an injective linear map between spaces of the same finite dimension — \(\dim{\mathbb{V}^{\ast}}=\dim{\mathbb{V}}=n\) was established after Equation (5.87) — is surjective.

\((2)\Rightarrow(3)\): in the basis \(\set{e_{i}}\) and its dual basis \(\set{f^{j}}\) of Equation (5.86),

because both sides agree on every \(e_{k}\): the left gives \(B_{ik}\) and the right gives \(\sum_j B_{ij}\delta^{j}_{\ k}=B_{ik}\). So the matrix of \(\flat_{B}\) in these bases is \(\left[B\right]\transpose\); a linear map between spaces of equal finite dimension is bijective exactly when its matrix is invertible, that is, when its determinant does not vanish, and \(\det\left(\left[B\right]\transpose\right)=\det\left[B\right]\).

\((3)\Rightarrow(1)\): if \(B(u,v)=0\) for all \(v\) then in components \(\sum_{i}u^{i}B_{ij}=0\) for every \(j\), a homogeneous linear system whose matrix is invertible, so \(u^{i}=0\) for all \(i\); the other half is the same argument on the other index. Independence of the basis follows from \(\det\left[B'\right]=(\det\left[A\right])^{2}\det\left[B\right]\) with \(\det\left[A\right]\neq0\).

Derives Theorem 5.116. Existence. By induction on \(n\). If \(B=0\) then \(p=q=0\), \(r=n\) and any basis will do. Otherwise there is a vector \(v\) with \(B(v,v)\neq0\): were \(B(u,u)=0\) for every \(u\), polarization would give

for all \(u,w\), using symmetry, and \(B\) would vanish identically. Set \(e_{1}=v/\sqrt{\abs{B(v,v)}}\), so that \(B(e_{1},e_{1})=\pm1\), and let

which is a subspace, being defined by a linear condition. Every \(u\in\mathbb{V}\) decomposes as

whose second term lies in \(\mathbb{V}_{1}\), as substituting it into \(B(e_{1},\ )\) shows; and \(\gen{\set{e_{1}}}\cap\mathbb{V}_{1} =\set{0_{\mathbb{V}}}\), since \(B(e_{1},\lambda e_{1}) =\lambda B(e_{1},e_{1})=0\) forces \(\lambda=0\). Hence \(\mathbb{V}=\gen{\set{e_{1}}}\oplus\mathbb{V}_{1}\) and \(\dim{\mathbb{V}_{1}}=n-1\). Applying the induction hypothesis to the restriction of \(B\) to \(\mathbb{V}_{1}\) — which is again symmetric — and reordering the resulting basis gives Equation (5.142).

Uniqueness. \(r\) is basis-independent because \(n-r\) is the rank of \(\left[B\right]\), and congruent matrices have equal rank (\(\left[A\right]\) being invertible). For \(p\), suppose two bases give \((p,q,r)\) and \((p',q',r)\) with, say, \(p>p'\). Let \(\mathbb{P}\) be the span of the \(p\) basis vectors of the first basis on which \(B\) is \(+1\): on \(\mathbb{P}\setminus\set{0_{\mathbb{V}}}\) the form is strictly positive, since \(B\left(\sum_{i\leq p}a^{i}e_{i},\sum_{j\leq p}a^{j}e_{j}\right) =\sum_{i\leq p}(a^{i})^{2}>0\). Let \(\mathbb{M}'\) be the span of the \(q'+r=n-p'\) vectors of the second basis on which \(B\) is \(-1\) or \(0\): on \(\mathbb{M}'\) the form satisfies \(B(u,u)\leq0\), by the same computation with the signs of Equation (5.142). Now

so \(\mathbb{P}\cap\mathbb{M}'\) contains a non-zero vector \(u\) — two subspaces whose dimensions sum to more than \(n\) cannot meet only in \(0_{\mathbb{V}}\), since \(\dim(\mathbb{P}+\mathbb{M}')\leq n\). That vector would satisfy \(B(u,u)>0\) and \(B(u,u)\leq0\) at once. Hence \(p\leq p'\), and by symmetry \(p'\leq p\); with \(r\) fixed, \(q=n-p-r\) is determined too.

Derivation. Derives Proposition 5.119. Antisymmetry first gives \(\omega(u,u)=-\omega(u,u)\), hence \(\omega(u,u)=0\) for every \(u\), the field being \(\R\). The construction is then an induction on \(n\) — a symplectic Gram–Schmidt, in which vectors are paired off instead of being normalized one at a time.

If \(\mathbb{V}=\set{0_{\mathbb{V}}}\) there is nothing to prove and \(m=0\). Otherwise choose any \(e_{1}\neq0_{\mathbb{V}}\). Non-degeneracy (Definition 5.113) supplies a \(w\) with \(\omega(e_{1},w)\neq0\), and

These two are linearly independent: were \(f_{1}=\lambda e_{1}\) we should have \(\omega(e_{1},f_{1})=\lambda\,\omega(e_{1},e_{1})=0\). Put \(W=\gen{\set{e_{1},f_{1}}}\) and

a subspace, being the intersection of the kernels of two functionals.

The two split \(\mathbb{V}\). Given \(v\in\mathbb{V}\), set

Using \(\omega(e_{1},e_{1})=\omega(f_{1},f_{1})=0\) and \(\omega(f_{1},e_{1})=-1\),

so \(v'\in W^{\omega}\) and \(v=\left(\omega(v,f_{1})e_{1} -\omega(v,e_{1})f_{1}\right)+v'\in W+W^{\omega}\). The sum is direct: if \(v=ae_{1}+bf_{1}\) lies in \(W^{\omega}\) then \(0=\omega(v,f_{1})=a\) and \(0=\omega(v,e_{1})=-b\). Hence \(\mathbb{V}=W\oplus W^{\omega}\) and \(\dim{W^{\omega}}=n-2\).

The restriction is again non-degenerate. Let \(v\in W^{\omega}\) satisfy \(\omega(v,u)=0\) for every \(u\in W^{\omega}\). It also annihilates \(e_{1}\) and \(f_{1}\), hence the whole of \(W\); by the direct sum \(\omega(v,\ )\) vanishes on \(\mathbb{V}\), and \(v=0_{\mathbb{V}}\) by non-degeneracy of \(\omega\).

The induction hypothesis applied to \(W^{\omega}\) therefore makes \(n-2\) even — so \(n\) is even, \(n=2m\) — and supplies vectors \(e_{2},\ldots,e_{m},f_{2},\ldots,f_{m}\) of \(W^{\omega}\) satisfying Equation (5.143) among themselves. All products of \(e_{1}\) or \(f_{1}\) with one of them vanish, since those vectors lie in \(W^{\omega}\), so the whole collection satisfies Equation (5.143) and, being a basis of \(W\) together with a basis of \(W^{\omega}\), is a basis of \(\mathbb{V}\). Listing the \(e_{i}\) first and the \(f_{i}\) second turns Equation (5.143) into Equation (5.144). That two such forms are congruent now follows because each is carried to Equation (5.144) by a change of basis, and Equation (5.139) composes.

Derivation. Derives Proposition 5.125. Every identity is read off the definition Equation (5.74) of the matrix representation, here with \(\mathbb{W}=\mathbb{V}\) and \(B_{2}=B_{1}=B\), so that \(T(e_{i})=\sum_{j}T^{j}_{\ i}e_{j}\).

Linearity, Equation (5.152). By the definition of the linear combination of two LT's,

and by the uniqueness established in Proposition 5.45 the coefficients displayed are the matrix entries of \(\alpha S+\beta T\).

Composition, Equation (5.153). Apply the two transformations in turn:

so that \(\left(S\circ T\right)^{k}_{\ i}=S^{k}_{\ j}T^{j}_{\ i}\) in the summation convention — precisely the row-by-column rule for the matrix product \(\left[S\right]\left[T\right]\). Note that the order is preserved: the transformation applied first stands to the right in both notations.

Bracket, Equation (5.154). Combine the two:

Bijectivity is Proposition 5.45: distinct endomorphisms have distinct matrices, and every array arises from the endomorphism it defines through Equation (5.74).

Structure constants. Let \(A\subset\mathcal{L}(\mathbb{V})\) be a subalgebra with basis \(\set{X_{a}}\) and structure constants defined by Equation (5.150), \(\comm{X_{a}}{X_{b}}=C^{c}_{\ ab}X_{c}\). Taking matrix representations of both sides and using Equations (5.152) and (5.154),

with the same coefficients \(C^{c}_{\ ab}\). Moreover the \(\left[X_{a}\right]\) are linearly independent, since \(T\longmapsto\left[T\right]\) is injective and linear, so they are a basis of the image and the two algebras have the same structure constants in corresponding bases. Any identity following from the axioms — the antisymmetry Equation (5.157) and the Jacobi identity Equation (5.158) of a Lie algebra, for instance — therefore holds for the matrices exactly when it holds for the operators.

Derives Proposition 5.130. Equation (5.161) says \(TT\transpose=\identity\); in finite dimension a one-sided inverse is the inverse (Proposition 5.49), so \(T\transpose T=\identity\) as well, and the two conditions are interchangeable. Given them, \(\braket{Tu}{Tv}=u\transpose T\transpose Tv=u\transpose v\), which is Equation (5.163). Conversely, taking \(u\) and \(v\) to be the standard basis vectors in Equation (5.163) reads off \(\left(T\transpose T\right)_{ij}=\delta_{ij}\). The norm statement is Equation (5.44) with \(u=v\), and the angle statement then follows because both the numerator and the denominator of Equation (5.18) are unchanged.

Derives Lemma 5.132. Fix \(p\) and let \(R^{(p)}\) be the diagonal array with \(+1\) in the \(p\)-th place and \(-1\) in the other two: it satisfies \(R\transpose R=\identity\) and has determinant \((+1)(-1)(-1)=+1\), so it lies in \(\SO(3)\) by Equation (5.162). Being diagonal, \(R^{(p)}_{ij}=s_{i}\delta_{ij}\) with \(s_{p}=1\) and \(s_{q}=-1\) for \(q\neq p\), so Equation (5.164) collapses to

with no summation, \(N-m_{p}\) being the number of indices different from \(p\). If \(N-m_{p}\) is odd the component equals its own negative.

Derives Theorem 5.133. The converse first. \(\delta_{ij}\) is isotropic, because \(R_{ia}R_{jb}\delta_{ab}=R_{ia}R_{ja}=\delta_{ij}\) is \(RR\transpose=\identity\) (Proposition 5.130); products of Kronecker deltas are then isotropic index group by index group; and \(\varepsilon_{ijk}\) is isotropic by Equation (5.59), since \(\det R=1\) on \(\SO(3)\).

Two further elements of \(\SO(3)\) do the rest of the work. The cyclic array \(C\), with \(C_{21}=C_{32}=C_{13}=1\) and every other entry zero, is orthogonal and has determinant \(+1\), a \(3\)-cycle being an even permutation; substituted in Equation (5.164) it gives

that is, the components are unchanged when every index value is relabelled cyclically. The quarter turn \(Q\) about the third axis, with \(Q_{21}=1\), \(Q_{12}=-1\), \(Q_{33}=1\), is likewise in \(\SO(3)\) and gives

\(m_{1}\) being the number of indices equal to \(1\).

Rank one. Lemma 5.132 asks every \(m_{p}\) to be odd while \(m_{1}+m_{2}+m_{3}=1\), which is impossible; every component vanishes.

Rank two. Every \(m_{p}\) must be even and they sum to \(2\), so one of them is \(2\) and the others vanish: \(T_{ij}=0\) unless \(i=j\). By Equation (5.169), \(T_{11}=T_{33}=T_{22}\), and Equation (5.166) holds with \(a\) that common value.

Rank three. Every \(m_{p}\) must be odd and they sum to \(3\), so each equals \(1\): only components with three distinct indices survive. Equation (5.169) gives \(T_{123}=T_{312}=T_{231}\) and \(T_{213}=T_{132}=T_{321}\), and Equation (5.170) applied to \((1,2,3)\), where \(m_{1}=1\), gives \(T_{123}=-T_{213}\). The six surviving components are therefore \(\pm T_{123}\) according to the parity of the index arrangement, which is Equation (5.167) with \(a=T_{123}\).

Rank four. Every multiplicity must be even and they sum to \(4\), so they are either \((4,0,0)\) or \((2,2,0)\) in some order: the surviving components are the three \(T_{pppp}\) and, for each ordered pair of distinct values \(p\neq q\), the three arrangements \(T_{ppqq}\), \(T_{pqpq}\), \(T_{pqqp}\). Equation (5.169) makes \(T_{1111}=T_{3333}=T_{2222}\), a common value \(\alpha\), and carries the ordered pair \((p,q)\) to \((\sigma(p),\sigma(q))\), whose orbits are \(\set{(1,2),(3,1),(2,3)}\) and \(\set{(2,1),(1,3),(3,2)}\); Equation (5.170) applied to \((1,1,2,2)\), \((1,2,1,2)\) and \((1,2,2,1)\) — each with \(m_{1}=2\), hence with sign \(+1\) — gives \(T_{1122}=T_{2211}\), \(T_{1212}=T_{2121}\) and \(T_{1221}=T_{2112}\), joining the two orbits to each other. Hence the three numbers

do not depend on which ordered pair \(p\neq q\) is used, and Equation (5.168) reproduces every surviving component except possibly \(T_{pppp}\), where it returns \(a+b+c\) instead of \(\alpha\).

The last relation. Let \(R(\theta)\) be the rotation by \(\theta\) about the third axis, \(R_{11}=R_{22}=\cos{\theta}\), \(R_{21}=-R_{12}=\sin{\theta}\), \(R_{33}=1\). Only the values \(1\) and \(2\) occur in \(R_{1j}\), so Equation (5.164) for the component \(T_{1111}\) involves only components whose indices lie in \(\set{1,2}\), each carrying a factor \(\cos{\theta}\) for an index \(1\) and \(-\sin{\theta}\) for an index \(2\). Those with an odd number of \(2\)'s vanish by Lemma 5.132, and the rest give, writing \(\gamma=\cos{\theta}\) and \(s=\sin{\theta}\),

the six mixed components \(T_{1122},T_{1212},T_{1221},T_{2112}, T_{2121},T_{2211}\) each carrying the factor \(\gamma^{2}s^{2}\) and summing to \(2(a+b+c)\). Since \(\gamma^{4}+s^{4}=1-2\gamma^{2}s^{2}\), this reads \(2\gamma^{2}s^{2}\left(a+b+c-\alpha\right)=0\), and \(\theta=\pi/4\) makes \(\gamma^{2}s^{2}=1/4\), so \(\alpha=a+b+c\).

Derives Proposition 5.136. Invertibility. Taking determinants in Equation (5.174) and using \(\det\left(\bigl(M\bigr)\transpose\right)=\det{M}\), which is immediate from the Leibniz expansion Equation (5.19) on relabelling \(\sigma\longmapsto\sigma^{-1}\),

and \(\det{J}\neq0\) because \(J\) is invertible by Equation (5.173). Hence \(\left(\det{M}\right)^{2}=1\), so \(\det{M}=\pm1\neq0\) and \(M\in\GL(2n,\R)\) by Equation (5.81).

Closure. If \(M\) and \(N\) both satisfy Equation (5.174) then

the first line using \(\bigl(MN\bigr)\transpose =\bigl(N\bigr)\transpose\bigl(M\bigr)\transpose\).

Inverses. Transposing \(MM^{-1}=M^{-1}M=\identity_{2n}\) gives \(\bigl(M^{-1}\bigr)\transpose\bigl(M\bigr)\transpose =\bigl(M\bigr)\transpose\bigl(M^{-1}\bigr)\transpose=\identity_{2n}\), so \(\bigl(M^{-1}\bigr)\transpose\) is the inverse of \(\bigl(M\bigr)\transpose\) — the uniqueness of Proposition 5.48. Multiplying \(\bigl(M\bigr)\transpose J\,M=J\) on the left by \(\bigl(M^{-1}\bigr)\transpose\) and on the right by \(M^{-1}\) therefore yields

so \(M^{-1}\) is symplectic as well. Associativity and the identity \(\identity_{2n}\), which trivially satisfies Equation (5.174), are inherited from \(\GL(2n,\R)\).

Derives Lemma 5.137. First, \(\Omega\) vanishes whenever two of its arguments are equal: exchanging those two leaves it unchanged and, by hypothesis, reverses its sign, so \(\Omega=-\Omega\) and \(2\Omega=0\) in \(\R\).

Expanding each argument by \(N\)-linearity,

and every term in which two indices coincide vanishes by the previous paragraph. The surviving tuples are exactly those of the form \(i_{j}=\sigma(j)\) for a permutation \(\sigma\) of \(N\) symbols; writing \(\sigma\) as a product of transpositions and applying the sign change once per transposition gives \(\Omega\left(b_{\sigma(1)},\ldots,b_{\sigma(N)}\right) =\sgn(\sigma)\,\Omega\left(b_{1},\ldots,b_{N}\right)\). Hence

whose sum is the Leibniz expansion Equation (5.19) of \(\det{\left[v^{i}_{\ j}\right]}\). This is Equation (5.175).

For Equation (5.176), note that the components of \(Mv_{j}\) are the entries of the array \(\left[M\right] \left[v^{i}_{\ j}\right]\), since a matrix acts on a column of components by Proposition 5.45. Applying Equation (5.175) twice and using the multiplicativity of the determinant,

Derives Theorem 5.138. Build from \(\omega\) the \(2n\)-argument function

the sum running over all permutations \(\sigma\) of \(2n\) symbols.

\(\Omega\) is \(2n\)-linear. In each term every argument occurs in exactly one of the \(n\) factors and occurs there once, and \(\omega\) is linear in each of its slots.

\(\Omega\) is alternating. Let \(\tau\) be the transposition exchanging two of the argument positions. Replacing the arguments by their \(\tau\)-permutation and substituting \(\sigma'=\tau\sigma\) in the sum — which merely reindexes it, \(\sigma\longmapsto\tau\sigma\) being a bijection of the permutations onto themselves — gives back Equation (5.177) with \(\sgn(\sigma)=\sgn(\tau) \sgn(\sigma')=-\sgn(\sigma')\), that is, with the opposite sign.

\(\Omega\) does not vanish identically. Evaluate it on the canonical basis of Equation (5.143), taken in the interleaved order \((e_{1},f_{1},e_{2},f_{2},\ldots,e_{n},f_{n})\). A term survives only if each factor pairs some \(e_{i}\) with the corresponding \(f_{i}\), that is, only if the \(n\) unordered pairs \(\set{\sigma(2i-1),\sigma(2i)}\) are exactly the \(n\) pairs \(\set{2i-1,2i}\) of argument positions; there are \(n!\) ways to order those pairs among the factors and \(2\) ways to order each pair internally, so \(2^{n}n!\) terms survive. Each contributes \(+1\): reversing one pair internally multiplies its factor \(\omega(e_{i},f_{i})=1\) by \(-1\), by antisymmetry, and multiplies \(\sgn(\sigma)\) by \(-1\) as well, while permuting whole pairs is an even permutation, being a product of transpositions of two-element blocks. Dividing by \(2^{n}n!\),

Conclusion. If \(M\) is symplectic then \(\omega(Mu,Mv)=\omega(u,v)\), so every factor in Equation (5.177) is unchanged when each argument is replaced by its image, and \(\Omega(Mv_{1},\ldots,Mv_{2n})=\Omega(v_{1},\ldots,v_{2n})\). But Equation (5.176) says the left-hand side equals \(\det{M}\,\Omega(v_{1},\ldots,v_{2n})\). Evaluating both on the canonical basis, where \(\Omega=1\), gives \(\det{M}=1\).

For \(n=1\) a direct computation settles the converse inclusion: for any real \(2\times2\) array \(M\) one has \(\bigl(M\bigr)\transpose J\,M =\det{M}\,J\), so Equation (5.174) reduces to \(\det{M}=1\) and \(\Sp(2,\R)=\SL(2,\R)\).

Derives Proposition 5.139. It is a subspace and a subalgebra. The defining condition Equation (5.178) is linear in \(X\), so the set is a subspace. For the bracket, let \(X,Y\in\mathfrak{sp}(2n,\R)\), so that \(\bigl(X\bigr)\transpose J=-JX\) and \(\bigl(Y\bigr)\transpose J=-JY\). Then

so \(\comm{X}{Y}\in\mathfrak{sp}(2n,\R)\), and the two Lie-algebra axioms Equations (5.155) and (5.156) hold because the commutator satisfies them on the whole matrix algebra.

The dimension. Put \(S=J\,X\), so that \(X=J^{-1}S=-J\,S\) by Equation (5.173); the assignment \(X\longmapsto S\) is a linear bijection of \(\mathcal{M}_{2n\times2n}\left[\R\right]\) onto itself. Using \(\bigl(J^{-1}\bigr)\transpose=\bigl(-J\bigr)\transpose=J\) and \(J^{2}=-\identity_{2n}\),

so the condition \(\bigl(X\bigr)\transpose J+JX=0\) becomes \(-\bigl(S\bigr)\transpose+S=0\): precisely the statement that \(S\) is symmetric. The symmetric arrays of order \(2n\) form a subspace of dimension \(2n(2n+1)/2=n(2n+1)\), a basis being given by the arrays with \(1\) in a diagonal entry and by those with \(1\) in a pair of mirror-image off-diagonal entries. This is Equation (5.179).

That the group has the same dimension is the general correspondence between a Lie group and its algebra, established in Lie Groups, Lie Algebras, and Fibre Bundles; the check \(n=1\) is reassuring, for \(n(2n+1)=3\) agrees with the three parameters of \(\SL(2,\R)\) — four entries subject to one determinant condition — as Theorem 5.138 requires.

Derives Lemma 5.150. Let \(w\in\mathbb{S}^{\perp}\) and \(a\in G\); we must show \(U(a)w\in\mathbb{S}^{\perp}\), that is, \(\braket{s}{U(a)w}=0\) for every \(s\in\mathbb{S}\). Because \(U\) is a group homomorphism into \(\GL(\mathbb{V})\), the operator \(U(a)\) is invertible with \(U(a)^{-1}=U(a^{-1})\) by Equation (4.25), so we may write \(s=U(a)U(a^{-1})s\) and use Equation (5.183):

the last step because \(a^{-1}\in G\) — a group contains the inverse of each of its elements — so \(U(a^{-1})s\in\mathbb{S}\) by the invariance of \(\mathbb{S}\), and \(w\) is orthogonal to every vector of \(\mathbb{S}\). As \(s\) was arbitrary, \(U(a)w\in\mathbb{S}^{\perp}\).

Derives Proposition 5.151. We prove the contrapositive: if \(U\) is reducible then it is totally reducible. Suppose \(U\) is reducible, so that by Definition 5.148 there exists a non-trivial invariant subspace \(\mathbb{S}\). By Lemma 5.150 the orthogonal complement \(\mathbb{S}^{\perp}\) is invariant too, and that is exactly the condition Definition 5.149 imposes for \(U\) to be totally reducible.

Hence no unitary representation can be reducible without being totally reducible, and the stated implication follows.

Derives Proposition 5.153. Part 1. That \(\mathbb{S}^{\perp}\) is invariant is Lemma 5.150, which used no finiteness at all. That the two subspaces exhaust \(\mathbb{V}\),

is Corollary 5.29, and this is where the finite dimension enters: it is what guarantees the orthonormal basis of \(\mathbb{S}\) from which the projection was built. Every vector of \(\mathbb{V}\) therefore splits uniquely as \(v=s+w\) with \(s\in\mathbb{S}\) and \(w\in\mathbb{S}^{\perp}\), and since \(U(a)s\in\mathbb{S}\) and \(U(a)w\in\mathbb{S}^{\perp}\), the operator \(U(a)\) acts on the two summands separately:

In a basis adapted to the splitting the matrix \(D(a)\) is block diagonal for every \(a\) simultaneously, which is what totally reducible means operationally.

Each restriction is itself a representation: it is linear; it is invertible on its subspace, because \(U(a^{-1})\) maps that subspace into itself and \(U(a)U(a^{-1})=U(e)=\id\); and it inherits \(U(ab)=U(a)U(b)\) from \(U\). It is unitary, because Equation (5.183) holds in particular for vectors of the subspace.

Part 2. Induction on \(n=\dim{\mathbb{V}}\). If \(n=1\) the only subspaces are the trivial ones, so \(U\) is irreducible by Definition 5.148 and the decomposition has one summand. Let \(n>1\) and suppose the claim proved for all dimensions smaller than \(n\). If \(U\) is irreducible we are again done. If not, there is a non-trivial invariant subspace \(\mathbb{S}\), and by Part 1

the strict inequalities holding because \(\mathbb{S}\) is neither the null space nor the whole of \(\mathbb{V}\). The restrictions of \(U\) to \(\mathbb{S}\) and to \(\mathbb{S}^{\perp}\) are unitary representations on spaces of strictly smaller dimension, so the induction hypothesis applies to each and decomposes it into irreducible summands; collecting the two lists gives Equation (5.184). The recursion terminates after at most \(n\) steps because the dimension drops by at least one each time, and the summands are mutually orthogonal by construction.

Derives Proposition 5.155. Step 1: Equation (5.185) is an inner product. Each summand is the original inner product evaluated on the images of \(u\) and \(v\), so the five axioms transfer term by term.

• Linearity in the second argument Equations (5.34) and (5.35): \(U(g_{i})\) is linear, so \(U(g_{i})(av+bw)=aU(g_{i})v+bU(g_{i})w\), and Equations (5.34) and (5.35) applied inside each summand give the same for the average. The conjugate rules Equations (5.36) and (5.37) follow in the same way.

• Hermiticity Equation (5.33): each summand satisfies

  \[ \braket{U(g_{i})u}{U(g_{i})v}^{\ast}=\braket{U(g_{i})v}{U(g_{i})u}\ec \]

  and \(1/n_{G}\) is real, so the sum inherits the property.

• Positive definiteness Equation (5.32): let \(u\neq0_{\mathbb{V}}\). Since \(U(g_{i})\in\GL(\mathbb{V})\) is invertible, \(U(g_{i})u\neq0_{\mathbb{V}}\), so every summand \(\braket{U(g_{i})u}{U(g_{i})u}\) is strictly positive by Equation (5.32); a sum of \(n_{G}\) strictly positive reals divided by \(n_{G}\) is strictly positive. Equation (5.31) follows, the direction (\(\Leftarrow\)) from Equation (5.38).

Note where the finiteness of \(G\) has been used: the sum has finitely many terms, and \(n_{G}\) is invertible in \(\mathbb{K}\).

Step 2: invariance, Equation (5.186). Using the homomorphism property \(U(g_{i})U(a)=U(g_{i}a)\),

the third line because right multiplication by \(a\) is a bijection of \(G\) onto itself: it is injective, since \(g_{i}a=g_{j}a\) implies \(g_{i}=g_{j}\) on multiplying by \(a^{-1}\), and therefore surjective on a finite set, so \(\set{g_{i}a}\) is a relabelling of \(\set{g_{j}}\) and the summands are the same numbers in another order. This is the whole of the trick: an average over the group cannot tell which element of the group it is looking at. With Equation (5.186) in hand, \(U\) satisfies Equation (5.183) for the product \(\avg{\ ,\ }_{G}\), that is, it is a unitary representation on the inner-product space \((\mathbb{V},\avg{\ ,\ }_{G})\).

Step 3: the equivalence Equation (5.187). What remains is bookkeeping: to say “unitary” with respect to the original product rather than the new one. By Proposition 5.28 applied twice, choose a basis \(\set{e_{i}}^{n}_{i=1}\) orthonormal for \(\braket{\ }{\ }\) and a basis \(\set{\tilde{e}_{i}}^{n}_{i=1}\) orthonormal for \(\avg{\ ,\ }_{G}\), and let \(S\) be the linear map determined by \(S\tilde{e}_{i}=e_{i}\), which is invertible because it carries a basis to a basis. For \(u=\sum_{i}u^{i}\tilde{e}_{i}\) and \(v=\sum_{i}v^{i}\tilde{e}_{i}\), orthonormality of each basis for its own product gives

that is, \(S\) carries the averaged product to the original one. Therefore, for all \(x,y\in\mathbb{V}\),

so \(U'\) satisfies Equation (5.183) for \(\braket{\ }{\ }\) and is a unitary representation, equivalent to \(U\) by construction.

Derives Theorem 5.158. Both hypotheses on \(\mathbb{V}\) are used at once, and only once: because \(\dim{\mathbb{V}}\) is finite and non-zero, \(A\) has, in any basis, a characteristic polynomial \(\det\left(\left[A\right]-\lambda\identity\right)\) of degree \(\dim{\mathbb{V}}\geq1\), and because \(\C\) is algebraically closed that polynomial has a root \(\lambda\). For that \(\lambda\) the operator \(A-\lambda\identity\) is not invertible, so its kernel

is a subspace different from the null space: an eigenvector exists, which is precisely what a real or an infinite-dimensional space would not guarantee.

\(\mathbb{E}\) is invariant under the representation. Indeed, for \(v\in\mathbb{E}\) and any \(a\in G\),

so \(U(a)v\in\mathbb{E}\).

Irreducibility (Definition 5.148) leaves an invariant subspace only the two trivial possibilities, and \(\mathbb{E}\) is not the null space; hence \(\mathbb{E}=\mathbb{V}\). That is \(Av=\lambda v\) for every \(v\in\mathbb{V}\), which is Equation (5.196). Note that nothing was assumed about \(A\) beyond linearity — in particular it was not assumed diagonalizable, and no eigenbasis was used, only one eigenvalue.

Derives Theorem 5.160. We proceed in two steps.

• First note that the image and the kernel of the operator \(A\) are the subspaces

  \begin{align*} \im(A) & = \left\{\ket{v}\in\mathbb{V}\mid\ket{v}=A\ket{v'}\right\} \leq\mathbb{V}\ec\\ \ker(A) & = \left\{\ket{v'}\in\mathbb{V'}\mid A\ket{v'}=\ket{0},\ 0\in\mathbb{V}\right\}\leq\mathbb{V}'\ep \end{align*}

  The image of \(A\) is an invariant subspace with respect to \(U\), since

  \begin{align*} U(a)\ket{v} & = U(a)A\ket{v'}\\ & = AU'(a)\ket{v'}\ec \end{align*}

  and since \(U'(a)\ket{v'}\in\mathbb{V}'\), we have \(AU'(a)\ket{v'}\in\im(A)\), and hence \(U(a)\ket{v}\in\im(A)\). As \(U\) is irreducible, \(\im(A)\) is a trivial invariant subspace: the image of \(A\) would be the null space if and only if \(A=0\); otherwise one may have \(\im(A)=\mathbb{V}\).

  On the other hand, the kernel of \(A\) is an invariant subspace with respect to \(U'\). This is because

  \begin{align*} AU'(a)\ket{v'} & = U(a)A\ket{v'}\\ & = 0 \in\mathbb{V}\ec \end{align*}

  so \(U'(a)\ket{v'}\in\ker(A)\). Since we are assuming that \(U'\) is irreducible, the kernel of \(A\) is a trivial invariant subspace: again, the kernel of \(A\) would be the whole space \(\mathbb{V}'\) if and only if \(A=0\); otherwise the kernel of \(A\) is the null space.

  From these observations we may summarize: either \(A=0\), or the kernel of \(A\) is the null space and the image of \(A\) is the whole codomain, which tells us that \(A\) is a bijection establishing an isomorphism between the vector spaces \(\mathbb{V}'\) and \(\mathbb{V}\).

• Moreover, since \(A\) is bijective its inverse exists. Therefore, operating on the right with \(A^{-1}\),

  \begin{align*} AU'(a) & = U(a)A\\ AU'(a)A^{-1} & = U(a)\ec \end{align*}

  so \(U(a)\) and \(U'(a)\) are equivalent.

Derives Proposition 5.161. Consider an irreducible representation \(U:G\longrightarrow\GL(\mathbb{V})\) of an abelian group \(G\), with \(\mathbb{V}\) complex and \(0<\dim{\mathbb{V}}<\infty\). Let \(a,b\in G\). Applying \(U\) to \(ab=ba\),

By Schur's first lemma (Theorem 5.158), applied to the operator \(U(a)\), which commutes with every \(U(b)\),

so that

and thus

This does not mean that all the vectors of \(\mathbb{V}\) are eigenvectors, but rather that the irreducible representations of an abelian group are one-dimensional.

Derives Proposition 5.164. \(\ker\rho\) is a subspace because \(\rho\) is linear. Let \(X\in\ker\rho\) and \(Y\in A\). Then, by Equation (5.215),

so \(\comm{X}{Y}\in\ker\rho\); taking linear combinations, \(\comm{\ker\rho}{A}\subset\ker\rho\), which is Definition 5.124. If \(A\) is simple the ideal \(\ker\rho\) is either \(\set{0_{A}}\), and \(\rho\) is injective, that is, faithful, or the whole of \(A\), and \(\rho\) sends everything to \(\Theta\).

Derivation. Derives Equation (5.218). Linearity in \(X\) is manifest, both terms being linear. For Equation (5.215), write \(\sigma(X)\) for the right-hand side of Equation (5.218) and use that operators acting on different factors commute, \(\comm{P\otimes\identity}{\identity\otimes Q}=0\), together with \((P\otimes\identity)(P'\otimes\identity)=PP'\otimes\identity\):

the two cross terms of the first line having cancelled in pairs.

Derives Proposition 5.167. Each \(\ad_{X}\) is a linear transformation of \(L\) by right distributivity Equation (5.146) and the scalar rule Equation (5.147), so \(\ad_{X}\in\mathcal{L}(L)\); and \(X\longmapsto\ad_{X}\) is itself linear by left distributivity Equation (5.145). What must be verified is Equation (5.215), namely \(\ad_{\comm{X}{Y}}=\comm{\ad_{X}}{\ad_{Y}}\). Applying both sides to an arbitrary \(Z\),

so the adjoint map is a representation — and note what the derivation shows: the Jacobi identity is exactly the statement that \(\ad\) is a representation, neither more nor less.

For the matrices, apply Equation (5.219) to a basis vector and compare with Equation (5.74) and the definition Equation (5.150) of the structure constants:

whose coefficients are, by Proposition 5.45, the matrix entries claimed in Equation (5.220).

Finally \(\ad_{X}=\Theta\) means \(\comm{X}{Y}=0_{L}\) for every \(Y\), which is the definition of the centre; by Proposition 5.164 the centre is an ideal, as it must be.

Derives Corollary 5.168. Immediate from Proposition 5.167: the kernel is the centre, and a representation is faithful exactly when its kernel is trivial. For a simple \(L\) the centre is an ideal, hence \(\set{0_{L}}\) or \(L\); the second would make \(L\) abelian and is excluded. The matrices are those of Equation (5.220), and Proposition 5.125 guarantees they satisfy the same algebra.