Since our general discussion of commutativity involves the use of invariant subspaces, we opted for not directly using the Spectral Theorem in this subsection. This will make more clear at what point self-adjointness is indeed necessary (see also Remark B.1.2).
We begin by showing (Proposition B.1.2) that every A ∈ L(E,R) has an invariant subspace of dimension 1 or 2. Equivalently, either there exists a non-null vectoru∈E such that Au=λu or there exist linearly independent u, v∈ E such that Au and Av are both linear combinations of u andv, i.e., Au=αu+βv,Av=γu+δv.
To show this, we first prove Lemma B.1.1, which states there exists an irreducible monic polynomialpof degree 1 or 2 such that the Ker(p(A)) is non-empty (a monic polynomial is a polynomial whose coefficient of the highest order term is 1). The proof of Lemma B.1.1 makes use of the Fundamental Theorem of Algebra, which implies that every monic real polynomial is decomposable as the product of irreducible monic polynomials of the first and second degrees. Here, one should remember that an irreducible second degree polynomial does not have real roots.
Denote p(x) =a0+a1x+...+anxn, and p(A) =a0I+a1A+...+anAn.
1 or 2 and a non-null vector v such that p(A)v= 0.
Proof. The space L(E,R) has dimension n2, and therefore the operators I, A, ..., An2 are linearly dependent. This means there existα0, α1, ..., αn2, of which at least one is not zero, such that
α0I +α1A+...+αn2An 2
= 0.
Letαmbe the highest-indexed non-zero coefficient. If we setβi =αi/αm, we obtain a monic
polynomial
q(x) :=β0+β1x+...+βm−1xm−1+xm
such that q(A) = 0. By the Fundamental Theorem of Algebra, we can factor q(x) =
q1(x)...qk(x), where eachqi(x) is a monic irreducible polynomial of degree 1 or 2. Therefore,
q(A) =q1(A)... qk(A) = 0,
which implies there exists i∈ {1, ..., k} such thatqi(A) is not invertible. Therefore, there
exists a non-nullv such thatqi(A)v= 0. To finish the proof, just setp=qi.
Proposition B.1.2. Any A∈ L(E,R) has an invariant subspace of dimension 1 or 2. Proof. Let p be the polynomial given by Lemma B.1.1. If p(x) = x−λ, then p(A)v = (A−Iλ)v= 0, and thus we obtain a 1-dimensional invariant subspace.
Alternatively, if p is of degree 2, then we can write it as p(x) =x2+ax+b, a, b∈ R.
This means that p(A)v = A2v+aAv+bv = 0, and thus, A(Av) = −a(Av)−bv. Thus,
the subspace generated by v and Av is invariant by A. Furthermore, this subspace must be 2-dimensional. In fact, assume by contradiction that v and Av are linearly dependent. Then, there existsλ∈Rsuch that Av=λv, and thus
0 =A2v+aAv+bv=λ2v+aλv+bv= (λ2+aλ+b)v,
which implies λ2 +aλ+b = 0. This is impossible, since the irreducible second-degree polynomial phas no real root.
Although Proposition B.1.2 proves the existence of a 1- or 2-dimensional invariant sub- space for an operator A, it is not clear whetherA has an eigenvector basis. This is where self-adjointness comes into play. We now prove a simple fact about self-adjoint operators.
Lemma B.1.2. Let E be a vector space with inner product, and let A ∈ L(E,R) be self- adjoint. Ifλandλ0 are two distinct eigenvalues of A, their respective eigenvectors v andv0 are orthogonal.
Proof. This follows by self-adjointness and the fact thatλ−λ0 6= 0, since
(λ−λ0)hv, v0i=hλv, v0i − hv, λ0v0i=hAv, v0i − hAv, v0i= 0.
The next proposition shows the existence of an orthonormal basis of eigenvectors for a self-adjointAin the case whereE is 2-dimensional. Note that the existence of an invariant subspace as stated in Proposition B.1.2 is, in fact, necessary for the argument to work.
Proposition B.1.3. Let E be a 2-dimensional vector space with inner product, and let
A∈ L(E,R) be a self-adjoint operator. There exists an orthonormal basis {u1, u2} ⊆E of eigenvectors of A.
Proof. Let {v, w} be an arbitrary orthonormal basis of E. Due to the symmetry of the matrix representation of A, we have
Av =av+bw and Aw=bv+cw.
Thus, the eigenvalues ofAare the roots of the polynomialp(λ) =λ2−(a+c)λ+(ac−b2). If
the discriminant is zero, thenb= 0, a=c and thus A=aI, which implies that every non- null vector inE is an eigenvector ofA. If the discriminant is greater than zero, thenλ1 and
λ2are real and distinct roots. Thus,A−λ1IandA−λ2Iare both non-invertible. Therefore,
there exist eigenvectors u1, u2 of A, i.e., Au1 = λ1u1 and Au2 = λ2u2 (without loss of
generality, we can assumeu1 andu2 have norm 1). Since the eigenvectors corresponding to distinct eigenvalues of a self-adjoint operator are orthogonal (Lemma B.1.2), {u , u } ⊆E
is an orthonormal basis of eigenvectors of A.
Proposition B.1.4. LetE be a vector space with inner product. Every self-adjoint operator
A∈ L(E,R) has an eigenvector.
Proof. By Proposition B.1.2, there exists a 1- or 2-dimensional subspace V ⊆ E which is invariant by A. If dim(V) = 1, then every non-null vectorv∈V is an eigenvector of A. If dim(V) = 2, then by applying Proposition B.1.3 to the restriction A:V → V of A to the invariant subspaceV, we obtain an eigenvector of A.
Remark B.1.1. What Proposition B.1.4 ensures is the existence of an eigenvector when we are restricted to the field R(for instance, in this context a rotation inSO(2)\{I,−I} does not have an eigenvector, although Proposition B.1.2 still holds). Over the fieldC, the exis- tence of an eigenvector is an immediate consequence of applying the Fundamental Theorem of Algebra to the polynomial det(A−λI), and does not depend on specific assumptions on
A such as self-adjointness.
Proposition B.1.5. Let E be a vector space with inner product. If the subspaceV ⊆E is invariant by the linear operator A∈ L(E,F), then V⊥ is invariant by the adjoint A∗. Proof. Let u ∈ V, v ∈ V⊥. Note that hA∗v, ui = hv, Aui = 0, since V is invariant by A.
Thus,A∗v∈V⊥.
Proposition B.1.5 yields the following result.
Proposition B.1.6. Let E be a vector space with inner product, and letA∈ L(E,R) be a self-adjoint operator. If the subspace V is invariant by A, then so is V⊥.
We can now prove the main result of this subsection.
Theorem B.1.1. Let E be a vector space with inner product, and let A, X ∈ L(E,R)
be self-adjoint, linear operators. A and X commute if and only if there exists a basis of common eigenvectors.
Proof. Letu1, ..., un be a basis of common eigenvectors ofA andX. LetλA1, ..., λAn be their
respective (possibly repeated)Aeigenvalues, and letλX1 , ..., λXn be their respective (possibly repeated)X eigenvalues. Take a vectorv=Pni=1αiui ∈E,αi ∈R, and write
XAv =XA ³Xn i=1 αiui ´ = ³Xn i=1 αiλXi λAi ui ´ = ³Xn i=1 αiλAi λXi ui ´ =AX ³Xn i=1 αiui ´ =AXv.
For the converse, as a consequence of Proposition B.1.4, there exists an eigenspace Eλ1
ofAwith associated eigenvalueλ1. Now assumeAandX commute. By Proposition B.1.1,
Eλ1 is invariant byX. By Proposition B.1.4, X has an eigenvector w ∈Eλ1, which must also be an eigenvector ofA. Thus, wis a common eigenvector ofAandX. By Proposition B.1.6, the subspace span(w)⊥ ⊆ E is invariant by both A and X, so the argument can
repeated to obtain a new common eigenvector in this subspace. So, by repeatedly applying Proposition B.1.6, we obtain a basis of common eigenvectors.
From the matrix perspective, Theorem B.1.1 states that two self-adjoint linear operators
A,Xcommute if and only if there is a basis from which we can construct a matrixO∈O(n) that simultaneously diagonalizes Aand X, i.e.,
A=ODAO∗ and X=ODXO∗.
As in the more general case of diagonalizable operators, the commutativity ofA and X is related to the fact that diagonal matrices commute.
Remark B.1.2. The Spectral Theorem for E with inner-product and field R states that
A∈ L(E,R) is self-adjoint if and only if there exists an orthonormal basis of eigenvectors of
A. So, one could have proved Theorem B.1.1 by directly employing the Spectral Theorem in place of Proposition B.1.4.
of normal operators. Of course, this involves dealing with complex vector spaces. See, for instance, Gantmacher (1959), chapter 9.