NoteNextra-origin/content/Math429/Math429_L24.md

Lecture 24

Chapter V Eigenvalue and Eigenvectors

5E Commuting Operators

Definition 5.71

• For T,S\in\mathscr{L}(V), T and S commute if ST=TS.
• For A,B\in \mathbb{F}^{n,n}, A and B commute if AB=BA.

Example: For p,q\in \mathscr{P}(\mathbb{F}), p(T) and q(T) commute

• Partial Derivatives \frac{d}{dx},\frac{d}{dy}:\mathscr{P}_m(\mathbb{R}^2)\to \mathscr{P}_m(\mathbb{R}^2), \frac{d}{dy}\frac{d}{dd}=\frac{d}{dx}\frac{d}{dy}
• Diagonal matrices commute with each other

Proposition 5.74

Given S,T\in \mathscr{L}(V), S,T commute if and only if M(S), M(T) commute.

Proof: ST=TS\iff M(ST)=M(TS)\iff M(S)M(T)=M(T)M(S)

Proposition 5.75

Suppose S,T\in \mathscr{L}(V) commute and \lambda\in \mathbb{F}, then the eigenspace E(\lambda, S) is invariant under T.

Proof: Suppose V\in E(\lambda, S), S(Tv)=(ST)v=(TS)v=T(Sv)=T\lambda v=\lambda Tv

Tv is an eigenvector with eigenvalue \lambda (or Tv=0) Tv\in E(\lambda, S)

Theorem 5.76

Let S,T\in \mathscr{L}(V) be diagonalizable operators. Then S,T are diagonalizable with respect to the same basis (simultaneously diagonalizable) if and only if S,T commute.

Proof:

\Rightarrow

diagonal matrices commute

\Leftarrow

Since S is diagonalizable, V=E(\lambda, S)\oplus...\oplus E(\lambda_m,S), where \lambda_1,...,\lambda_m are the (distinct) eigenvalues of S. consider T\vert_{E(\lambda_k,S)} (Theorem 5.65) this operator is diagonalizable because T is diagonalizable. We can chose a basis of E(\lambda_k,S) such that T\vert_{E(\lambda_k,S)} gives a diagonal matrix. Take the basis of V given by concatenating the bases of E(\lambda_k,S) the elements of this basis eigenvectors of S and T so S and T are diagonalizable with respect to this basis.

Proposition 5.78

Every pair of commuting operators on a finite dimensional complex nonzero vector spaces has at least one common eigenvectors.

Proof: apply (5.75) and the fact that operator on complex vector spaces has at least one eigenvector.

Theorem 5.80

If S,T are commuting operators, then there is a basis where both have upper triangular matrices.

Proof:

Induction on n=dim\ V.

n=1, clear

n>1, use (5.78) to find v_1 and eigenvector for S and T. Decompose V as V=Span(v_1)\oplus W then defined a map P:V\to W,P(av_1+w)=w define \hat{S}:W\to W as \hat{S}(w)=P(S(w)) similarly \hat{T}(w)=P(T(w)) now apply the inductive hypothesis to \hat{S} and \hat{T}. get a basis v_2,...,v_n where they are both upper triangular and then exercise: S,T are upper triangular with respect to the basis v_1,...,v_n.

Theorem 5.81

For V finite dimensional S,T\in \mathscr{L}(V) commuting operators then every eigenvalue of S+T is a sum of an eigenvector of S and an eigenvalue of T; every eigenvalue of S\cdot T is a product of an eigenvector of S and an eigenvalue of T.

Proof:

For upper triangular matrices

\begin{pmatrix}
    \lambda_1 & & *\\
    & \ddots & \\
    0 & & \lambda_m
\end{pmatrix}+
\begin{pmatrix}
    \mu_1 & & *\\
    & \ddots & \\
    0 & & \mu_m
\end{pmatrix}=
\begin{pmatrix}
    \lambda_1+\mu_1 & & *\\
    & \ddots & \\
    0 & & \lambda_m+\mu_m
\end{pmatrix}