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In the following text https://www.math.stonybrook.edu/~mde/534F_00/PS4.pdf, I am reading the proof of -

8. Let A, B be commuting linear operators: AB = BA. Prove that
(a) they have a common eigenvector.
(b) they have a common invariant flag, i.e., there exists a basis in
which both A and B are upper-triangular.
(c) the eigenvalues of AB are products of eigenvalues of A and B.
(d) Which of these statements still hold if AB 6= BA?
Idea of proof: (a) Let λ be an eigenvalue of A, and V_λ = Ker(A−λ) the
space of eigenvectors. We claim that V_λ is invariant under B. Indeed:
if v ∈ V_λ, then A(Bv) = BAv = Bλv = λBv and thus, Bv is an
eigenvector for A with eigenvalue λ.
Consider the restriction of B to V_λ. This restricted operator has at
least one eigenvector (say, w) in V_λ. On the other hand, every vector
in V_λ is an eigenvector for A, so w is an eigenvector for both A and B.

I don't see two points in the proof of (a). First, I see how B(V_λ) ⊂ V_λ, but I don't see how B(V_λ) ⊃ V_λ, meaning I don't see B(V_λ) = V_λ, and hence I don't see how V_λ is invariant under B.

Second, I don't see why the comment ...This restricted operator has at least one eigenvector (say, w) in V_λ is true. Why, indeed, must the restriction of B onto V_λ have at least one eigenvector in V_λ?

Thanks in adavance.

Joe Shmo
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1 Answers1

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  • The definition of $V_\lambda$ being invariant under $B$ is $B(V_\lambda)\subseteq V_\lambda$, not $B(V_\lambda)=V_\lambda$,.
  • The restriction of $B$ to $V_\lambda$ is just another linear operator on $V_\lambda$, so it has an eigenvector (presumably you are working over $\mathbb{C}$). Another way to see this is that if you expand a basis for $V_\lambda$ to a basis for the entire vector space, then with respect to this larger basis, $B$ is block diagonal, and the restriction of $B$ to $V_\lambda$ is the block matrix corresponding to the basis elements of $V_\lambda$, i.e. it is another complex matrix so it has an eigenvector.
angryavian
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