4

I am going through this interesting paper A simultaneous triangularization result by Dan Shemesh. The Theorem 2 read

If $\textbf{A}[\textbf{A},\textbf{B}]=\textbf{0}$ and $\textbf{B}[\textbf{A},\textbf{B}]=\textbf{0}$, then matrices $\textbf{A}$ and $\textbf{B}$ are simultaneously triangularizable.

However, the compactness of the proof (attached below as pic) makes it hard to follow for someone like me who is not an expert in the field.

In particular:

  1. What is meant by $\textbf{C}$-invariant?
  2. Why is $[\textbf{C},\textbf{D}]$ is not ivertible?
  3. How is it that $\textbf{C}\textbf{D}x_0 =\textbf{D}\textbf{C}x_0=0=\lambda \textbf{D} x_0$ implies $\textbf{D}x \in ker(\textbf{C}-\lambda \textbf{I})$?
  4. Similarly the last line "And by $[\textbf{C},\textbf{D}]\textbf{D}=\textbf{0} \dots$" is not clear to me.

enter image description here

Mike
  • 288
  • 1
    (1.) a vector space $V$ has a T-invariant subspace subspace $W\subseteq V$ when $TW\subseteq W$. When $\dim W=1$ we call $W$'s generator an eigenvector; most proof based introductions to Linear Algebra will walk you through this. (2.) $\big[C,D\big]\neq \mathbf 0$... I don't know why you wrote this but I suggest you re-read the proof. Now the simpler, standard simultaneous triangularization argument is for the case $\big[C,D\big]=\mathbf 0$ which I suggest you learn first. The above is a generalization of the simpler case but I suspect the simpler case will be a challenge. – user8675309 Apr 12 '23 at 16:10
  • Thanks @user8675309, in (2.) I meant "why $[C, D]$ is not invertible". – Mike Apr 13 '23 at 08:39
  • $\det\Big(\big[C,D\big]\Big)\neq 0\implies C= \big[C,D\big]^{-1} \mathbf 0=\mathbf 0\implies \big[C,D\big]=\big[\mathbf 0,D\big]=\mathbf 0\implies \det\Big(\big[C,D\big]\Big)= 0$ which is a contradiction. Where did you get stuck when you considered the implications of $\big[C,D\big]$ being invertible? – user8675309 Apr 13 '23 at 15:28

0 Answers0