Suppose $\mathscr{A}$ and $\mathscr{B}$ are linear transformations over the vector space $V$, which also satisfy $\mathscr{A}\mathscr{B}$=$\mathscr{B}\mathscr{A}$, can we get: if $W$ is a subspace of $V$ and $\mathscr{A}(W)\subseteq W$, then $\mathscr{B}(W)\subseteq W$. I think of this question because when $W=\mathrm{ker(\mathscr{A})}$ or $W=\mathrm{Im(\mathscr{A})}$ then it is right. I cannot prove or give a counterexample, but I myself think it is not true for general case. Is there any counterexample, or is it true? Thank you!
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You can see the answer here: https://math.stackexchange.com/questions/236212/simultaneously-diagonalizable-proof – Emilio Novati Sep 11 '17 at 19:14
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Consider the case $\mathscr A$ the identity, $\mathscr B$ anything else, $W$ any subspace with $\mathscr B(W)\not\subseteq W$
Hagen von Eitzen
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