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Question:Let $V$ be a finite-dimensional vector space over $\mathbb C$ with $\dim(V) = n$ and $T,S$ are linear operator(i.e. linear transformation from $V$ to itself) such that$$TS-ST=T$$prove that $T,S$ have an eigenvector in common.

My attempt to the question:I want to solve it by induction,so i try to find a suitable invariant space by considering the space below:$$U:=span\{v,Sv,..,S^{m-1}v\}$$Where $m$ is the biggest positive integer such that $v,Sv,..,S^{m-1}v$ are independent and $v$ is an eigenvector of $T$

Noting that S is invariant in $U$,and since $\forall 0\le k\le m-1$ $$TS^kv=(E+S)TS^{k-1}v=..=(E+S)^kTv=\lambda (E+S)^kv$$Therefore $T$ is also invariant in $U$.($E$ is the identity matrix)

By induction to dimension of $V$,and if $\dim U< \dim V$,we can conclude that $T_{|U},S_{|U}$ has an eigenvector in common.

But how can i deduce this when $\dim U=\dim V$?I am stuck here.
Any help would be appreciated.

Qo Ao
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    The title needs fixed -- they have an eigenvector in common not an eigenvalue. Also you can directly check $S\cdot \ker T\subseteq \ker T$ and $S_{\vert \ker T}$ must have an eigenvector (fundamental theorem of algebra). – user8675309 Nov 01 '24 at 16:26
  • Oops!Thank you for pointing out! – Qo Ao Nov 02 '24 at 11:12
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    to close the loop, we know $\dim \ker T \geq 1$ since $T$ is nilpotent, per https://math.stackexchange.com/questions/724294/if-a-ab-ba-is-a-nilpotent – user8675309 Nov 02 '24 at 15:29

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Let $E_{\lambda}$ the eigenspace of $S$ with eigenvalue $\lambda$. As $TS-ST=T$ if $x\in E_{\lambda}$ then $S(Tx)=(\lambda -1) Tx$. Therefore $T$ maps $E_{\lambda}$ in $E_{\lambda -1}$. And for every integer $T$ maps $E_{\lambda -k}$ in $E_{\lambda -k-1}$. AS $E$ is finite dimensional some $E_{\lambda -k}$ is reduce to $0$. This means that the restriction of $T$ to $E_{\lambda -k+1}$ is the zero map, and any vector in this space is in the kernel of $T$ and an eigenvector of $S$.

Thomas
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