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Let T be a linear map on a finite-dimensional vector space V , and let $\lambda$ be an eigenvalue of T with corresponding eigenspace and generalized eigenspace $E_{\lambda}$ and $K_{\lambda}$. Let U be an invertible operator on V that communutes with T(i.e. TU=UT) Prove that $U(E_{\lambda})=E_{\lambda}$ and $U(K_{\lambda})=K_{\lambda}$.

Theorem:Let T be a linear map on a finite-dimensional vector space V such that the characteristic polynomial of T splits. suppose that $\lambda$ is an eigenvalue of T with multiplicity m. Then $dim(K_{\lambda}) \leq m$ and $K_{\lambda}=N((T-\lambda I)^m)$

Theorem: Let T be a linear map on a finite-dimensional vector space V, and let $\lambda$ be an eigenvalue of T, then $K_{\lambda}$ is a T-invariant subspace of V containing $E_{\lambda}$ (the eigenspace of T corresponding to $\lambda$).

Since U is an inverse linear oeprator on V and TU=UT=I and since U is linear, then we have $U(E_{\lambda})=E_{\lambda}$ and $U(K_{\lambda})=K_{\lambda}$.

I am also thinking about the fact that if $v \in ker T$, then $U(v) \in Ker T$ and that U commutes with $(T-\lambda I)^k$ for all $k \geq 0$. I am not sure how to finish this proof.

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    $\begingroup$ $U$ is invertible, but it is not the inverse of $T$. $\endgroup$ – Exodd May 29 at 17:09
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The relation $UT=TU=I$ is false, since you don't know whether $U$ is the inverse of $T$.

If $v\in E_\lambda$, then $Tv=\lambda v$ and $$ T(U(v)) = U(T(v)) = \lambda U(v) $$ so $U(v)\in E_\lambda$. This is enough to say $U(E_\lambda)\subseteq E_\lambda$ and since $U$ is invertible, it must necessarily hold $U(E_\lambda)=E_\lambda$.

There exists an index $m$ for which $K_\lambda = N((T-\lambda I)^m)$. Notice that $$ UT^k = T^kU \quad\forall k\implies U(T-\lambda I)^m = (T-\lambda I)^mU $$ so if $v\in K_\lambda$, then $$ 0 = U(T-\lambda I)^m(v) = (T-\lambda I)^m(U(v)) $$ and $U(v)\in N((T-\lambda I)^m)=K_\lambda$, so $U(K_\lambda)\subseteq K_\lambda$ and since $U$ is invertible, it must necessarily hold $U(K_\lambda)=K_\lambda$.

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