Prove that $U(E_{\lambda})=E_{\lambda}$ and $U(K_{\lambda})=K_{\lambda}$.

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.

• $U$ is invertible, but it is not the inverse of $T$. May 29 '20 at 17:09

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 $$U(K_\lambda)=K_\lambda$$.