Finding eigen value of $(T- \lambda I)^{-1}$

Question

Let $V$ be a finite dimensional complex vector space and $T \in \mathfrak L(V),$ the set of all continuous linear operators from $V$ to $V.$ Fix $\mu \notin \sigma(T).$

I have to show

1. $1/(\lambda- \mu)$ is an eigen value of $(T- \mu I)^{-1}$ for each $\lambda \in \sigma(T).$

2. $E(1/(\lambda- \mu),( T- \mu I)^{-1}) = E(\lambda, T).$

I basically know the definition/properties of the terms used here but was unable to solve this one. Any thoughts. Thank you.

Answer 1

Because $V$ is finite-dimensional, all elements of the spectrum are eigenvalues.

So assume $Tv=\lambda v$. We have $$\tag1 (T-\mu I)v=Tv-\mu v=\lambda v-\mu v=(\lambda -\mu)v. $$ Because $T-\mu I$ is invertible, applying $(T-\mu I)^{-1}$ to both sides and dividing by $\lambda-\mu$ (which is nonzero since $\lambda\in\sigma(T)$ and $\mu\not\in\sigma(T)$), $$\tag2 \tfrac1{\lambda-\mu}\,v=(T-\mu I)^{-1}v. $$ Thus we have shown that $\tfrac1{\lambda-\mu}\in\sigma((T-\mu I)^{-1})$, and that $E(\lambda, T)\subset E(\tfrac1{\lambda-\mu},(T-\mu I)^{-1})$.

Now, if $w\in E(\tfrac1{\lambda-\mu},(T-\mu I)^{-1})$, we may undo the above: start from $$ \tfrac1{\lambda-\mu}\,w=(T-\mu I)^{-1}w, $$ rewrite it as $$ (T-\mu I) w=(\lambda -\mu) w, $$ and simplify to $Tw=\lambda w$. So $E(\lambda, T)\supset E(\tfrac1{\lambda-\mu},(T-\mu I)^{-1})$.