Let $V$ be a finite dimensional vector space and $T \in \mathcal L (V).$ Then for any $\lambda, z \in \mathbb C$ consider the following identity $:$ $$-(T – \lambda I) = (z I – T) – (z – \lambda) I.$$
This shows that $\lambda$ is an eigenvalue of $T$ if and only if $(z – \lambda)$ is an eigenvalue of $(z I – T).$ Raising both sides of the above equation to the $\dim V$ power and taking null spaces of both sides shows that the multiplicity of $\lambda$ as an eigenvalue of $T$ coincides with the multiplicity of $(z – \lambda)$ as an eigenvalue of $(z I – T).$ The bold faced argument is given in Sheldon Axler's book Linear Algebra Done Right which I am unable to follow. Could anyone please provide some insight on it?
Thanks for your time.
Best Answer
The desired result follows from the sentence above in boldface and the book's definition of the multiplicity of $\lambda$ as an eigenvalue of $T$ to be dim null $(T - \lambda I)^{\dim V}$.