If is an eigenvalue of A, then for any x in the eigenspace corresponding to
Now apply to both sides of that equation.
(b) is trivial if you think about what "diagonaizable" means.
Let T be an invertible linear operator on a finite-dimensional vector space V.
a) Recall that for any eigenvalue λ of T, λ^-1 is an eigenvalue of T^-1. Prove that the eigenspace of T corresponding to λ is the same as the eigenspace of T^-1 corresponding to λ^-1.
b) Prove that if T is diagonalizable, then T^-1 is diagonalizable