Diagonalizability of a linear transformation

Question

I'm trying to prove the following:

Let $ T: V -> V $ be a linear transformation and let $ \lambda_1 , ... , \lambda_k $ be distinct eigenvalues of T. Suppose the characteristic polynomial of T is $ (x-\lambda_1)^{n_1} ...(x-\lambda_k)^{n_k}$. Then $T$ is diagonalizable iff $rank(T-\lambda_iId) = rank(T-\lambda_iId)^2 $ $\forall $ $i$.

Thanks in advance.

Answer 1

Since you indicated that you're familiar with the Jordan normal form of matrices, you know that $(T-\lambda_iI)^{n_i}$ has nullity $n_i$. Therefore, to say $\operatorname{rank}(T-\lambda_iI)=\operatorname{rank}(T-\lambda_iI)^2$ is to say $$\operatorname{null}(T-\lambda_iI)=\operatorname{null}(T-\lambda_iI)^2 = \ldots = \operatorname{null}(T-\lambda_iI)^{n_i},$$ of dimension $n_i$, which is to say that $T$ has $n_i$ linearlly independent eigenvectors belonging to $\lambda_i$, which in turns is equivalent to diagonlizability.