diagonalizability and the Null space of eigenvector

Question

If $A\in{\mathcal{C}^{n\times n}}$, and for each eigenvalue $\lambda$ of $A$, $N((A-\lambda I)^2)=N(A-\lambda I)$, prove that $A$ is diagonalizable.

Answer 1

Since it is over $\mathbb{C}$, we can write the characteristic polynomial of $A$ $$ \chi_A(X)=\det(XI-A)=\prod_{\lambda\in \sigma(A)}(X-\lambda)^{k_\lambda} $$ where $\lambda $ runs over $\sigma(A)$ the spectrum of $A$, where the eigenvalues are not repeated.

By Cayley-Hamilton $\chi_A(A)=0$, we obtain the decomposition $$ \mathbb{C}^n=\bigoplus_{\lambda \in\sigma(A)}N((A-\lambda I)^{k_\lambda}). $$ as always when the characteristic polynomial splits.

Now it follows from the assumption that $N((A-\lambda I)^k)=N(A-\lambda I)$ for every $k\geq 1$ and every eigenvalue $\lambda$. Therefore $$ \mathbb{C}^n=\bigoplus_{\lambda \in\sigma(A)}N(A-\lambda I) $$ which means precisely that $A$ is diagonalizable. The converse is clearly true.

Key point 1: for every endomorphism $T\in L(V)$ on a finite-dimensional $K$ vector space $V$, the chain $\{0\}=\ker T^0\subseteq \ker T\subseteq \ker T^2\subseteq\ldots $ is strictly increasing until it becomes stationary. Your assumption says it happens at most at the second step for every $T=A-\lambda I$ in your case.

Key point 2: if $T$ is annihilated by $p\in K[X]$ and if $p=\prod_{j=1}^np_j$ where the $p_j$'s are relatively prime, then $V=\ker p(T)=\bigoplus_{j=1}^n\ker p_j(T)$. In your case, the relatively prime factors are the factors $(X-\lambda)^{k_\lambda}$.