Prove that $A$ is diagonalizable iff rank$(A-tI)=$rank$((A-tI)^2)$ for all $t\in\mathbb C$.
I have no idea for both directions. The only thing I can think of is
rank$(A-tI)=$rank$((A-tI)^2)$ iff nullity$(A-tI)=$nullity$((A-tI)^2)$.
Please me some idea. Thanks.
I'll give an idea, without Jordan, for the direction $A$ diagonalizable $\implies \text{rank}(A-tI)=\text{rank}( (A-tI)^2) $ for all $t \in \mathbb{C}$. Suppose $A$ is an $n$ dimensional square matrix. Since $A$ is diagonalizable we may write $A=PDP^{-1}$, where $P$ is invertible and $D=\text{diag}(\lambda_1,\lambda_2, \dots \lambda_n$) and the diagonal entries of $D$ are the eigenvalues of $A$. For every $t \in \mathbb{C}$ we can write: $\text{rank}(A-tI)=\text{rank}(PDP^{-1}-tI))=\text{rank}(P(D-tI)P^{-1})=\text{rank}(D-tI)=\text{rank}(\text{diag}(\lambda_1-t,\lambda_2-t, \dots ,\lambda_n-t))$. Now, what's the rank of a diagonal matrix? Perform a simila calculation for $\text{rank}((A-tI)^2)$ and compare.