I need to prove: matrix A is diagonalizable iff $\exp(A)$ is diagonalizble. exp means exponent function.
I know to prove that if $A$ is diagonalizable so $\exp(A)$ is diagonalizable, but have a problem with the other side.
can I write $P^{-1}.\exp(A).P=D$ (since $\exp(A)$ is diagonalizable ) and operate log on both sides of the equation? if yes I'm done, or any other hints?
Since you already know that "$A$ is diagonalisable" implies that "$\exp(A)$ is diagonalisable", we show that if $A$ is not diagonalisable, then neither is $\exp(A)$.
Let $A$ be non-diagonalisable and let $A=XJX^{-1}$ be its Jordan form, where $J$ is a direct sum of elementary Jordan blocks: $J=\oplus_{i=1}^l J_i$. The function $\exp(A)$ can then be written in the form $\exp(A)=X\exp(J)X^{-1}$, where $\exp(J)$ is again a direct sum of exponentials of elementary Jordan blocks: $\exp(J)=\oplus_{i=1}^l \exp(J_i)$. Since $A$ is non-diagonalisable and so is $J$, $J$ contains at least one non-trivial Jordan block $$ \tilde{J}=\lambda I + N\in\mathbb{C}^{k\times k}, \quad k\geq 2, $$ where $\lambda\in\mathbb{C}$ is an eigenvalue of $A$ and $N$ is the nilpotent matrix $$ N=\begin{bmatrix} 0 & 1 & & & \\ & 0 & 1 & & \\ & & \ddots & \ddots & \\ & & & 0 & 1 \\ & & & & 0 \end{bmatrix}. $$ If we show that $\exp(\tilde{J})$ of such a non-trivial block is non-diagonalisable, we are done. Looking again here how $\exp(J)$ looks like, we see that $$ \exp(\tilde{J})=\exp(\lambda)\exp(N) =\exp(\lambda)\sum_{i=0}^{k-1}\frac{1}{i!}N^i. $$ Note that $1$ is the only eigenvalue of the upper triangular matrix $\exp(N)$. So $\exp(N)$ is diagonalisable iff the kernel of the $k\times k$ matrix $\exp(N)-I$ is $\mathbb{C}^{k}$. This is, however, not the case as $\exp(N)-I$ is not a zero matrix. Therefore, $\exp(N)$ and consequently $\exp(\tilde{J})$ and consequently $\exp(A)$ is not diagonalisable.