Let $A\in\mathcal {M}_n (\mathbb {R})$ diagonalizable matrix. We suppose that $e^A $ is diagonal matrix. Show that $A $ is diagonal matrix.
Hint : if $A$ and $B $ are diagonalizables matrix such that $e^A =e^B$ then $A=B $.
My effort Let $P\in GL_n {(\mathbb {R})}$ and $\lambda_i\in \mathbb {R} $ such that $D=diag (\lambda_1,...\lambda_2)$. Then $$D =P^{-1}AP$$ then $e^D=P^{-1}e^AP$.
Can we say that if $e^A $ is diagonal matrix then $e^A =e^D$?
Not quite, only up to a permutation of the eigenvalues.
Perhaps a better way to formulate this is as follows: A diagonalizable matrix is diagonal in some basis if and only if it is an eigenbasis. So if $e^A$ is diagonal, then the standard basis is an eigenbasis of $e^A$. But eigenspaces of $A$ and $e^A$ are the same when $A$ is diagonalizable. So the standard basis is an eigenbasis of $A$, i.e. $A$ is diagonal.