Find a matrix $X$ such that $\exp(X) = M$

Question

Let $M=\begin{bmatrix} 4 & -2 \\ -1 & 5 \end{bmatrix}$. Find a matrix $X$ such that $e^X = M$, or show this is impossible.

Attempt. I think that this is not possible because of the convergence problem of the logarithm function. But I couldn't prove it. I've computed that $M=PDP^{-1}$ where $P=\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$ and $D=\begin{bmatrix} 3 & 0 \\ 0 & 6 \end{bmatrix}$. Thanks!

Answer 1

Hint: Consider $e^{PXP^{-1}}$.

Answer 2

Notice that $P^{-1} M P= D(3,6)$ and $\exp X=M \implies X=\ln M,$ We have $$M=P D P^{-1} \implies X=\ln M= P~ \ln D ~P^{-1},$$ $\ln D = \begin{bmatrix} \ln 3 & 0 \\ 0 & \ln 6 \end{bmatrix}.$ Here we have used the result that$$f(M)=P ~f(D) P^{-1}$$ and $$f(D)=D(\ln \lambda_1, \ln \lambda_2).$$