$ A=\begin{pmatrix} 0 & 2 & 0\\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix}$, calculate $e^A$ I have encounter a question in my book , it was For $ A=\begin{pmatrix}
0 & 2 & 0\\
0 & 0 & 3 \\
0 & 0 & 0
\end{pmatrix}$, calculate $e^A$
My solution way : I tried to find its eigenvalues , so i found that the only eigenvalue is $0$ and the eigenspace is $(1,0,0)$ .Hence , it cannot be diagonalized.
Then , i tried to use taylor exponential and it gives me $ A=\begin{pmatrix}
1 & 2 & 0\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .However the answer is $ A=\begin{pmatrix}
1 & 2 & 3\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .
What am i missing ,can you help me?

The structure of $A$ tells you immediately that $A$ is not diagonalizable without any calculation.
The matrix $A$ is a Jordan block that is nilpotent; there is an integer $N$ such that $A^N=0$. ($N$ is small in your particular case.) So you have $A^{m}=0$ for all $m>N$.
By definition, to calculate $e^A$, you need to know all the powers of $A$: $e^A:=\sum_{k=0}^\infty\frac{1}{k!}A^k$. But now you only need to calculate finitely many.
