$ 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?
 A: Note that $$A^2=\begin{pmatrix}
0 & 0 & 6\\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},$$ which is where the $3$ in the upper-right corner comes from.
Since $A^3$ is just the $3\times 3$ matrix of zeroes, $e^A=I+A+\frac12 A^2.$
A: 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.
A: The minimal polynomial of your matrix is $x^3$. To find $f(A)$ for a holomorphic function $f$ defined on some neighbourhood of $\sigma(A) = \{0\}$, you need to find a polynomial $p \in \Bbb{C}[x]$ of degree $< 3$ such that
$$p(0) = f(0), \quad p'(0) = f'(0), \quad p''(0)=f''(0)$$
and then $f(A)=p(A)$.
Trying out $p(x) = ax^2+bx+c$, we find that $p(x) = \frac12 x^2+x+1$ is the desired polynomial if $f$ is the exponential function.
Therefore
$$e^A = p(A) = \frac12A^2+A+I = \begin{pmatrix}
1 & 2 & 3\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}.$$
