The matrix $e^A$ is defined by $e^A=\Sigma_{k=0}^{\infty}\frac {A^k}{k!}$ Suppose M=$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$. Calculate $e^M$ The matrix $e^A$ is defined by $e^A=\Sigma_{k=0}^{\infty}\frac {A^k}{k!}$ Suppose M=$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$. Calculate $e^M$.
I did some calculating with real values and I got the iteration of values: $\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$, $\begin{bmatrix}2 & 1\\0 & 2\end{bmatrix}$, $\begin{bmatrix}\frac 5 2 & 2\\0 & \frac 5 2 \end{bmatrix}$, etc.
I believe that the solution is $\begin{bmatrix}e & e\\0 & e\end{bmatrix}$, but can't prove it. 

Observe
\begin{align}
M=
\begin{pmatrix}
1& 0\\
0 & 1
\end{pmatrix}
+
\begin{pmatrix}
0& 1\\
0 & 0
\end{pmatrix}=: I+ X
\end{align}
where $[X, I] = 0$ (of course since identity commutes with anything). Hence it follows
\begin{align}
\exp[M] = \exp[I]\exp[X].
\end{align}
Next, observe
\begin{align}
X^2 = 0 \ \ \Rightarrow \ \ \exp[X] = I + X+ \frac{1}{2!}X^2+\ldots = I+X
\end{align}
which means
\begin{align}
\exp[M] = eI \ast(I+X).
\end{align}
Thus, we have the desired solution.
