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. 
 A: 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.
A: If you want a direct solution, write $M$ as $M = I + X$ with
$$ X = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. $$
It is easy to prove by induction that $M^k = (I + X)^k = I + kX$ and so
$$ e^M = \sum_{k=0}^{\infty} \frac{M^k}{k!} = \sum_{k=0}^{\infty} \frac{I + kX}{k!} = \left( \sum_{k=0}^{\infty} \frac{1}{k!} \right) I + \left(\sum_{k=1}^{\infty} \frac{1}{(k-1)!} \right)X = eI + eX = eM.$$
A: It is easy to prove , first expand the given expression.
In 1 row and 1column you will get the sum 1+1÷1!+1÷2!+1÷3!+  so on  upto 1÷infinity! which is equal to  e.  Since we know e^x as 1+x÷1!+x÷2!+ so on upto infinite terms.substitute 1 in place of x you get answer as infinity. 
Simillarly solve all rows and columns.
