Exponentiating Matrices - Am I doing it correctly? I recently learned how to exponentiate matrices by extending the power series of $e^x$ to matrices:
$$e^A=A^0+A^1+\frac{A^2}{2!}+\frac{A^3}{3!}+...$$
I thought that this was pretty cool, so I decided to try it with the matrix
$$A=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
I then realized that I would have to find a general formula for $A^n$. I ended up getting
$$A^n=\begin{pmatrix}2^{n-1}&2^{n-1}\\2^{n-1}&2^{n-1}\end{pmatrix}$$
for $n\ge 1$, and for $n=0$, I have that $A^0=I$. So my final answer would be in the form
$$e^A=\begin{pmatrix}\alpha&\alpha-1\\\alpha-1&\alpha\end{pmatrix}$$
where
$$\alpha=1+2^0+\frac{2^1}{2!}+\frac{2^2}{3!}+...$$
$$\alpha=\frac{1}{2}\bigg(1+2^0+2^1+\frac{2^2}{2!}+\frac{2^3}{3!}+...\bigg)$$
$$\alpha=\frac{1}{2}(e^2+1)$$
and so my answer would be
$$e^A=\begin{pmatrix}\frac{e^2+1}{2}&\frac{e^2-1}{2}\\\frac{e^2-1}{2}&\frac{e^2+1}{2}\end{pmatrix}$$
Is this correct?
 A: Your method is correct, as you've been told. But here's another way of doing this that may interest you. Let$$M=\frac1{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}.$$Then$$M^{-1}=\frac1{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}\text{ and }M^{-1}.A.M=\begin{pmatrix}2&0\\0&0\end{pmatrix}.$$Therefore$$M^{-1}.e^A.M=e^{\begin{pmatrix}2&0\\0&0\end{pmatrix}}=\begin{pmatrix}e^2&0\\0&1\end{pmatrix}.$$So,$$e^A=M.\begin{pmatrix}e^2&0\\0&1\end{pmatrix}.M^{-1}=\begin{pmatrix}\frac{e^2+1}{2}&\frac{e^2-1}{2}\\\frac{e^2-1}{2}&\frac{e^2+1}{2}\end{pmatrix}.$$
A: It's not clear what you mean when you ask whether you are "doing it correctly."
If you're asking whether you computed the exponential for this matrix correctly, the answer is yes.
If you're asking whether this is a general method to use for an arbitrary matrix, the answer is a resounding no: it will be impossible to determine a formula for $A^n$ just by inspection or multiplying out a few terms (which I assume you did here). 
One general method for diagonalizable matrices is to first diagonalize $A$, because the exponential of a diagonal matrix is easy. (This is what José did.)
If the matrix isn't diagonalizable, you can do the next best thing and put it in Jordan normal form; it is also easy to work out the exponential of Jordan blocks.
Computationally, however, the problem is very delicate. See this paper, "Nineteen Dubious Ways to Calculate the Exponential of a Matrix," for example.
