Compute $e^{Ax}$ for the $2 \times 2$ matrix $A$ Consider the real $2 \times 2$ matrix $A$  $$A=\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$$
I want to compute $e^{Ax}$. 
I can see that $A^2=I$ which looks like it should be useful.
Using the definition of the matrix exponential $$e^{Ax}=I+Ax+\frac{(Ax)^2}{2!}+ \cdots$$ I find that $$e^{Ax}=I\left(1+\frac{x^2}{2!}+ \cdots\right)+A\left(x+\frac{x^3}{3!}+ \cdots\right)$$
Is there anything I can do from here?
 A: We have $$e^{Ax}=\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}\left(1+\frac{x^2}{2!}+ \cdots\right)+\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}\left(x+\frac{x^3}{3!}+ \cdots\right)$$ This can be written as $$\begin{pmatrix} 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+... & 0 \\ 0 & 1+(-x)+\frac{x^2}{2!}+\frac{(-x)^3}{3!}+...\end{pmatrix}$$ so $$\boxed{e^{Ax}=\begin{pmatrix} e^x & 0 \\ 0 & e^{-x}\end{pmatrix}}$$
A: Note that$$(Ax)^n=\begin{pmatrix}x^n&0\\0&(-x)^n\end{pmatrix}$$and therefore\begin{align}e^{Ax}&=\operatorname{Id}+A+\frac1{2!}(Ax)^2+\frac1{3!}(Ax)^3+\cdots\\&=\begin{pmatrix}e^x&0\\0&e^{-x}\end{pmatrix}.\end{align}
A: Your matrix  is diagonalizable you can use that to compute your matrix with raised to the power eigenvalues  at the diagonal 
A: You could note that $$1+\frac{x^2}{2}+\dots=\sum_{k=0}^\infty\frac{1+(-1)^k}{2}\frac{x^k}{k!}=\frac{e^x+e^{-x}}2,$$and similarly for the other sum.
Realizing that $A^2=I$ allows you to compute the exponential is admirable. There's another approach, that makes the exponential of any diagonal matrix trivial. Write down a general diagonal matrix $D$. Calculate $D^2$ and $D^3$. Clearly $D^k$ is just ..., which shows that $e^D$ is just ...
A: When you have a diagonal matrix
$$\mathsf{D} = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$
Then the matrix exponential is simply
$$e^{\mathsf{A}x} = \begin{pmatrix} e^{\lambda_1 x} & 0 \\ 0 & e^{ \lambda_2 x} \end{pmatrix}$$
This also holds for $n\times n$ diagonal matrices.
