How to prove that $\exp\left(\mathbf{A}\right)=-\mathbf{I}$ for given $\mathbf{A}$? I watched this nice video and I would like to prove that $\exp\left(\mathbf{A}\right)=-\mathbf{I}$ where
$$\mathbf{A}=\begin{bmatrix}0 & -\pi \\ \pi  &0\end{bmatrix}$$
The video actually proves this, but it does it in a very complicated way IMO. It first computes
$$\exp\left(\begin{bmatrix}0 & -1 \\ 1  &0\end{bmatrix}t\right)\ \forall t \in \mathbb{R}$$
and then plugs $t = \pi$ in the resulting expression). Are there simpler ways? What is the simplest way to compute the eigenvectors of $\mathbf{A}$?
 A: "Simple" will depend on your background and what you are allowed to assume a priori.
I immediately see that the answer is $-I$ from the facts that

*

*your matrix is skew-symmetric, and thus an infinitesimal rotation (element of the Lie algebra $\mathfrak{so}(2)$);

*this infinitesimal rotation has magnitude $\pi$;

*the matrix exponential computes the Lie group exponential map;

and thus $\exp(A)$ is the element of $SO(2)$ that rotates by $\pi$ radians, i.e., the 180-degree rotation $-I$.
Of course, this argument is not elementary.
Any elementary proof is going to need to work directly with the definition of $\exp$ as a power series, as in time ~18:00 in your video. I don't think there's a substantial shortcut.
A: Hint:
$A$ is conjugate to the diagonal matrix
$$
D=\begin{pmatrix} i\pi & 0\\
0 & -i\pi
\end{pmatrix}
$$
i.e., there is an invertible matrix $P$ such that $A=P^{-1}DP$.
Thus
$$e^A=\sum_n\frac{1}{n!}(P^{-1}DP)^n=P^{-1}\Big(\sum_n\frac{1}{n!}D^n\Big)P$$
can you finish from this?
A: Matrix $\rm A$ rotates by $\frac{\pi}{2}$ and scales by $\pi$. Hence, ${\rm A}^2 = - \pi^2 {\rm I}_2$ and, thus, matrix ${\rm M} := \frac{{\rm A}}{i \pi}$ is involutory, i.e., ${\rm M}^2 = {\rm I}_2$. Using Euler's formula,
$$\begin{aligned} \exp({\rm A}) = \exp \left( i \pi {\rm M} \right) &= \cos \left( \pi {\rm M} \right) + i \sin \left( \pi {\rm M} \right)\\ &= \underbrace{\cos \left( \pi \right)}_{=-1} {\rm I}_2 + i \underbrace{\sin \left( \pi \right)}_{=0} {\rm M} = \color{blue}{{-\rm I}_2 }\end{aligned}$$
where $\cos \left( \pi {\rm M} \right) = \cos \left( \pi \right) \, {\rm I}_2$ and $\sin \left( \pi {\rm M} \right)=  \sin \left( \pi \right)\, {\rm M}$ because ${\rm M}$ is involutory.

Related: Matrix exponential via Euler's formula and For which classes of matrix can the matrix exponential be easily computed?
