TITLE: Euler's identity in matrix form
QUESTION [10 upvotes]: I assume everyone is familiar with the famous mathematical identity due to L. Euler:
$$ e^{i \, \pi} + 1 = 0,$$ where $i^2 = -1$ and $e$ is the base of natural logarithms. I was wondering if this identity can be extended in a natural way to squared matrices as follows:
$$ e^{i \, \mathbf{\Pi}} + \mathbf{I} = \mathbf{0}, $$ where $\mathbf{\Pi} = \pi \, \mathbf{I}$, $\mathbf{I}$ is the $n\times n$ identity matrix, $\mathbf{0}$ is the $n\times n$ null matrix and now $e^\square$ stands for matrix exponentiation. I was able to check this identity for some small values of $n$, i.e., 2, 4, 10, etc. with the help of Mathematica. Is applying the definition of matrix exponential the way to prove this identity? Indeed, the identity tells us:
$$ \sum_{t=0}^\infty \frac{(i \pi \, \mathbf{I})^t}{t!} + \mathbf{I}  = \mathbf{I} \sum_{t=0}^\infty \frac{(i \pi )^t}{t!} + \mathbf{I} = e^{i \pi} \mathbf{I} + \mathbf{I} = (e^{i \pi} + 1) \, \mathbf{I} = 0 \, \mathbf{I} = \mathbf{0},$$ where I have made use of: $\mathbf{I}^k = \mathbf{I}$, $k \in \mathbb{N} \cup \{0\}$, the definition of $e^x$ and the distributive property of matrix multiplication. Is my approach correct? Furthermore, has it any application?
Cheers!

REPLY [5 votes]: The rotation matrix in $\mathbb{R}^2$ can be written as
\begin{equation}
R(\theta)=\left(\begin{matrix}
    \cos\theta & -\sin\theta  \\
    \sin\theta & \cos\theta\\
    \end{matrix}\right)=\cos\theta \left(\begin{matrix}
    1 & 0  \\
    0 & 1\\
    \end{matrix}\right)+\sin\theta\left(\begin{matrix}
    0 & -1\\
    1 & 0\\
    \end{matrix}\right).
\end{equation}
If we denote
\begin{equation}
I=\left(\begin{matrix}
    1 & 0\\
    0 & 1\\
    \end{matrix}\right),~\mathbb{I}=\left(\begin{matrix}
    0 & -1\\
    1 & 0\\
    \end{matrix}\right),
\end{equation}
it is easy to check that
$$I^2=I, I\mathbb{I}=\mathbb{I}I=\mathbb{I}, \mathbb{I}^2=-I,$$ and
\begin{eqnarray}
e^{\mathbb{I}θ}&=&I+\mathbb{I}θ+\frac{1}{2!}(\mathbb{I}θ)^2+\frac{1}{3!}(\mathbb{I}θ)^3+\frac{1}{4!}(\mathbb{I}θ)^4+\frac{1}{5!}(\mathbb{I}θ)^5+\ldots\\
&=&\left(1-\frac{1}{2!}θ^2+\frac{1}{4!}θ^4-\ldots\right)I+\left(θ-\frac{1}{3!}θ^3+\frac{1}{5!}θ^5+\ldots\right)\mathbb{I}\\
&=&\cos\theta I+\sin\theta\mathbb{I}
\end{eqnarray}
or,
$$e^{\mathbb{I}θ}=R(\theta).$$
Text Reference
Stillwell, J. Naive Lie Theory, Undergraduate Texts in Mathematics, Springer, 2008.