Euler's identity in matrix form 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!
 A: 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.
A: Yes, this matrix exponential analogue is exactly correct, and it verifies that $e^{i\pi}+1=0$, in this matrix setting, but it is not in itself an axiomatic proof of $e^{i\Pi}+I=0$.
I.e. it is correct, but it relies entirely on the fact that $e^{i\pi}+1=0$.
Very interesting and nice result though!
A: Perhaps more interesting, in the matrix context, is that if you look at $2 \times 2 $ matrices of the form 
$$
\begin{bmatrix}
a & -b \\
b & a
\end{bmatrix}
$$
then they are in 1-1 correspondence with complex numbers, where $a + b\mathbf i$ corresponds to the matrix I've just written. Furthermore, addition of complex numbers and addition of matrices match under this correspondence, and so do multiplication of complex numbers and matrix multiplication. In other words, you could simply declare that this set of matrices is $\mathbf C$, since it has all the properties that $\mathbf C$ is supposed to have. 
The nice thing is that the element $\mathbf i \in \mathbf C$ corresponds to the matrix 
$$
\mathbf {M_i} = \begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
$$
(whose square is, indeed, the negative identity). You can then use matrix exponential to verify that $$\exp(\pi \mathbf {M_i}) = -\mathbf I,$$ as you'd expect, and this might be regarded as a kind of proof of Euler's identity. 
