Euler's identity in matrix form

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.

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.

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!