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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!

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I think the more interesting identity is $$ e^{\pi \pmatrix{0&-1\\1&0}} + I = 0 $$ –  Omnomnomnom Apr 28 at 21:31
    
Of course. This is also explained by John in its answer. However this identity would be valid solely for the particular case of $2\times 2$ matrices, wouldn't it? –  Dmoreno Apr 28 at 21:37
    
Ah, I missed his answer. You could extend it if you want to; but what really makes this interesting is that you can get away with saying that $i$ means the matrix $\pmatrix{0&-1\\1&0}$. –  Omnomnomnom Apr 28 at 21:43
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If you wanted to extend this identity to larger matrices, you could always take $$ M_i = \pmatrix{ 0&-1&0&\cdots&0\\ 1&0&0&\cdots&0\\ 0&0&1&&\vdots\\ \vdots&&&\ddots&\\ 0&0&\cdots&0&1} $$ –  Omnomnomnom Apr 28 at 22:19
    
Thanks for your answers Omnomnomnom! I will take a closed look to the last one tomorrow. –  Dmoreno Apr 28 at 22:24

2 Answers 2

up vote 10 down vote accepted

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.

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Thanks John! Nice answer. I was familiar with this correspondence between such matrices and $\mathbb{C}$ but I couldn't have ever related this to my demonstration. Thanks again! –  Dmoreno Apr 28 at 21:33

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!

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I see the point. Thank you for your answer, ellya! This was in my mind since a long time ago. –  Dmoreno Apr 28 at 21:27

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