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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 $\mathbf{A}$?

Rodrigo de Azevedo
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DeltaIV
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    You can compute the characteristic polynomial and find eigenvalues, eigenvectors of A as usual. Then diagonalize it to compute $\exp\left(\mathbf{A}\right)$. – mathmathmath Apr 04 '21 at 21:34
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    Do you mean $-\mathbf{I}$? – thorimur Apr 04 '21 at 21:36
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    Two ways: 1) to just make totally sure I wasn't wrong, I plugged it into Mathematica. :) 2) I used the fact mentioned in the video about this particular calculation being related to viewing complex numbers as matrices. This is absolutely not generalizable, since only certain matrices correspond to complex numbers, but makes the above equation equivalent to $e^{i\pi} = -1$, since $\mathbf{A}$ corresponds to $i\pi$ and $-1$ to $\mathbf{I}$! – thorimur Apr 04 '21 at 21:52
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    Eigendecomposition of $A$ is the standard approach. This will make calculating arbitrary powers of $A$ much simpler. – Aaron Hendrickson Apr 04 '21 at 21:56
  • DeltaIV: as Oliver said above, you don't have to compute the eigenvectors if you know that matrix A is diagonalizable. – mathmathmath Apr 04 '21 at 22:05
  • @OliverDiaz done! I'll proceed to cancel my comments so that the question is more readable. – DeltaIV Apr 05 '21 at 14:13

3 Answers3

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"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

  1. your matrix is skew-symmetric, and thus an infinitesimal rotation (element of the Lie algebra $\mathfrak{so}(2)$);
  2. this infinitesimal rotation has magnitude $\pi$;
  3. 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.

user7530
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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?

Mittens
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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?

Rodrigo de Azevedo
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