3
$\begingroup$

Let $A_{nXn}(\mathbb{R})$ Skew-symmetric matrix $A=-A^t$ prove that $e^A(e^A)^t=I$

while: $e^A=\sum_{i=0}^{\infty} \frac{A^n}{n!}$

I tried this: $A=-A^t \Rightarrow A$ is Diagonalizable with orthogonal basis over $\mathbb{C}$ $\Rightarrow A=PDP^*$and $P^*=P^{-1}$ and D is: \begin{pmatrix} \lambda_1 &0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{pmatrix} we know that :$\lambda_i=0$ or $\lambda_i=ib$ for $b \in R$ because A is Skew-symmetric matrix. so: $e^A=Pe^DP^t$ so if I will prove that $\lambda_i=0$ we will get $e^A=PP^t=I$. Is it possible?

$\endgroup$
7
$\begingroup$

We can see easily that the linear map

$$\mathcal M_n(\Bbb R)\rightarrow \mathcal M_n(\Bbb R),\quad A\mapsto A^t$$ is continuous hence we have

$$\left(e^A\right)^t=\left(\lim_{n\to\infty}\sum_{k=0}^n\frac{A^k}{k!}\right)^t=\lim_{n\to\infty}\left(\sum_{k=0}^n\frac{A^k}{k!}\right)^t=e^{A^t}=e^{-A}$$ and the result follows from the fact: if $AB=BA$ then $$e^Ae^B=e^{A+B}$$

$\endgroup$
  • $\begingroup$ actually it's looks amazing but we didn't learn that this map is continuous(doesn't part of the course). so I think that I need to prove it with more "algebraic" way. Do you have other suggestion ? what about the way I mentioned in the Question? tnx! $\endgroup$ – GroundIns Jun 15 '14 at 16:29
  • $\begingroup$ I didn't see an alternative answer. For the way you mentioned notice that the diagonalization of the skew-symmetric matrix is more complicated than you wrote. See this. $\endgroup$ – user63181 Jun 15 '14 at 16:49
  • $\begingroup$ fixed it, now it is over C $\endgroup$ – GroundIns Jun 15 '14 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.