exp and calculation

alekk

easy question that an examinator asked me a few days ago to begin an examination:
Does there exist a matrix $A \in M_n(\mathbb{R})$ such that $exp(A)=-I$ ?
If it exists, comute it .

**Posted:** Fri Jul 15, 2005 11:38 pm

_________________
They misunderestimated me!

grobber

$A$ should be diagonalizable in $\mathcal M_n(\mathbb C)$ , I believe. Also, it's eigenvalues must be $\pm \pi i$ . It doesn't exist if $n$ is odd, and it exists and it's similar (in $\mathcal M_n(\mathbb R)$ ) to the matrix having $\frac n2$ blocks of the form $\begin{pmatrix}0&1\\-\pi^2&0\end{pmatrix}$ on the main diagonal.

Is it Ok?

**Posted:** Sat Jul 16, 2005 12:16 am

Kent Merryfield

I would have gone for

$\begin{bmatrix}0&-\pi\\ \pi&0\end{bmatrix}$

But it all works.

$\exp\left(t\begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\right)= \begin{bmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{b...$

**Posted:** Mon Jul 18, 2005 10:53 am

alekk

**Posted:** Mon Jul 18, 2005 11:49 am

_________________
They misunderestimated me!

yassinus

thank you alekk for the idea, but can you explain more , i hope succes in ENS the previous years ( i think you are from Louis le grand or Saint ... Razz

) i didnt see how to use this countinios morphism , but i think that a recurence can show this diagonalization by block

**Posted:** Thu Aug 25, 2005 11:34 am

alekk

oops, sorry for this late answer (hollydays...)

$f$ is a continuous morphism, so if $z=a+ib$ then
$f(1+z+z^2/2+..+z^k/k!+..)=f(1) + f(z) + f(z)^2/2! + ... + f(z)^k/k! + ... = exp(f(z))$
Hence
$f(exp(z))=exp(f(z))$
$f(cos(t) + isin(t))=exp(f(a+ib))$
(the examinator was quite happy to see this sln ..)

If you have any other question, don't hesitate Smile

**Posted:** Sat Sep 10, 2005 3:21 am

_________________
They misunderestimated me!