Invertible matricies are the exponential of another matrix

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Another one that I am guessing is fairly easy

Every invertible n x n matrix A can be written as A=e^X

for some $X \in M_n(\mathbb{C})$ .

The hint is to start with $A = PBP^{-1}$ where

is a block diagonal matrix in which each block is of the form $\lambda I + N_\lambda$ and $N_\lambda$ is nilpotent.

I know that if

is unipotent then $\log A$ is nilpotent and if

is nilpotent then $\log A$ is unipotent

So certainly $e^{N_\lambda}$ exists, and is unipotent.

I'm not really sure where to go with this one.

Any hints?

outermeasure · **Posted:** Thu, 1 Dec 2011 06:07:59 UTC

qwirk wrote:

Another one that I am guessing is fairly easy

Every invertible n x n matrix A can be written as A=e^X

for some $X \in M_n(\mathbb{C})$ .

The hint is to start with $A = PBP^{-1}$ where

is a block diagonal matrix in which each block is of the form $\lambda I + N_\lambda$ and $N_\lambda$ is nilpotent.

I know that if

is unipotent then $\log A$ is nilpotent and if

is nilpotent then $\log A$ is unipotent

So certainly $e^{N_\lambda}$ exists, and is unipotent.

I'm not really sure where to go with this one.

Any hints?

No, you want to construct your nilpotent

from a unipotent matrix, not starting with N nilpotent and show e^N

is unipotent.

Recall:

$\exp(A+B)=\exp(A)\exp(B)$ when .
$\exp(PAP^{-1})=P\exp(A)P^{-1}$
$\log(I+A)=A-\frac{1}{2}A^2+\frac{1}{3}A^3-\dots$ is (for those A such that the series is defined) a right inverse of $\exp(B)=I+B+\frac{1}{2}B^2+\frac{1}{3!}B^3+\dots$ .

So conjugating to decompose $\mathbb{C}^n$ into (irreducible) invariant subspaces of

, and obviously A|subspace commutes with A|{another subspace}. So you are down to constructing log when it is of the form $\lambda I+N$ , nilpotent N, by upper-triangularising (e.g. Jordan normal form).

The $\lambda$ is easy to get rid of --- just use $\log(\lambda) I$ since it commutes with everything, so you are down to taking log of I+N'

, nilpotent $N'=\lambda^{-1}N$ . That's where point (3) comes in.

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Thanks again outermeasure

S.O.S. Mathematics CyberBoard

Invertible matricies are the exponential of another matrix

Who is online