- #1

jackmell

- 1,807

- 54

## Homework Statement

Let ##G=GL_n(F)## for ##F## a field, and define an equivalance relation by ##A\sim B## iff ##A## and ##B## are conjugate, that is, iff ##A=PBP^{-1}## for some ##P\in GL_n(F)##. Does ##\sim## respect multiplication?

## Homework Equations

The equivalency respects multiplication if for ##A=PBP^{-1}## for some ##P\in GL_n(F)##, we also have ##AC=PBCP^{-1}## and ##CA=PCBP^{-1}## for some ##P\in GL_2(F)##.

## The Attempt at a Solution

Claim ##A\sim B\Rightarrow \exists P\in GL_2(F): A=PBP^{-1}## does not respect multiplication.

Now my argument for proposing this is a practical one: I checked a lot of linear algebra texts about the matter and found that similar matrices have the same determinant, trace, rank, even similar polynomials ##F(A)=F(PBP^{-1})## but could find no relation regarding products of similar matrices. Thus I suspect there is none. To test this, I propose the following:

First, find a case in ##GL_2(\mathbb{R})##:

Show that if ##AC## is a diagonal matrix, then can find a ##BC## with non-independent eigenvectors such that ##A\sim B## and thus ##AC\neq P BC P^{-1}\;\forall\; P\in GL_2(F)##, that is, ##BC## cannot be diagonalized. Then scale-up to ##GL_3(\mathbb{R})## and then perhaps I can see a trend, and then generalize it to ##GL_n(\mathbb{R})## and then finally to ##GL_n(F)##. For example, let:

##

A=\begin{bmatrix}0 & 1 \\ 1 & 1

\end{bmatrix}

##

and

##

C=\begin{bmatrix}1 & -1 \\ -1 & 0

\end{bmatrix}

##

so that ##AC=\begin{bmatrix}0 & 1 \\ 1 & 1

\end{bmatrix}\begin{bmatrix}1 & -1 \\ -1 & 0

\end{bmatrix}=\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}.

##

Now need to find a similar matrix to ##A##, that is, ##A=PBP^{-1}## such that ##BC## is non-diagonalizable, i.e., has linearly-dependent eigenvectors. I'm not sure though how to find ##BC## meeting this requirement.

So I was hoping that if someone could just help me find a counter-example in ##GL_2(\mathbb{R})##, then I could work out the rest.

Ok, thanks for reading,

Jack