# Do similar matrices respect multiplication

1. Sep 18, 2015

### jackmell

1. The problem statement, all variables and given/known data
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?

2. Relevant 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 $P\in GL_2(F)$.

3. 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.

Jack

2. Sep 18, 2015

### RUber

I look at it this way. If A is in a vector space X, then B is in a similar vector space Y.
The transformation from A to B is $P^{-1}A P = B$ and the inverse gives $P B P^{-1} = A$.
Multiplying by a matrix C in X should be the same as multiplying by the matrix similar to C in Y. In this way the relationship respects multiplication.

Let $P^{-1}C P = D$ and $P D P^{-1} = C$.
Then CA is similar to DB.
$CA = P D P^{-1}P B P^{-1} = P D I B P^{-1} = P D B P^{-1}.$

3. Sep 18, 2015

### RUber

For a counterexample, you simply need to show that P does not necessarily have a special structure that would allow CP = PC.
One example is similar to what you had above. Let (using matlab matrix notation - sorry)
$B = [2,0; 0,3], \, P = [1,1;1,0], \,P^{-1} = [0,1;1,-1] \text{ and } A = [ 3, -1; 0,2 ]$
You can pick just about any matrix C other than the identity to show that $CP \neq PC$.

If $CP = PC$, then $P^{-1} C P =P^{-1} P C = C$ If this were to hold for any C, then the transformation between your 'similar' spaces is the identity transformation, which means P would be a scalar multiple of the identity.

4. Sep 18, 2015

### jackmell

Ok thanks Ruber. I understand that if $A=PBP^{-1}$ and $C=PDP^{-1}$ then in this particular case, we have $AC=PBDP^{-1}$. Afraid though I don't understand the significance of showing $P$ does not have the structure that would allow $CP=PC$. Could you please explain this a little further to me?

Thanks,
Jack

5. Sep 19, 2015

### Ray Vickson

My interpretation of the question seems different from yours. To me, it is asking: if $A_1 \sim A$ and $B_1 \sim B$, is it necessarily true that $A_1 B_1 \sim A B$? In other words, does the existence of invertible $P, Q$ with $A_1 = P A P^{-1}$ and $B_1 = Q B Q^{-1}$ imply the existence of $R$ such that $A_1 B_1 = R\, AB \,R^{-1}$?

6. Sep 19, 2015

### jackmell

Hi Ray,

In order for the equivalence relation to respect multiplication, then if $A=PBP^{-1}$, then for all $C\in GL_n(F)$, there exists some $X\in GL_n(F)$ such that $AC=XBCX^{-1}$ (and also $CA=YCBY^{-1}$). That is, $X$ can be different for each $C$ as long as $AC\sim BC$.

Someone in class mentioned that similar matrices have similar traces and so this could be proved very easily in 2 or so lines but I couldn't do it. I'm pretty sure the equivalence does not respect multiplication and so would have had to find a counter example but didn't know how.

7. Sep 19, 2015

### vela

Staff Emeritus
Suppose A = PBP-1 and AC = PBCP-1. Insert the identity in the form P-1P in between B and C.

8. Sep 19, 2015

### jackmell

I understand that: If $A=PBP^{-1}$ and $AC=PBCP^{-1}$ then we have:
\begin{aligned} AC&=\left(PBP^{-1}\right)\left(PCP^{-1}\right) \\ AC&=APCP^{-1} \\ C&=PCP^{-1} \\ \end{aligned}
That would imply $PC=CP$ which in general is not the case.

However, that's not how I understand what the problem is asking and maybe I'm wrong. It's saying if $A=PBP^{-1}$ for some $P\in GL_2(F)$ then in order to respect left-multiplication, there would need to exist elements in $GL_2(F)$ (perhaps different for each $C$) such that $AC_1=XBC_1X^{-1}$, $AC_2=X_2BC_2X_2^{-1}$, $AC_3=X_3BC_3X_3^{-1}$, etc., for all $C_i\in GL_n(F)$ so that we could not insert $PP^{-1}$ like we did above.

Am I not understanding the problem then?

9. Sep 19, 2015

### vela

Staff Emeritus
I thought you didn't see where RUber got the condition that C and P have to commute.

It's not completely clear to me what you mean by "respect multiplication." The definition you gave in the original post uses P to relate both A to B and AC to BC.

10. Sep 19, 2015

### jackmell

Sorry about that. I re-read it and it's ambiguous. To be honest, I'm not so sure myself. But if I had to take sides then I would say it means that if $A=PBP^{-1}$ and the equivalence respects multiplication (say left-multiplication for now), then for each $C\in GL_n(F)$, there exists an $X\in GL_n(F)$ such that $AC=XBCX^{-1}$ (sometimes $X=P$, sometimes not though).

Also I didn't understand what Ruber meant until you explained it. So thanks!