Do similar matrices respect multiplication

In summary: 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)##, then the equivalence relation respects multiplication.
  • #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
 
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  • #2
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
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
RUber said:
For a counterexample, you simply need to show that P does not necessarily have a special structure that would allow CP = PC.

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
jackmell said:

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

Ok, thanks for reading,
Jack

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
Ray Vickson said:
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}##?

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
jackmell said:
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
Suppose A = PBP-1 and AC = PBCP-1. Insert the identity in the form P-1P in between B and C.
 
  • #8
vela said:
Suppose A = PBP-1 and AC = PBCP-1. Insert the identity in the form P-1P in between B and C.

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
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
vela said:
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.

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!
 

What is a similar matrix?

A similar matrix is a matrix that can be transformed into another matrix by performing a series of elementary row and column operations.

What does it mean for two matrices to respect multiplication?

Two matrices respect multiplication if their product is equal to the product of their individual elements.

Do similar matrices always respect multiplication?

Yes, similar matrices always respect multiplication since the operations performed to transform one matrix into another do not change the product of their elements.

Can two matrices be similar but not respect multiplication?

No, if two matrices are similar, that means they can be transformed into each other without changing the product of their elements. Therefore, they must also respect multiplication.

What is the significance of similar matrices respecting multiplication?

When two matrices are similar and respect multiplication, it means that they represent the same linear transformation. This is useful in various applications, such as solving systems of linear equations, finding eigenvalues and eigenvectors, and diagonalizing matrices.

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