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Do similar matrices respect multiplication

  1. Sep 18, 2015 #1
    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.

    Ok, thanks for reading,
    Jack
     
  2. jcsd
  3. Sep 18, 2015 #2

    RUber

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    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}.##
     
  4. Sep 18, 2015 #3

    RUber

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    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.
     
  5. Sep 18, 2015 #4
    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
     
  6. Sep 19, 2015 #5

    Ray Vickson

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    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}##?
     
  7. Sep 19, 2015 #6
    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.
     
  8. Sep 19, 2015 #7

    vela

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    Suppose A = PBP-1 and AC = PBCP-1. Insert the identity in the form P-1P in between B and C.
     
  9. Sep 19, 2015 #8
    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?
     
  10. Sep 19, 2015 #9

    vela

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    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.
     
  11. Sep 19, 2015 #10
    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!
     
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