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Similar matricies

  1. Jun 27, 2010 #1
    Let A and B be 2x2 real matricies, and suppose there exists an invertible complex 2x2 matrix P such that B = [P^(-1)]AP.

    Show that there exists a real invertible 2x2 matrix Q such that B = [Q^(-1)]AQ.


    A and B are similar when thought of as complex matricies, so they represent the same linear transformation on C2 for appropriately chosen bases, and share many other properties:

    same trace, same determinant, same characteristic equation , same eigenvalues.


    If I take Q = (1/2)(P + P bar) (the "real part" of P),
    then I can show QB = AQ, and so B = [Q^(-1)]AQ if Q is invertible, but this Q may not be invertible.

    I also noticed that each of A and B may be triangularized (since each of A and B has an eigenvalue), but I don't know where to go from there...
     
  2. jcsd
  3. Jun 28, 2010 #2

    CompuChip

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    You know that P is invertible, right?
    Then if Q = 1/2(P + P*), you can show that det Q = 1/2(det P + (det P)*) so if det P is non-zero, then so is det Q and you are done.
     
  4. Jun 29, 2010 #3
    What if det P = i. Then det P + (det P)* = 0 (I'm assuming that x* means the complex conjugate of x). Also, if each entry of P is pure imaginary,then P + P* = 0.

    I was unable to derive the formula det Q = 1/2(det P + (det P)*. Computing det Q explicitly in terms of the entries in P I got det Q = 1/2(det P + (det P)* + a) where a is a possibly nonzero real number.
     
  5. Jul 7, 2010 #4
    Think more simpler.

    How about [itex] Q = \alpha P [/itex] for any real nonzero [itex]\alpha[/itex]?
     
  6. Jul 7, 2010 #5
    Let [tex]X[/tex] and [tex]Y[/tex] be real and imaginary parts of [tex]P[/tex] (take the real and imaginary parts of each entry), so [tex]P= X+ iY[/tex].

    We can rewrite the equation [tex]B = P^{-1}A P[/tex] as [tex]PB = AP[/tex], or, using the above notation [tex](X+iY)B = A(X+iY)[/tex].

    Comparing real and imaginary parts we get that [tex]X B = A X [/tex] and [tex]Y B = A Y [/tex]. Therefore [tex](X+\alpha Y)B = A(X+\alpha Y)[/tex] for all real (and complex) [tex]\alpha[/tex].

    So, we prove the statement we only need to find a real [tex]\alpha[/tex], such that the matrix [tex]Q:= X + \alpha Y[/tex] is invertible.
    To see that it is possible, consider a function [tex]f(z) = \operatorname{det} (X+ zY) [/tex]. If all matrices are [tex]n\times n[/tex], it is a polynomial of degree at most [tex]n[/tex].

    We know that [tex]f(i) \ne 0[/tex] (because [tex]P[/tex] is invertible), so [tex]f(z)[/tex] is not identically zero. Therefore it has at most n roots, so any [tex]\alpha[/tex] avoiding these roots works.
     
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