Solving Similar Matricies: Find Real Invertible 2x2 Matrix Q

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Homework Statement



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.

Homework Equations


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.




The Attempt at a Solution


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...
 
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does it help to know as A & B are real, then:
B* = B
A* = A

which gives
B = (P^{-1})^*AP^* = P^{-1}AP
 
I don't know if this helps, since P* may not be real.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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