Solving Similar Matricies: Find Real Invertible 2x2 Matrix Q

[samkolb]

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...

 

[lanedance]

does it help to know as A & B are real, then:
[tex] B* = B [/tex]
[tex] A* = A[/tex]

which gives
[tex]B = (P^{-1})^*AP^* = P^{-1}AP [/tex]

 

[samkolb]

I don't know if this helps, since P* may not be real.