Prove Existence of Real Invertible Matrix Q for A & B 2x2 Similar Matricies

[samkolb]

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...

 

[CompuChip]

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.

 

[samkolb]

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.

 

[trambolin]

Think more simpler.

How about Q = \alpha P for any real nonzero \alpha?

 

[Hawkeye18]

Let X and Y be real and imaginary parts of P (take the real and imaginary parts of each entry), so P= X+ iY.

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

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

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

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