Real matrix Q for similar 2x2 matrices with complex similarity

[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.