I'm trying to solve this problem: Given complex matrices A and B, prove there's a nonsingular real matrix Q such that $A=QBQ^{-1}$, if and only if there's a nonsingular complex matrix S such that $A=SBS^{-1}$ and $\overline{A}=S\overline{B}S^{-1}$

The forward direction is fine since a real matrix is also complex. I'm having trouble with the backwards direction.

The only clue I have is a theorem that says given 2 real matrices that are similar over complex numbers... they must also be similar over the real numbers.

$A\overline{A} = SB\overline{B}S^{-1}$

does this somehow imply S is a real matrix?

Also: $A = SBS^{-1}$

$A = \overline{S}B\overline{S^{-1}}$

Does A being similar to B by two different matrices somehow imply they are equal... thereby implying $S = \overline{S}$ ?

Appreciate any help. Thanks.


We get:

$A + \overline{A} = S(B+\overline{B})S^{-1}$

Since $A+\overline{A}$ and $B+\overline{B}$ are real, we can use the theorem to show that there's a real matrix T such that:

$A + \overline{A} = T(B+\overline{B})T^{-1}$

Similarly, $i(A-\overline{A}) = Si(B-\overline{B})S^{-1}$

Since $i(A-\overline{A})$ and $i(B-\overline{B})$ are real, we can use the theorem to show that there's a real matrix G such that:

$i(A - \overline{A}) = Gi(B-\overline{B})G^{-1}$ so

$(A - \overline{A}) = G(B-\overline{B})G^{-1}$

Can we use real matrices G and T to somehow show $A = QBQ^{-1}$ for a real matrix Q ?

Ok. I think I've found the solution now. The theorem also says that if one pair of real matrices are similar via a complex matrix S... then they're similar via a real matrix T... and if another pair of real matrices are also similar via S, they are also similar via the same real matrix T. Hence in my exposition above G = T.

So we get (by adding the two equations):

$2A = 2TBT^{-1}$

$A = TBT^{-1}$


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