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Prove that there is a nonsingular $T \in M_n(R)$ such that $A = TBT^{−1}$ if and only if there is a nonsingular $S \in M_n$ such that both $A = SBS^{−1}$ and $\bar{A} = S\bar{B}S^{−1}$. (Here $M_n$ is the set of $n \times n$ complex matrices).

If direction: We can take $S = T$ i.e. imaginary part being zero. Then, $A = SBS^{-1}$ obviously. Moreover, $\bar{A} = \overline{SBS^{-1}} = \bar{S}\bar{B}\bar{S^{-1}} = S\bar{B}S^{-1}$.

The only if direction: Let $S = H+iC$ and $\bar{S} = H-iC$ where $H,C \in M_n(R)$. Then, $A = SBS^{-1} = (H + iC)B(H+iC)^{-1}, \bar{A} = \bar{S}\bar{B}\bar{S^{-1}} = (H-iC)\bar{B}(H-iC)^{-1} = S\bar{B}S^{−1} =(H + iC)\bar{B}(H+iC)^{-1}$.

I guess I should show that $S = \bar{S}$ i.e. $H+iC = H - iC$. Because if complex conjugate of $S$ is equal to $S$ then $S \in M_n(R)$ so, we will complete the proof.

user13
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    It is possible to have $A = SBS^{-1}$ and $\bar A = S\bar BS^{-1}$ without also having $S = \bar S,$ so your approach doesn't work – Ben Grossmann Apr 01 '21 at 17:32
  • It's not quite the same question, but this post regarding matrices over arbitrary fields is related (and potentially interesting to the readers of this post). – Ben Grossmann Apr 01 '21 at 18:06
  • @IgorRivin I think it answers my question. However, I am struggling to fully comprehend it. Actually, I found that question earlier but I thought that there may be some 'simpler' solution to this problem. – user13 Apr 01 '21 at 18:23

1 Answers1

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The only direction you need help with here is the "only if" direction. We are given that $$ A = SBS^{-1}, \quad \bar A = S\bar BS^{-1}. $$ Rearranging these equations and taking the conjugate of both sides on the right brings us to the equations $$ AS = SB, \quad A \bar S = \bar S B. $$ We can therefore conclude that $H$ and $C$ (the real and imaginary parts of $S$) satisfy $AH = HB$ and $AC = CB$. The catch, however, is that we need an invertible real matrix $T$ for which $AT = TB$, and we have no guarantee that either $H$ or $C$ is invertible.

With that in mind, consider the polynomial $$ p(t) = \det(H + Ct). $$ We know that $p(i) \neq 0$, which means that $p$ cannot be the zero-polynomial. Thus, $p$ has at most finitely many zeros. Thus, there necessarily exists a $t \in \Bbb R \subset \Bbb C$ for which $p(t) \neq 0$. If we define $T = H + Ct$, then we see that $T$ is real, $T$ is invertible, and $AT = TB$. We can rearrange this last equation to get $$ A = TBT^{-1}, $$ which was what we wanted.

Ben Grossmann
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  • Thank you. I saw a similar solution to yours. However, the problem is, as I wrote, $A,B$ are complex $n \times n$ matrices too. So, we can't say that $AH = HB, AC = CB$. It would rather be something like $(A_0 + iA1)(H+iC) = (A_0H - A_1C) + I(A_1H + A_0C), (H+ iC)(B_0+iB_1) = (HB_0 - CB_1) + I(HB_1 + CB_0)$ and $(A_0H - A_1C) = (HB_0 - CB_1) , (A_1H + A_0C) = (HB_1 + CB_0)$. – user13 Apr 01 '21 at 18:28
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    You can say that $AH = HB$ and $AC = CB$. Take the equations $AS = SB$ and $A\bar S = \bar S B$; you can add them together and simplify to find that $AH = HB$. Similarly, you can subtract the equations and simplify to find that $AC = CB$ – Ben Grossmann Apr 01 '21 at 18:35
  • I see it. Thank you very much. – user13 Apr 01 '21 at 18:55