Could you please help me with the following problem?
Let $A$ be an $n$$\times$$n$ complex matrix. Prove that $A$ is similar to $B$, which is an $n$ $\times$ $n$ real matrix, if and only if $A$ is similar to its conjugate transpose.
I have been trying to solve it for quite a long time, I tried to expand the definition of similarity but I did not manage to get anything useful. I have no better idea.

  • $\begingroup$ @AlgebraicPavel: I guess I do not understand you exactly. Why should I prove that $A$ is similar to $A$$*$, if that is a given condition? $\endgroup$
    – Berciboy
    Jan 22 '15 at 12:06
  • $\begingroup$ Sorry for the not exactly understandable sentence then. Anyway, it is true, the only question is how to be proven, $\endgroup$
    – Berciboy
    Jan 22 '15 at 12:16
  • $\begingroup$ Yes, that is what I am willing to prove. As far as I am concerned, that is the task. Thanks for your help. I ask a friend of mine whether I misunderstood the task. $\endgroup$
    – Berciboy
    Jan 22 '15 at 12:26
  • 1
    $\begingroup$ @AlgebraicPavel: Not every (complex) matrix is similar to its conjugate transpose; for instance $\mathbf iI$ is not. But every (real or otherwise) matrix is similar to its transpose. I think the real question is to prove $A\sim B\;\Leftrightarrow\; A\sim B^*$, where $B^*=B^\top$ since $B$ is supposed real; this is of course trivial if you already know $B\sim B^\top$ . The question is just badly formulated: "its" is unclear. $\endgroup$ Jan 22 '15 at 12:59
  • $\begingroup$ @MarcvanLeeuwen I see, thanks for correcting me :) $\endgroup$ Jan 22 '15 at 13:02

The following facts are useful:

  • (1) For any $A\in\mathbb{C}^{n\times n}$, $A\sim A^T$.
  • (2) For any $A\in\mathbb{C}^{n\times n}$, $A\sim B$ for some $B\in\mathbb{R}^{n\times n}$ if and only if $A\sim\bar{A}$.

So if $A\sim B$ for some real $B$, then $$A\stackrel{(2)}\sim\bar{A}\stackrel{(1)}\sim\bar{A}^T=A^*.$$ On the other hand, if $A\sim A^*$, then $$ A\sim A^*=\bar{A}^T\stackrel{(1)}\sim\bar{A}\stackrel{(2)}\sim B $$ for some real $B$.

Fact (1) can be shown by establishing a similarity transformations between elementary Jordan blocks and their transposes. If $J_{\lambda}$ is an elementary Jordan block and $R$ is the "reversal" matrix (an identity with flipped columns (or rows)), then $R^{-1}J_{\lambda}R=J_{\lambda}^T$ (note that $R^{-1}=R^T$).

For fact (2), the tricky part is to show that if $A\sim\bar{A}$, then $A$ is similar to a real matrix (the other direction of the equivalence is easy). If $A\sim\bar{A}$, then both $A$ and $\bar{A}$ have the same Jordan form $J$. But since both $A$ and $\bar{A}$ are similar to $J$, for each elementary Jordan block $J_{\lambda}\in\mathbb{C}^{k\times k}$, there must be a "matching" Jordan block $J_{\bar{\lambda}}$. The tricky part is to show that the "paired" Jordan blocks $$ \pmatrix{J_{\lambda}&0\\0&J_{\bar{\lambda}}} $$ are similar to a real matrix of the form $$ \pmatrix{K&I_2&&\\ &K&I_2&\\ &&\ddots&\ddots\\ &&&K&I_2\\&&&&K}\in\mathbb{R}^{2k\times 2k}, \quad K:=\pmatrix{\Re\lambda&\Im\lambda\\-\Im\lambda&\Re\lambda}, $$ where $I_2$ is the $2\times 2$ identity (this leads to the so-called real Jordan form).

  • $\begingroup$ Is there an easy proof for (1)? $\endgroup$
    – Berci
    Jan 22 '15 at 14:57
  • $\begingroup$ @Berci Yes. It uses simply the reversal matrix. $\endgroup$ Jan 22 '15 at 15:10
  • $\begingroup$ @AlgebraicPavel: Thank you very much for your help.That is a great proof! $\endgroup$
    – Berciboy
    Jan 22 '15 at 15:31
  • $\begingroup$ @Berciboy You are welcome! $\endgroup$ Jan 22 '15 at 15:33

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