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The problem statement:

Let $A$ be a real $9\times 9$ matrix with transpose $B$. Prove that the matrices are real equivalent in the following sense: There exists a real invertible $9\times 9$ matrix $H$ such that $AH=HB$.

Unfortunately, I don't have much of an attempt at a solution. My first thought was to notice that $A$ and $B$ have the same Jordan normal form so they're similar. But the Jordan form may have complex entries as the eigenvalues may be complex. Also confusing is why the matrix is $9\times 9$. How could this matter? Nine is a perfect square and it's odd, but so are a lot of other numbers.

Any help is welcomed.

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up vote 6 down vote accepted

The key is to prove the following:

Theorem. Let $A$ and $B$ be matrices with real coefficients. If $A$ and $B$ are similar over $\mathbb{C}$, then they are similar over $\mathbb{R}$.

This problem appears, among other places, in Berkeley Problems in Mathematics, by de Souza and Nuno Silva (Problem 7.7.10 in the second edition; the problem has been used many times in the Prelim exam at Berkeley). The book offers two answers, though I think that one of the solutions (using the rational canonical form) is a bit sloppy.

Here is the other approach. Assume $A = UBU^{-1}$, with $U$ a complex matrix. Write $U=K+iL$, where $K$ and $L$ are real matrices, and assume that $L\neq 0$. Then $$A(K+iL) = AU = UB = (K+iL)B$$ and so the real parts and imaginary parts of these matrices are equal. That is, $AK = KB$ and $AL = LB$ (the problem is we don't know that $K$, by itself, is invertible).

Now, for any complex number $z$ we have $A(K+zL) = (K+zL)B$. Let $p(z) = \det(K+zL)$. Prove that $p$ is not identically zero, and take a real number that is not a root.

Once you have this theorem, your observation that $A$ and $B$ are similar over $\mathbb{C}$ implies the result you want.

(For what it is worth, the other argument given by the book is that the Rational canonical forms of $A$ and $B$ have to be the same, since they are the same over $\mathbb{C}$, and so $A$ and $B$ are similar over $\mathbb{R}$; the reason this is sloppy is that the rational canonical form is usually defined, for uniqueness purposes, in terms of the irreducible factors of the minimal/characteristic polynomials, and the rational blocks are companion matrices of powers of these irreducible factors. But the irreducible factors over $\mathbb{R}$ are not necessarily the same as the irreducible factors over $\mathbb{C}$, so that the "uniqueness" of the rational canonical form does not directly apply; an argument needs to be made that the equality over $\mathbb{C}$ implies an equality when the irreducible factors are taken over $\mathbb{R}$ instead, and this argument is missing in the book).

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You don't need to know the degree of $p$. Since $p(i) \ne 0$, $p$ is not identically $0$, so there are real non-roots. – Robert Israel Jul 24 '12 at 19:11
@RobertIsrael: Oh, right. And thanks for correcting the typo. – Arturo Magidin Jul 24 '12 at 19:11
Thanks for the excellent answer. – Derek Allums Jul 24 '12 at 19:20

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