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Everybody knows that a square matrix $A$ has the same eigenvalues as $A^T$. And it is clear that if $A^T=BAB^{-1}$ then $B$ maps eigenvectors of $A$ to those of $A^T$. But I have not found any discussion of the benefits of knowing $B$. Perhaps it is unusual to know $B$ exactly, without having analyzed $A$ completely. But I have some examples where this is known.

It seems at least to be nontrivial information. For example, with $A$ real and $B$ real and symmetric, you know an indefinite inner product $\langle x,y\rangle = x^HBy$ with respect to which $A$ is symmetric. This gives certain orthogonality relations (for example non-real eigenvalues have null eigenvectors) but it doesn't seem to lead to anything quantitative about eigenvalues.

The worst case may be for symmetric $A$, $A^T=IAI^{-1}$ tells you nothing.

So the question is

Have examples been studied, where knowing $B$ in $A^T=BAB^{-1}$ helped to find eigenvalues of $A$?

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If you replaced $A$ with $2A$, your $B$ would be unchanged, but your eigenvalues would double, right? – Cam McLeman Oct 13 '10 at 22:26
Your $B$ is not even unique. If $C$ is any matrix that commutes with $A$, then $BC$ is another matrix that conjugates $A$ to $A^T$. – Willie Wong Oct 13 '10 at 22:43
up vote 2 down vote accepted

Generally, over the algebraically closed field, say over $\mathbb{C}$, $A^T$ and $A$ are always conjugate, because they have the same Jordan form. So such $B$ always exists. But the knowledge of the concrete $B$ satisfying to this relation may clarify what the eigenspaces for $A$ are, because ${B^T}B$ always commutes with your A. As previously was mentioned, this condition $A^T=BAB^{-1}$ doesn't determine the eigenvalues of A uniquely, so it may help only with eigenspaces. And the amount of information it gives you depends on this particular $B$. For $B=I$ it gives you nothing, as you mentioned above.

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It doesn't matter if you're over an algebraically closed field. These two have the same rational canonical form as well, and this works over any field. – Dylan Wilson Oct 14 '10 at 0:29
I don't see why $B^TB$ should commute with $A$. – Denis Serre Oct 14 '10 at 5:38
@Denis: I guess N B means $(B^{-1})^T B$? – D. Savitt Oct 14 '10 at 5:45
Yes, $B^{-T}B$ does commute with $A$. This reminds me that a matrix $M\in M_n({\mathbb C})$ is normal if and only if $M^{*}=MU$ with $U$ unitary. In the real case, this means $M^T=MO$ with $O$ orthogonal. – Denis Serre Oct 14 '10 at 6:58
yes, you are right , it's $B^{-T}B$, so when, for example, $B^{-T}B=I$, that is, $B=B^T$ for non-degenerate $B$, it gives you nothing. Moreover, if $B^{-T}B$ takes values in the center of $GL_n$, it follows that $B$ is either symmetric or skew-symmetric matrix. So non-degenerate symmetric and skew-symmetric matrices $B$ are exactly those which don't give us any information via commuting $B^{-T}B$ and $A$. – N B Oct 14 '10 at 15:55

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