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Let $A$ be a square matrix over $\mathbb C$, and let $A^T$ denote its transpose.

It is not hard to see that $A$ and $A^T$ have the same set of eigenvalues, so given $Ax=\lambda x$ for some vector $x\in V$ and eigenvalue $\lambda\in\mathbb C$, we know that there must always be some other $y\in V$ such that also $$A^T y=\lambda y.$$

We also know that, if $Ax=\lambda x$ and $A^T y=\mu y$ with $\lambda\neq \mu$, then $\langle y^*,x\rangle=0$, where $y^*$ denotes the vector whose elements are complex conjugate of those of $y$, as follows from $$\langle y^*,Ax\rangle=\lambda \langle y^*,x\rangle=\mu\langle y^*,x\rangle.$$

The same argument, however, does not provide any information for the case $\mu=\lambda$. Is there relation holding in general for such a case?

More precisely, given $Ax=\lambda x$ and $A^T y=\lambda y$, is there any general relation between $x$ and $y$?

glS
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1 Answers1

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This is a classical result that any matrix $A$ is similar to its transpose.

Then, write $A^T=PAP^{-1}$, you have that if $A^Tx=\lambda x$, then $$A(P^{-1}x)=\lambda (P^{-1}x)$$

So that $E_\lambda(A^T)=P\cdot E_\lambda(A).$

The matrix $P$ isn't always easy to find, and indeed the "classical result" is not that easy to show. This can be done with a density argument when the field is for example $\mathbb C$. In a more general context, you can use Frobenius matrix reduction to see it.

But for an orthogonal matrix for example, it is easy to see that the eigenvectors are the same.

elidiot
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