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  1. For a square complex/real matrix $A$, $A$ and $A^T$ have the same set of eigenvalues, each with same algebraic multiplicities, since their characteristic polynomials are the same.

    I wonder for each eigenvalue, are its geometric multiplicities for $A$ and for $A^T$ the same?

  2. Similar question for $A$ and $A^H$, where $H$ means conjugate and transpose, and the relation between their eigenvalues is conjugate.

Thanks!

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Note that if $J$ is the Jordan form for $A$, then $A^t$ is similar to $J^t$. –  Gerry Myerson Nov 27 '12 at 3:14
    
@Gerry: Thanks! Although $J^t$ or $J^H$ may not be a Jordan form, it seems that geometric multiplicity stays the same for the same/conjugate eigenvalue? –  Tim Nov 27 '12 at 3:24
    
Think about why $J$ gives you information about multiplicities, and whether you can extract the same information from $J^t$ (or $J^h$). –  Gerry Myerson Nov 27 '12 at 4:49
    
Hmmm, I am not sure. –  Tim Nov 27 '12 at 5:12
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1 Answer

up vote 1 down vote accepted

Another attempt:

First note that for a square matrix $M$, we have $\dim \ker M = \dim \ker M^T =\dim \ker M^*$. Hence $\dim \ker (\sigma I-M) = \dim \ker (\sigma I-M)^T = \dim \ker (\sigma I-M^T)$, and $\dim \ker (\sigma I-M) = \dim \ker (\sigma I-M)^* = \dim \ker (\overline{\sigma} I-M^*)$.

The geometric multiplicity of an eigenvalue $\lambda$ of $A$ is $\dim \ker (\lambda I -A)$.

It follows that $A$ has an eigenvalue $\lambda$ of geometric multiplicity $k$ iff $A^T$ has an eigenvalue $\lambda$ of geometric multiplicity $k$.

It follows that $A$ has an eigenvalue $\lambda$ of geometric multiplicity $k$ iff $A^*$ has an eigenvalue $\overline{\lambda}$ of geometric multiplicity $k$.

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