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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!

  • 0
    Hmmm, I am not sure.2012-11-27

1 Answers 1

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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 $k$.