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From the definition of normal matrix, $AA^*=A^*A$, we know that $A$ and $A^*$ share the same eigenvectors, but my question is that do defective matrix $B$ and its conjugate transpose $B^*$ also have the same eigenvectors, although their eigenvectors are not complete. If not, is there a simple relation between the eigenvectors of $B$ and $B^*$?

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    How do you deduce directly from $AA^*=A^*A$ that $A$ and $A^*$ have the same eigenvectors? It's certainly true, but using the definition directly I only get that if $x$ is an eigenvector of $A$ then so is $A^*x$; if the eigenspaces aren't one-dimensional this doesn't immediately imply that $x$ is an eigenvector of $A^*$. Am I missing something?2012-07-16
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    @joriki My understanding is that two commutable matrices have the same eigenvectors if they are diagonalizable. So if you admit $A$ and $A^*$ are diagonalizable, then $A$ and $A^*$ share the same eigenvectors.2012-07-16
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    I see, OK -- I thought you were inferring it directly from the fact that they commute.2012-07-16

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Try $B = \pmatrix{0 & 1\cr 0 & 0\cr}$. Do it and $B^*$ share an eigenvector?

EDIT: The relationship is this. If $u$ is an eigenvector of $B$ for eigenvalue $\lambda$ and $v$ is an eigenvector of $B^*$ for eigenvalue $\mu$, and $\mu \ne \overline{\lambda}$, then $v^* u = 0$.

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    Thanks, but is there a simple relation between eigenvectors of $B$ and $B^*$?2012-07-16
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    Is there any general statement about the eigenvectors for eigenvalues $\mu = \bar{\lambda}$?2012-07-16
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    The eigenvalue $\mu$ for $B^*$ and the eigenvalue $\lambda = \overline{\mu}$ for $B$ have the same algebraic and geometric multiplicities. If those algebraic multiplicities are $1$, the corresponding eigenvectors are not orthogonal ($v^* u \ne 0$). I think that's about all you can say about them.2012-07-17