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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? – joriki Jul 16 '12 at 6:17
@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. – chaohuang Jul 16 '12 at 15:36
I see, OK -- I thought you were inferring it directly from the fact that they commute. – joriki Jul 16 '12 at 15:40

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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 $B^*$? – chaohuang Jul 16 '12 at 15:51
Is there any general statement about the eigenvectors for eigenvalues $\mu = \bar{\lambda}$? – chaohuang Jul 16 '12 at 20:52
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. – Robert Israel Jul 17 '12 at 2:30
Thanks Robert, accepted. – chaohuang Jul 17 '12 at 20:10

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