For a defective matrix $B$, do $B$ and $B^*$ have the same eigenvalues? 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 $B^*$?

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 $v^* u = 0$.
