$\begin{pmatrix}1&2&0&-a-1\\ 1&1&0&0\\ 0&1&1&a+2\\ 0&0&1&1\end{pmatrix}$ can not be diagonalizable in complex field, find $a$.
What's the idea? Intuititively, $a=-1$ or $a=-2$, so some entries of the matrix is zero$, but using mathematical software, it is diagonalizable!
What is needed for this type problem? I find no such ideas from the books, and though hard to find nothing. Help.
A square matrix is diagonalizable, if it is a normal matrix, that is, it commutes with its adjoint matrix, the matrix transposed with complex conjugate entries.
The example A yields a constant -4 at position (2,2) for the commutator, independent of the choice of a.
$$ A A^* - A^* A =\left( \begin{array}{cccc} (a+1) a^*+a+4 & 0 & 2-(a+1) \left(a^*+2\right) & 0 \\ 0 & -4 & 0 & a \\ 2-(a+2) \left(a^*+1\right) & 0 & (a+2) \left(a^*+2\right) & 0 \\ 0 & a^* & 0 & -(2 a+3) a^*-3 a-4 \\ \end{array} \right)$$