I know that any Heritian matrix is diagonalizable, however I am trying to find when a non-Hermitian matrix is diagonalizable.
Is it enough that there are $N$ linearly independent eigenvectors if the matrix is $N\times N$?
Also can the same method be applied to these non-hermitian matrices (those that are diagonalizable) as the hermitian matrices, i.e. writing $A = PDP^{-1}$ where $P$ is made of the eigenvectors etc.
Best Answer
It suffices to have $ n $ linearly independent eigenvectors for the matrix to be diagonalizable. As an example which is not Hermitian, consider the matrix
$$ \begin{bmatrix} 1 & 4 \\ 1 & 1 \end{bmatrix} $$
Solving $ (1 - \lambda)^2 = 4 $, we find that the eigenvalues are $ \lambda = 3 $ and $ \lambda = -1 $, and the corresponding eigenvectors $ (2, 1)^T $ and $ (2, -1)^T $, which are linearly independent. Therefore, we conclude that our matrix is diagonalized as
$$ \begin{bmatrix} 3 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} 1 & 4 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & 2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{1}{2} \\ \frac{1}{4} & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & 4 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & 2 \\ 1 & -1 \end{bmatrix} $$