Let $A$ be an $n\times n$ matrix.
If $A$ is invertible then one of it's eigenvalues is 0.
If $A$ is diagonalizable then it has $n$ linearly independent eigenvectors.
- Are these two statements true?
- Is there any connection between them? May I infer one from the other?
One does not imply the other.
This matrix is invertible and not diagonalizable:
$\begin{pmatrix} 1 & 1 \\ 0 & 1\\ \end{pmatrix}$
This matrix is diagonalizable (in fact it is already a diagonal matrix) but not invertible:
$\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix}$