Does diagonalizable imply invertible? The converse?
Solution
Neither implication holds.
1. A diagonalizable matrix need not be invertible.
For example, \[ A=\begin{pmatrix}1&0\\0&0\end{pmatrix} \] is diagonal, hence diagonalizable, but \(\det A=0\). More generally, a diagonalizable matrix is invertible exactly when none of its eigenvalues is zero.
2. An invertible matrix need not be diagonalizable.
Take \[ B=\begin{pmatrix}1&1\\0&1\end{pmatrix}. \] Then \(\det B=1\), so \(B\) is invertible. Its characteristic polynomial is \((t-1)^2\), while \[ \ker(B-I)=\ker\begin{pmatrix}0&1\\0&0\end{pmatrix} =\operatorname{span}\!\left\{\binom10\right\} \] has dimension \(1\). Thus the geometric multiplicity of the only eigenvalue is smaller than its algebraic multiplicity, so \(B\) is not diagonalizable.