Is $A = \left( \begin{matrix} \lambda & 0 \\ 0 & 0 \end{matrix}\right)$ diagonalizable if $\lambda$ is the only eigenvector of $A$?

Question

My book states the following lemma:

Suppose that $\lambda$ is the only eigenvalue of $A \in M_{2\times 2}(\mathbb{F})$. Then, $A$ is diagonalizable if and only if $A = \left( \begin{matrix} \lambda & 0 \\ 0 & \lambda \end{matrix}\right)$

But, isn't $A = \left( \begin{matrix} \lambda & 0 \\ 0 & 0 \end{matrix}\right)$ diagonalizable since it is already a diagonal matrix? Or is it impossible to have such a $2\times 2$ matrix with a single eigenvalue in a single diagonal entry and the other a $0$.

Answer 1

The problem is that the matrix $\left( \begin{matrix} \lambda & 0 \\ 0 & 0 \end{matrix}\right)$ has two eigenvalues: $\lambda$ and $0$.