Prove $A$ is scalar matrix

Question

Let $A\in M_n(F)$ and let's assume $A$ has only one eigenvalue. Also, $A$ is diagonalized. Prove that $A$ is a scalar matrix.

My Try:
$${P^{ - 1}}AP = \left( {\matrix{ \lambda & {} & 0 \cr {} & \ddots & {} \cr 0 & {} & \lambda \cr } } \right) \Leftrightarrow AP = P\left( {\matrix{ \lambda & {} & 0 \cr {} & \ddots & {} \cr 0 & {} & \lambda \cr } } \right) \Leftrightarrow A({v_1}, \ldots {v_n}) = (\lambda {v_1}, \ldots ,\lambda {v_n}) \Leftrightarrow A{v_i} = \lambda {v_i} \Leftrightarrow A = \lambda I$$

Is that right? I'm not 100% sure of it.

Answer 1

Simply: by the hypothesis $A$ is similar to $\lambda I_n$ i.e. there's $P\in\operatorname{GL}_n(\Bbb F)$ such that $$A=P(\lambda I_n)P^{-1}=\lambda I_n$$