If $A$ has two eigenvalues $\lambda _1, \lambda_2$ and $\dim (E_{\lambda_1})=n-1$, then $A$ is diagonalizable Suppose that $A  \in  M_{n\times n}(\Bbb F)$ has two distinct eigenvalues $\lambda_{1}$ and $\lambda_{2}$ and that $\dim (E_{\lambda_1})=n-1$ show that $A$ is diagnolizable.  

Hint: Since $\dim(E_{\lambda _1})=n-1$, there exist $v_1, \ldots , v_{n-1}$ linearly independent eigenvectors of $\lambda _1$. Let $v_n$ be an eigenvector of $\lambda _2$. Now consider the $n\times n$ matrix $P$ whose $i^{\text{th}}$column is $v_i$. The invertibility of $P$ follows from this.
Can you take it from here?
