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.  

$\dim (E_{\lambda_1})=n-1$ indicates there are $(n-1)$ linearly independent eigenvectors, forming the set $\beta$, corresponding to $\lambda_1$.
$\lambda_2$ is an eigenvalue for $A$, implying there are at least $1$ eigenvectors,  $v$, for $A$ corresponding to $\lambda_2$.
Then $\beta\cup\{v\}$ forms the set consisting of $n$ linearly independent vectors in $F^n$, and it can be the basis of $F^n$. 
Because $F^n$ has the basis consisting of $A$'s eigenvectors, A is diagonalizable. 
