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.  
 A: Being diagonalisable means that the sum of the different eigenspaces is equal to the whole space. It is a fundamental fact that this sum is always a direct sum, so its dimension is the sum of the dimensions of the eigenspaces; all that remains to show is that this sum attains the value$~n$. Now $\dim(E_{\lambda_1})=n-1$ is given, and $\dim(E_{\lambda_2})\geq1$ since otherwise $\lambda_2$ would not be an eigenvalue. This proves that $\dim(E_{\lambda_1})+\dim(E_{\lambda_2})\geq n$, which can only be equality, so one can also conclude $\dim(E_{\lambda_2})=1$.
A: 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?
A: $\dim (E_{\lambda_1})=n-1$ 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. 
