When eigenvectors for a matrix form a basis It is well known that if n by n matrix A has n distinct eigenvalues, the eigenvectors form a basis. 
Also, if A is symmetric, the same result holds. 
Consider 
$ A =\left[ {\begin{array}{ccc}
   1 & 2 & 3 \\
   0 & 1 & 2 \\
   0 & 0 & 1 \\
\end{array}}\right]
$
.
This matrix has single eigenvalue $\lambda=1$, and is not symmetric. 
But, the eigenvectors corresponding to $\lambda=1$,
($
v_1 =\left[ {\begin{array}{c}
   1  \\
   0  \\
   0  \\
\end{array}}\right]
$
, 
$
v_2 =\left[ {\begin{array}{c}
   0  \\
   1/2  \\
   0  \\
\end{array}}\right]
$
,
$
v_3 =\left[ {\begin{array}{c}
   0  \\
   -3/8  \\
   1/4  \\
\end{array}}\right]
$
)
form a basis. 
What sufficient conditions offer the above result?
 A: A square matrix is diagonalizable if and only if there exists a basis of eigenvectors. That is, $A$ is diagonalizable if there exists an invertible matrix $P$ such that $P^{-1}AP=D$ where $D$ is a diagonal matrix.
One can show that a matrix is diagonalizable precisely when the dimensions of each eigenspace correspond to the algebraic multiplicity of the corresponding eigenvalue as a root of the characteristic polynomial. 
If the dimension of an eigenspace is smaller than the multiplicity, there is a deficiency. The eigenvectors will no longer form a basis (as they are not generating anymore). One can still extend the set of eigenvectors to a basis with so called generalized eigenvectors, reinterpreting the matrix w.r.t. the latter basis one obtains a upper diagonal matrix which only takes non-zero entries on the diagonal and the 'second diagonal'. This is the Jordan normal form which captures the failure of the eigenvectors to form a basis.
A: Let $A$ be the  matrix of a linear transformation $\alpha: V\to V$, where $V$ is an $n$-dimensional vector space over the field $\mathbb{F}$. 
$V$ has a basis consisting of eigenvectors of $\alpha$ (or $A$ if you prefer) if, and only if, the minimal polynomial $m_\alpha(X)$ of $\alpha$ (or of $A$ if you prefer) is a product of distinct linear factors in $\mathbb{F}[X]$. 
The proof is  in any decent linear algebra textbook. 
