RE: singular matrix and eigenvectors RE:  I can't understand this matrix
$\begin{pmatrix}-1&1/3&0&0\\1&-1&2/3&0\\0&2/3&-1&1\\0&0&1/3&-1 \end{pmatrix}$
It has determinant 0, rank 3 out of a possible 4, and yet it has four linearly independent eigenvectors.
$\begin{pmatrix} 1\\3\\3\\1\end{pmatrix}$,
$\begin{pmatrix} 1\\-1\\-1\\1\end{pmatrix}$,
$\begin{pmatrix} -1\\-1\\1\\1\end{pmatrix}$,$\begin{pmatrix} -1\\3\\-3\\1\end{pmatrix}$
These are linearly independent, so they form a complete basis.  I don't understand this.  I thought that a matrix with determinant zero should not have a set of eigenvectors that make a complete basis.   Where I am going wrong?  I realize the first eigenvector has eigenvalue 0.
Thanks so much.
ps  When I put these four eigenvectors into a matrix, it has rank 4, so I know they are linearly independent.
 A: As said in the comments,
It's not a problem that these are independent. It only means that the matrix is diagonalizable, hence that $\mathbb R^4$ is decomposed into a direct sum $E_0\oplus E_\lambda\oplus E_\mu\oplus E_\kappa$, where $\lambda,\mu,\kappa$ are the non-zero eigenvalues.
Hence in the appropriate basis, your matrix is 
$$\begin{pmatrix}\kappa & 0 & 0 & 0\\0 & \mu & 0 & 0\\0 & 0 & \lambda & 0\\0 & 0 & 0 & 0\end{pmatrix}$$
which is still not invertible, but is diagonal.
A: An $n\times n$ matrix $A$ is diagonalizable if $A$ has $n$ linearly independent eigenvectors. 
An $n\times n$ matrix $A$ is invertible if there exists an $n\times n$ matrix $B$ such that $AB=I_n$. 
There are diagonalizable matrices that are not invertible. For example, the zero-matrix is diagonalizable but not invertible.
There are also invertible matrices that are not diagonalizable. For example, let
$$
A=
I_n+S_n
$$
where
$$
[S_n]_{ij}=
\begin{cases}
0 & j\neq i+1 \\
1 & j= i+1
\end{cases}
$$
Then $A$ is invertible but not diagonalizable.
The two examples above demonstrate that diagonalizability is unrelated to invertibility.
