Eigen value and eigen vectors I have given a matrix $$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 3 \end{pmatrix}. $$ 
We have to find 3 linearly independent eigen vector. I have calculated the eigen values that is 1 with multiplicity 3 . I have found the eigen  vector $ \begin{pmatrix}1 & 1 & 1 \end{pmatrix}^T$. But I can't find the other two linearly independent eigen vectors. 
Any help is appreaciating.
 A: The characteristic polynomial is
\begin{align}
\det\begin{pmatrix}
0-X & 1 & 0 \\
0 & 0-X & 1 \\
1 & -3 & 3-X
\end{pmatrix}
&=
-X\det\begin{pmatrix}-X & 1 \\ -3 & 3-X\end{pmatrix}
-\det\begin{pmatrix}0 & 1 \\ 1 & 3-X\end{pmatrix}
\\
&=-X(-3X+X^2+3)+1
\\[6px]
&=-X^3+3X^2-3X+1=(1-X)^3
\end{align}
There cannot exist three linearly independent vectors: the matrix would be diagonalizable and similar to
$$
\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}
$$
but the identity matrix is only similar to itself.
Actually, if we compute the rank of $A-I$ (where $A$ is the given matrix), we get, with Gaussian elimination,
$$
A-I=\begin{pmatrix}
-1 & 1 & 0 \\
0 & -1 & 1 \\
1 & -3 & 2
\end{pmatrix}
\to
\begin{pmatrix}
-1 & 1 & 0 \\
0 & -1 & 1 \\
0 & -2 & 2
\end{pmatrix}
\to
\begin{pmatrix}
-1 & 1 & 0 \\
0 & -1 & 1 \\
0 & 0 & 0
\end{pmatrix}
$$
so the rank is $2$ and the eigenspace relative to $1$ has dimension $3-2=1$. No set of two eigenvalues can be linearly independent.
A: Furthermore the generalized eigenspace can be found through:
$$(A-I_3)^{-1}[1,1,1]^T = [-1,0,1]^T$$
Which is then the vector which has image split onto self and first eigenvector.
Iterating:
$$(A-I_3)^{-1}([-1,0,1]^T+[1,1,1]^T) = [-1,-1,2]^T$$
Now we have our $\bf T$, or as in Jordan normal form often called $\bf S$:
$${\bf A = SJS}^{-1}$$
$${\bf J} = \left[\begin{array}{ccc}\color{red} 1&\color{blue}1&0\\0&\color{red} 1&\color{blue}1\\0&0&\color{red} 1\end{array}\right]\hspace{1cm}{\bf S} = \left[\begin{array}{ccc}
1&0&- \frac 1 3\\
1&1&-\frac 1 3\\
1&2&\frac{2}3
\end{array}\right]$$
The red values in our Jordan block is the eigenvalue and the blue numbers are the off diagonal ones (those will always be one and have nothing to do with the eigenvalue).
EDIT as amd points ut the equation solved above $(A-I_3)^{-1}v$ actually is a pseudo inverse or least squares solution since $A-I_3$ is singular.
