Jordan form of 3 x 3 repeated eigenvalue Consider the matrix
\begin{pmatrix}
        1 & -3   &1 \\
        1 & 5 &  -1 \\
         \
        2 & 6 &0 \\
        \end{pmatrix}
This has eigenvalue 2 with 3 multiplicity.
However, obtained two more eigenvector
(-3 1 0) and (1 0 1). How can I found the third one to form with the Jordan form with
\begin{pmatrix}
        2 & 0   &0 \\
        0 & 2 &  1 \\
         \
        0 & 0 &2 \\
        \end{pmatrix}
https://www.mathstools.com/section/main/jordan_form_3x3_triple_root#.YI_pgS_R1N0
here is doing the same thing but I dont understand..
 A: That matrix only has 2 independent eigenvectors, not 3.
In general, there is always exactly 1 eigenvector for each Jordan-Block, no matter how big the Jordan block is.
In your case , the eigenvalue $2$ has algebraic multiplicty = 3 (multiplicity of the root of the characteristic polynomial), but geometric multiplicity =2 (dimension of eigen-space).
You already found the eigenvectors $v_1,v_2$. It remains to find a suitable $v_3$ such that
\begin{align}
P&=\begin{pmatrix}v_1&v_2&v_3\end{pmatrix} \\
A&= PJP^{-1} \\
\Leftrightarrow AP&=PJ
\end{align}
with your Jordan-matrix $J$. From the last equation you only need the third column:
\begin{align}
A v_3 &= \begin{pmatrix}v_1 &v_2 &v_3\end{pmatrix}\begin{pmatrix}0\\1\\2\end{pmatrix} \\
&=v_2+2v_3 \\
\Rightarrow (A-2)v_3 &= v_2
\end{align}
This is a linear equation you should be able to solve for $v_3$.
Such a recursion relation like $(A-2)v_3 = v_2$ always holds if you need additional basis-vectors for a jordan-block for which you already know the eigenvector.
A: Take the last vector $v_3$ to be any vector that is not an eigenvector for $\lambda=2$, then take $v_2=(A-2I)v_3$ (which must be an eigenvector since $\operatorname{im}(A-2I)\subset\ker(A-2I)$), and finally take $v_1$ to be an eigenvalue linearly independent of $v_2$. For instance take $v_3=(1,0,0)$, $v_2=(-1,1,2)$, and $v_1=(1,0,1)$. This works because $Av_1=2v_1$, $Av_2=2v_2$, and $Av_3=v_2+2v_3$, by construction.
