How to find a eigenvector with a repeated eigenvalue? The eigenvalues of my matrix are $x_1= 1$ 
and 
$x_2=3$
I get an eigenvector
$V = t~[ 4~~~~~~ 3 ~~~~~1 ]^T $ but how can I diagonalize the matrix if I have the same column repeated twice. Should I just use different values for t since they are all eigenvectors of the same matrix?
matrix
$\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
8 & 6 &3 \\
\end{bmatrix}$
Edit: I realized I had the eigenvalues wrong
, now fixed

I will assume that you are capable of finding the eigenvector for the simple root, and only focus on the repeated eigenvalue $x_1=1$. Let $\mathbf{v}=\begin{bmatrix}
a \\
b \\
c \\ 
\end{bmatrix}$ and consider $(A-I)\mathbf{v}=\mathbf{0}$.
This yields $\begin{bmatrix}
0 \\
0 \\
8a+6b+2c \\
\end{bmatrix} = \begin{bmatrix}
0 \\
0 \\
0 \\ 
\end{bmatrix}$
in which case $8a+6b+2c=0$. Let $b$ and $c$ be free variables, giving us $\mathbf{v}=\begin{bmatrix}
-\frac{3}{4}b-\frac{1}{4}c \\
b \\
c \\ 
\end{bmatrix}=b\begin{bmatrix}
-\frac{3}{4} \\
1 \\
0 \\ 
\end{bmatrix}+c\begin{bmatrix}
-\frac{1}{4} \\
0 \\
1 \\ 
\end{bmatrix}$.  I hope this helps. 
