How to determine the eigenvectors of this matrix? I have some problems to determine the eigenvectors of a given matrix:
The matrix is:
$$
A = \left( \begin{array}{ccc}
1 & 0 &0 \\
0 & 1 & 1 \\
0 & 0 & 2
\end{array} \right)
$$
I calculated the eigenvalues first and got $$ \lambda_1  = 1, \lambda_2 = 2, \lambda_3 = 1$$
There was no problem for me so far. But I do not know how to determine the eigenvectors. The formula I have to use is
$$ (A-\lambda_i E)u=0, \lambda_i = \{1,2,3\}, u\ is\ eigenvector$$
When I determined the eigenvector with $ \lambda_2=2$ there was not a problem. I got the result that $x_3 = variable$ and $x_2 = x_3$, so:
$$ 
EV_2= \left( \begin{array}{ccc}
0 \\
\beta \\
\beta
\end{array} \right) \ \beta\ is\ variable,\ so\ EV = span\{\left( \begin{array}{ccc}
0 \\
1 \\
1
\end{array} \right)\}
$$
But when I used $ \lambda_1 = \lambda_3 = 1 $, I  had to calculate:
$$
\left( \begin{array}{ccc}
0 & 0 &0 \\
0 & 0 & 1 \\
0 & 0 & 1
\end{array} \right) * 
\left( \begin{array}{ccc}
x_1 \\
x_2 \\
x_3
\end{array} \right)
=0
$$
what in my opinion means that $x_3 = 0 $ and $x_1$ and $x_2$ are variable, but not necessarily the same as in the case above, so $ EV_{1,3} = \left( \begin{array}{ccc}
\alpha \\
\beta \\
0
\end{array} \right) $
What does that mean for my solution? is it
$$
EV_{1,3} = span\{\left( \begin{array}{ccc}
1 \\
0 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
0 \\
1 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
1 \\
1 \\
0
\end{array} \right)\}
$$
What exactly is now my solution in this case for the eigenvectors $ \lambda_1, \lambda_3 $? In university we just had one variable value in the matrix so I don't know how to handle two of them being different.
 A: Every linear combination of $EV_{1}=\pmatrix{1\\0\\0}$ and $EV_3=\pmatrix{0\\1\\0}$ is a eigenvector with eigenvalue $1$.
$EV_{1,3} = span\{\left( \begin{array}{ccc}
1 \\
0 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
0 \\
1 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
1 \\
1 \\
0
\end{array} \right)\}$
is the same as $EV_{1,3} = span\{\left( \begin{array}{ccc}
1 \\
0 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
0 \\
1 \\
0
\end{array} \right)\}$.
A: Update: I have undeleted my answer because I think it is fixed now. 
You got $$ V_{\lambda_2} = \left(\begin{array}{ccc} 0 \\ 1 \\ 1 \end{array} \right) $$
correct but then copied it down wrongly.(I think..)
Then you correctly wrote down the case $\lambda_1$. From 
$$ \left(\begin{array}{ccc } 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{array} \right) $$ 
you should easily conclude (I think you did) that $ z = 0 $ ,  x and y can be anything leading to 
$$ V_{\lambda_{1 \ or \ 3}} = \left(\begin{array}{ccc } 1 \\ 1 \\ 0 \end{array} \right) $$
Now , since the dimension of the nullspace is 2 we can decompose this into 2 seperate eigenvectors corresponding to the repeated eigenvalue of 1
$$ V_{ \lambda_1} = \left(\begin{array}{ccc } 1 \\ 0 \\ 0 \end{array} \right) \ \ , \ \  V_{ \lambda_2 } = \left(\begin{array}{ccc } 0 \\ 1 \\ 1 \end{array} \right) \ \ , \ \  V_{ \lambda_3} = \left(\begin{array}{ccc } 0 \\ 1 \\ 0 \end{array} \right) $$ 
You can check all three are independent and satisfy
$$AV_i = \lambda_iV_i$$ 
