Set of linear equations Find eigenvalues and eigenvectors of the matrix:
$\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix}$
$\begin{pmatrix} 1-\lambda & 0 & -2 \\ 1 & 3-\lambda & -1 \\ -1 & 0 & 2-\lambda \end{pmatrix}$
I have found the eigenvalues, which are: 0 and 3.
For the 0-eig I have found their corresponding eigenvector (-6,1,-3)
and now I try to find the eigenvector for eig = 3:
$\begin{pmatrix} -2 & 0 & -2 \\ 1 & 0 & -1 \\ -1 & 0 & -1 \end{pmatrix}$~$\begin{pmatrix} 1 & 0 & -1 \\ 0 & 0 & 1 \end{pmatrix}$
and I am then supposed to find this solution: (0,1,0)
First, I don't understand how to get from the first matrix to the second matrix.
Secondly, how to find the solution. 
May you help me please.
Inverse matrix:
First step:
$$
\left[
      \begin{array}{ccc|ccc}
        1&0&0&1&0&0\\
        -1&2&1&0&1&0\\
        0&0&1&0&0&1\\
      \end{array}
    \right]
$$
Second step:
$$\left[
      \begin{array}{ccc|ccc}
        1&0&0&1&0&0\\
        0&2&1&1&1&0\\
        0&0&1&0&0&1\\
      \end{array}
    \right]
$$
Last step:
$$\left[
      \begin{array}{ccc|ccc}
        1&0&0&1&0&0\\
        0&1&0&1/2&1/2&-1/2\\
        0&0&1&0&0&1\\
      \end{array}
    \right]
$$
 A: I would gladly write a comment but I have no enough rep. yet.
It seems that your matrix (3x3) has only 2 eigenvalues which means that you can not find the third eigenvector. You have to find a  generalized eigenvector, which seems to be your solution.
Regarding your second matrix, I believe that eig = 0 will give you (1 0 -1)' (its corresponding eigenvector transposed), and eig = 3 will give you (0 0 1)'
Again, apologize for using the answer window! 
A: First thing first. To find the eignvalues and the eigenvectors of a matrix you need to compute characteristic polynomial, which is given by the evaluation of
$$\mathcal{X}_A(\lambda) =det(\lambda - A) = det (A - \lambda) = 0$$
where  $$ A=\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix}$$
Once you have done this you should get $\mathcal{X}(\lambda) = \lambda(\lambda -3)^2$. From this equation you can see that you have three eigenvalues: $\lambda_1 = 0, \lambda_{2,3} = 3$. Now you are interested in finding the eigenvectors corresponding to these eigenvalues and you can do that by evaluating the eigenspaces of the eigenvalues:
$$E(A,\lambda_1):=Kern(A-\lambda_1) = Kern (A)$$
To find out $Kern(A)$ you need to find the vector $v_1$ such that $A \cdot v_1 = 0$ you can do that by solving the system
$$\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix} \cdot v_1= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$$ You should be able to solve this system of linear equation with the help of the gaussian algorithm. In our case we subtract the first line to the second line of our matrix and this is what we get:
$$\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix} \cdot v_1= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix} \overset{II - I}{\leadsto} \begin{pmatrix} 1 & 0 & -2 \\ 0 & 3 & 1 \\ -1 & 0 & 2 \end{pmatrix} \cdot v_1= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$$
The next step of the guassian algorithm is for example this one: you add the third line to the first one and you see that the third line cancels out:
$$\begin{pmatrix} 1 & 0 & -2 \\ 0 & 3 & 1 \\ -1 & 0 & 2 \end{pmatrix} \cdot v_1= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix} \overset{III + I}{\leadsto} \begin{pmatrix} 1 & 0 & -2 \\ 0 & 3 & 1 \\ 0 & 0 & 0 \end{pmatrix} \cdot v_1= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$$
What you can see is that the third component of the solution vector $v_1$ is a parameter, we set it to be $k$. In fact you have infinitely many solution to this system of equation, and when you evaluate the eigenspaces of a matrix you will always have infinitely many solutions! Those solutions are in fact the span of the eigenvectors. However if you set the third component of $v_1$ to be $k$ you will see that 
$$v_1 = \begin{pmatrix} 2k \\ \frac{k}{3} \\ k \end{pmatrix}= k \cdot \begin{pmatrix} 2 \\ \frac{-1}{3} \\ 1 \end{pmatrix}$$
Where $k \in \mathbb{R}\backslash \{0\}$ (you find your solution of $v_1$ by taking $k=3$). Hence $E(A,\lambda_1) = span \{ v_1 \}$
To find the other eigenvectors you do the same thing. You evaluate solution to the eigenspace
$$E(A, \lambda_{2,3}) = Kern (A-3\mathbb{1}) = Kern\left(\begin{pmatrix} -2 & 0 & -2 \\ 1 & 0 & -1 \\ -1 & 0 & -1 \end{pmatrix}\right)$$
I leave the calculation to you and the solution should be $v_2 = \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}$, which is exactly the solution you are after. I hope that now that i've shown you how to find such solution you are also able to find it, if not just write it in the comment section below and i'll help you.
There are a few remarks that you should keep in consideration. As you can see the dimension of you matrix is $3$, but we have only $2$ eigenvectors. This is due to the fact that your matrix is not diagonalizable. In fact a matrix is diagonalizable if and only if the sum of the dimension of the eigenspaces is equal to the dimension of the matrix. Probably you will see very soon the famous Jordan Normal Form of a matrix and once you have seen it i recommend you to redo this exercise and try to understand why the matrix $A$ has only two eigenvectors and where is the third eigenvector disappeared. I hope that you understand my calculations, if not please ask me :-)
