Finding Eigenvalues and Eigenvectors weird equations I have matrix:
$A = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$
And I have to find eigenvalues and coresponding eigenvectors.
I get $det(A-xI_3) = -x(x-1)^2$, so eigenvalues are $x_1 = 0$ and $x_2 = 1$
Now I want to find coresponding eigenvectors. I get:
$A - 0I_3 = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right] $ 
and
$A  - I_3= \left[\begin{array}{ccc}2&-1&-2\\2&-1&-2\\2&-1&-2\end{array}\right]$
To first equation I have solution $y = x, z = x$, but I dont know what eigenvectors are and to the second I dont even know the solution to equation $(A-I_3)[x,y,z]^T$. Could you help me?
 A: Since, from your previous problem, we know that $A^2=A$, that gives us that each column vector of $A$ is an eigenvector of $A$ with eigenvalue $1$. (Note the column matrices are not linearly independent.)
An eigenvector with eigenvalue $0$ can be found via inspection: $\begin{bmatrix}1\\1\\1\end{bmatrix}$
A: Since you have the eigenvalues, you can consider finding a basis of the eigenvectors associated to each eigenvalues, which means resolving (with Gauss method) this equation (for the 0 eigenvalue) :
$X=
\begin{pmatrix}x\\y\\z\end{pmatrix} \in E_0 \Leftrightarrow
\begin{cases}
3x-y-2z&=0 \\ 
2x-2z&=0 \\ 
2x-y-z&=0
\end{cases}\Leftrightarrow AX=0.X$
$\begin{cases}
3x-y-2z&=0 \\ 
2x-2z&=0 \\ 
2x-y-z&=0
\end{cases}
\Leftrightarrow
\begin{cases}
2x-y-z&=0 \\ 
-y+z&=0 \\ 
0&=0
\end{cases}
\Leftrightarrow
\begin{pmatrix}x\\y\\z\end{pmatrix}\in \{\begin{pmatrix}s\\s\\s\end{pmatrix},s\in\mathbb{K}\}$
Finally :
$E_0 = Vect(\begin{pmatrix}1\\1\\1\end{pmatrix})$
The case of the eigenvalue 1 starts exactly the same way  :
$X=
\begin{pmatrix}x\\y\\z\end{pmatrix} \in E_1 \Leftrightarrow
\begin{cases}
3x-y-2z&=x \\ 
2x-2z&=y \\ 
2x-y-z&=z
\end{cases}\Leftrightarrow AX=1.X$ and you find the above eigenvectors ;-).
So, you have to find a basis of the plane : $P: 2x-y-2z=0$.
One way is to parameter the problem :
$\begin{pmatrix}x\\y\\z\end{pmatrix}\in P \Leftrightarrow \begin{pmatrix}x\\y\\z\end{pmatrix}\in \lbrace\begin{pmatrix}\frac{1}{2}s+t\\s\\t\end{pmatrix},(s,t)\in\mathbb{K}\rbrace$
So you find immediately that $E_1 = Vect(\begin{pmatrix}\frac{1}{2}\\1\\0\end{pmatrix},\begin{pmatrix}1\\0\\1\end{pmatrix})$
This is the classic method of resolution.
