Finding eigenvectors of a matrix I want to find all eigenvalues and eigenvectors of the matrix $\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix}$. 
Here is how I find eigenvalues:
$$\begin{align*}
 \det(A - \lambda I) &= \det \Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}\lambda&0&0\\0&\lambda&0\\0&0&\lambda \end{bmatrix} \Bigg)\\
&= \det \Bigg(\begin{bmatrix} -\lambda&1&0 \\ 0&-\lambda&1 \\ -1&0&-\lambda \end{bmatrix} \Bigg)\\
&= -\lambda^3 - 1\\
\therefore \lambda =& -1
\end{align*}$$
Using eigenvalue that I found ($-1$), I want to find eigenvectors:
$$\begin{align*}
        (A - \lambda I)\vec{V} =& 0\\
        \Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\Bigg) \begin{bmatrix}x\\y\\z \end{bmatrix} =& \begin{bmatrix}0\\0\\0\end{bmatrix}\\
        \begin{bmatrix}1&1&0\\0&1&1\\-1&0&1\end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} = & \begin{bmatrix}0\\0\\0\end{bmatrix}\\
        \begin{bmatrix} x+y \\ y+z \\ -x+z \end{bmatrix} = & \begin{bmatrix}0\\0\\0\end{bmatrix}\\
    \end{align*}$$
But what I should do from now? What is really the eigenvectors? Does this means that I have unlimited eigenvectors and any number that satisfies three equations can be eigenvectors?
 A: Since your characteristic equation is:
$$
\lambda^3 = -1 \rightarrow \lambda = e^{\pi i + \frac{2n\pi}{3}i}
$$
and gives three distinct eigenvalues, there are exactly three eigenvectors only one of which has eigenvalue $\lambda = -1$.
$$
\begin{bmatrix} 1 & 1 & 0 \\
0 &1& 1 \\
-1 &0 & 1
\end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 0 \\
0 &1& 1 \\
0 &1 & 1
\end{bmatrix}
$$
Now the last two are degenerate (as we would expect) which gives:
$$
y = -z \\
x = -y = z \\
(z, -z, z) \rightarrow (1, -1, 1)
$$
So $\left\langle1, -1, 1\right\rangle$ or $\left\langle \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$ is the only eigenvector for $\lambda = -1$.
By only eigenvector, I mean that all eigenvectors for $\lambda = -1$ will be scalar multiples of the above.
A: Well, the polynomial $\lambda^3+1$ has two more (complex) roots, which means a rotation in a $2$ dimensional subspace.
In your last equation substitute $\lambda=-1$ and, say, $x=1$ to find one eigenvector. 
(You are right: there are infinitely many eigenvectors if there is one as they always form a subspace.)
A: For any scalar $k$, if $v$ is an eigenvector for the eigenvalue $\lambda$ ($Av=\lambda v$) then so is $kv$ ($A(kv)=kAv=k\lambda v=\lambda(kv)$).  The last line of your set of matrix equations is the homogeneous system $x+y=0$, $y+z=0$, $-x+z=0$.  This system is not linearly independent, but that's OK.  Doing a bit of solving gives you $x=z$ and $y=-z$, so $(x,y,z)=(z,-z,z)=z(1,-1,1)$.  Yes, you get infinitely many eigenvectors, but they are all scalar multiples of eigenvector $(1,-1,1)$.  This is normal -- you actually get a whole subspace of eigenvectors for any eigenvalue, with the dimension of the subspace corresponding to the multiplicity of the root in the characteristic polynomial.  The vector $(1,-1,1)$ provides a basis for the eigenspace associated with your eigenvalue $\lambda=1$.  (You should also note that, in your example, you also get two complex roots -- if your base field is $\mathbb C$ then you get eigenspaces for those values, too.) [Edit: fixed a typo]
A: We could put the solution in a cleaner form:
Let $\lambda =-1$ be our eigenvalue. Then we have
$$
\begin{align}
M&=A-\lambda I\\&= \begin{bmatrix}1&1&0\\0&1&1\\-1&0&1\end{bmatrix}
\end{align}
$$
Let the corresponding eigenvector be $\overrightarrow{v}$:
$$
\overrightarrow{v}=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}
$$
such that $M\overrightarrow{v}=\overrightarrow{0},\ \overrightarrow{v}\ne0$. Then we have
$$
v_1+v_2=0\\ v_2+v_3=0\\-v_1+v_3=0
$$
Solving the system of linear equations, we have:
$$
ker(M)=\left\{\;t\begin{pmatrix}1\\-1\\1\end{pmatrix},\; t\in\mathbb{R}\backslash \left\{0\right\} \right\}
$$
Hence the corresponding eigenvector is 
$$
\overrightarrow{v}=\begin{bmatrix}1\\-1\\1\end{bmatrix}
$$
