<s>[INST] 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?
 [/INST] 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 
$$
\overrightarrow{v}=\begin{bmatrix}1\\-1\\1\end{bmatrix}
$$
</s>