<s>[INST] find vector $v$ , $v \in R^4$ that satisfy $Av\neq0$ and $A^2v=0$ $A=\begin{bmatrix}1&0&2&1\\0&-1&1&-2\\2&1&1&0\\1&1&0&1\end{bmatrix}$
here after row reduction
$\begin{bmatrix}1&0&2&1\\0&1&-1&2\\0&0&1&2\\0&0&0&0\end{bmatrix}$
clearly determinant is zero
but how can I find vector $v$ , $v \in R^4$ that satisfy $Av\neq0$ and $A^2v=0$ 
$A^2=\begin{bmatrix}6&3&4&2\\0&0&0&0\\4&0&6&0\\2&0&3&0\end{bmatrix}$
so I try to row reduce $A^2$ using gauss elimination
$A^2=\begin{bmatrix}1&0&\frac{3}{2}&0\\0&1&\frac{-5}{3}&\frac{2}{3}\\0&0&0&0\\0&0&0&0\end{bmatrix}$ 
so $x_3$ and $x_4$ is free variable, but in this question I'm not sure what I need to find and should I find Eigen value first?
 [/INST] No need for row reduction.
From $$A^2=\begin{pmatrix}6&3&4&2\\0&0&0&0\\4&0&6&0\\2&0&3&0\end{pmatrix}$$ you can immediately see that $A^2 e_2 = 3e_1$ and $A^2e_4 = 2e_1$.
Therefore $$A^2e_2 = \frac32 A^2 e_4 \implies 0 = A^2\left(e_2 - \frac32 e_4\right) = A^2 \begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix}$$
so $\begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix} \in \ker A^2$. On the other hand, we have that $$A \begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix} =\begin{pmatrix}1&0&2&1\\0&-1&1&-2\\2&1&1&0\\1&1&0&1\end{pmatrix}\begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix} = \begin{pmatrix}-\frac32 \\ 2 \\ 1 \\ -\frac12\end{pmatrix} \ne 0$$
so $\begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix} \notin \ker A$.
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