find vector $v$ , $v \in R^4$ that satisfy $Av\neq0$ and $A^2v=0$

Question

$A=\begin{bmatrix}1&0&2&1\\0&-1&1&-2\\2&1&1&0\\1&1&0&1\end{bmatrix}$ 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?

Answer 1

HINT

Take a vector in $\ker(A^2)$ in the general form $v=sv_1+tv_2$ then solve for $Av\neq 0 $ to find coefficients $s$ and $t$.

Answer 2

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 $\begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix} \notin \ker A$.