# 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?

• You need to find 1 vector or the general solution? – user Feb 27 '18 at 10:14
• @gimusi 1 vector that satisfy $Av\neq0$ and $A^2v=0$, should I find Eigen value first(?) – fiksx Feb 27 '18 at 10:17

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 $t$.

• $\begin{bmatrix} 3\\ 0\\-2\\0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -2\\0\\3 \end{bmatrix}$ is this right?? , but why I should find vector in $ker(A^2)$? is it mean I find span in column space of $A^2$? – fiksx Feb 27 '18 at 10:49
• take a linear combination of them $$v=t\begin{bmatrix} 3\\ 0\\-2\\0 \end{bmatrix}+s\begin{bmatrix} 0 \\ -2\\0\\3 \end{bmatrix}$$ and take $Av$ – user Feb 27 '18 at 11:03
• You should find 4 equations in t and s, for example t=s=1 should work and maybe it works for all t,s $\neq$ 0 – user Feb 27 '18 at 11:14
• okay so here I found null space of reduce row echelon form of $A^2$, I'm not sure what this want $Av\neq0$ , is it mean I need to find vector in column space of A? , for $Av$ here I just need to multiply A and v? so I multiplied it and got $s=\frac{2}{3}t$ ,$-2s=t$ ,$t=\frac{1}{2}s$ , $t=-\frac{1}{3}s$ (?) – fiksx Feb 27 '18 at 12:05
• all those are not equality it must be $\neq$ in each one, thus it suffices choose s and t to fulfil $Av\neq0$ – user Feb 27 '18 at 12:07

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

• thanks but how can you immediately see that $A^2 e_2 = 3e_1$ and $A^2e_4 = 2e_1$? Eigen value are 2,3 though – fiksx Feb 27 '18 at 14:34
• @fiksx The $j$-th column of any matrix $T$ is $Te_j$. Therefore the second column of $A^2$ is precisely $A^2e_2$ and the fourth column is $A^2e_4$. – mechanodroid Feb 27 '18 at 15:06
• is this because you chose column that has lots of zero? is it possible that I choose $A^2e_1$? how do you know if you choose second and fourth column you will get scalar multiple of $e_1$? is this related to Eigen value too? – fiksx Feb 28 '18 at 4:51
• @fiksx Nothing to do with eigenvalues, I just took two columns which both have only the first coordinate $\ne 0$. This means that they are both scalar multiples of $e_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\end{pmatrix}$, so in particular they are scalar multiples of each other. – mechanodroid Feb 28 '18 at 10:23
• $A^2 \begin{pmatrix}3 \\ 0\\ 0 \\ 0\end{pmatrix} = 3 \begin{pmatrix}6\\ 0 \\ 4 \\ 2\end{pmatrix}$ or $e_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\end{pmatrix}$ ? I was so confuse about the notation $e_3$ and $e_1$ here – fiksx Mar 2 '18 at 5:18