Find a nonzero matrix $A\in\mathbb{R}^{3\times 3}$ satisfying both conditions at the same time:
$\text{Ker}(A)=\text{span}{((1, -1, -1)^T,(1, 1, -3)^T)}$
$\text{Ker}(A^T)=\text{span}{((1, 2, 4)^T,(0, 1, 1)^T)}$
I have absolutely zero clue on how to solve this question, and it was on my final exam paper last week from linear algebra. Any help is much appreciated!
What I have tried so far:
Writing out $\text{Ker}(A)=\text{span}{((1, -1, -1)^T,(1, 1, -3)^T)}$ as
$x_1 = α + β$
$x_2 = β - α$
$x_3 = α -3β$
But I didn't know how to continue from that. I thought about making a product table, but I don't know how to implement it.
If you really haven't yet studied orthogonal complements and etc. it maybe they want you to do it the basic, pretty algebraic (and rather tiring and slightly boring), way: suppose
$$A=\begin{pmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{pmatrix}\implies\begin{pmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{pmatrix}\begin{pmatrix}\;\;1\\-1\\-1\end{pmatrix} ;\begin{pmatrix}\;\;1\\\;\;1\\-3\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix} $$
Now carry on the first multiplication, then the second and equal them (they both equal the zero vector):
$$\begin{pmatrix}a_{11}-a_{12}-a_{13}\\a_{21}-a_{22}-a_{23}\\a_{31}-a_{32}-a_{33}\end{pmatrix}=\begin{pmatrix}a_{11}+a_{12}-3a_{13}\\a_{21}+a_{22}-3a_{23}\\a_{31}+a_{32}-3a_{33}\end{pmatrix}$$
Denote by $\;R_k\;$ the $\;k\,-$the row, so we get comparing the different rows:
$$R_k: \implies a_{k2}=a_{k3}\;,\;\;k=1,2,3 \implies a_{k1}-2a_{k2}=0\implies a_{k1}=2a_{k2}$$
So substituting $\;a_{12}=x\;,\;\;a_{22}=y\;,\;\;a_{32}=z\;$ ,we get that our matrix looks like
$$A=\begin{pmatrix}2x&x&x\\ 2y&y&y\\ 2z&z&z\end{pmatrix}\;,\;\;\;x,y,z\in\Bbb R$$
Well, now you use the fact that $\;A^t\begin{pmatrix}1\\2\\4\end{pmatrix}=A^t\begin{pmatrix}0\\1\\1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\;$
and find and non-zero example of such a matrix $\;A\;$