Find one matrix $A\in M_{3,4}(\mathbb{R})$ such that $Ker(A)=L(\begin{bmatrix}
-3\\
1\\
0\\
0
\end{bmatrix},\begin{bmatrix}
-2\\
0\\
-6\\
0
\end{bmatrix})$ and then describe all matrices with this properity.
What I tried to do so far is the following:
Let $Ker(A)=V$. Then $V^{\perp }$ can be determined with the system:$$\begin{matrix}
-3x_{1} & + x_{2} & =0\\
-2x_{1}& – 6x_{3} & =0
\end{matrix}$$
Now, since $V=(V^{\perp })^{\perp }$, I find that the system that has $V$ as its set of solution is the system:
$$\begin{matrix}
3x_{1}-9x_{2}-x_{3}=0 & & \\
x_{4}=0 & &
\end{matrix}$$, so one matrix is:$A=\begin{bmatrix}
3 & -9 & -1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 &0
\end{bmatrix}$.
My question is how can I describe all matrices that have $V$ as its kernel.
Best Answer
The rows of your matrix should span the complement $V^\perp$. So if you take a basis for $V^\perp$, like $\{(3,-9,-1,0),(0,0,0,1)\}$, than every row vector should be a linear combination of these two. Hence every row is of the form $(3a,-9a,-a,b)$, with $a,b \in \mathbb{R}$.