Find a $D$ matrix such the linear subspace $S$ is equal to $\ker(D)$

Question

Find a $D$ matrix such the linear subspace S is equal to $\ker(D)$

$$S = ((1,2,0,0,1,2),(1,1,-1,3,0,1),(2,3,1,2,1,3))$$

I understand that A is the solution of homogeneous sistem $Dx = 0$, i had a idea of put 36 variables in the matrix and get equations from $Dx = 0$, but now i think that is non-viable, some idea? , thanks for reading :))

Answer 1

Consider those vectors $(a,b,c,d,e,f)\in\Bbb R^3$ such that:

• $(a,b,c,d,e,f).(1,2,0,0,1,2)=0$;
• $(a,b,c,d,e,f).(1,1,−1,3,0,1)=0$;
• $(a,b,c,d,e,f).(2,3,1,2,1,3)=0$;.

Finding such vectors is equivalent to solving the system$$\left\{\begin{array}{l}a+2b+e+2f=0\\a+b-c+3d+f=0\\2a+3b+c+2d+e+3f=0.\end{array}\right.$$The solutions are the vectors of the form$$\left(-5d+e,\frac52d-e-f,\frac12d,d,e,f\right).$$So take, for instance, the matrix $D$ such that its first line is the solution that you get if $(d,e,f)=(1,0,0)$, that its second line is the solution that you get if $(d,e,f)=(0,1,0)$, and that its third line is the solution that you get if $(d,e,f)=(0,0,1)$, in other words,$$D=\begin{bmatrix}-5 & \frac{5}{2} & \frac{1}{2} & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 0 & 1\end{bmatrix}.$$