Finding a base of $S^{\bot}$ is finding a solution for $Ax=0$ but for which matrix?

Question

Let $S$ is subspace of $\mathbb R^4$ and $L((1,2,2,3),(1,3,3,2))=S$

a) Find one base of subspace $S^{\bot}$

b) Finding that base is the same finding solution for $Ax=0$ but for which matrix?

a) It is easy I find a base $L((-5,1,0,1),(0,-1,1,0))$

b) For matrix which $\ker(A)=L((-5,1,0,1),(0,-1,1,0))$ and they are orthogonal to every member of row space of matrix $A$. Is this the full answer?

Answer 1

We have $$\begin{bmatrix} 1 & 2 & 2 & 3 \\ 1 & 3 & 3 & 2\end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 2 & 3 \\ 0 & 1 & 1 & -1\end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 5 \\ 0 & 1 & 1 & -1\end{bmatrix}$$

so the basis for $S^\perp$ is $\{(0,-1,1,0), (-5,1,0,1)\}$, as you said.

To find $S^\perp$ we can solve $Ax = 0$ if and only if the matrix $A$ satisfies $\ker A = S^\perp$, or equivalently $R(A^T) = S$. Therefore the rows of $A$ must span $S$.