Find a matrix $A$ such that $X$ generates the subspace $W$ (the solution space of the system $AX=0$.)

Question

Consider $W$ a subspace of $\mathbb{R}^{5}$ generated by: \begin{align*} X=\left \{(1,-1,0,5,1), (1,0,1,0,-2),(-2,0,-1,0,-1)\right \} \end{align*} Find a system of linear equations $AX=0$ such that W be the solution space of the system.

I understand that what I've to do is to find a matrix $A_{5x5}$ such that:

\begin{align*} A \begin{pmatrix} 1\\ -1\\ 0\\ 5\\ 1 \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 0 \end{pmatrix} \ \text{ , } \ A \begin{pmatrix} 1\\ 0\\ 1\\ 0\\ -2 \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 0 \end{pmatrix} \ \text{ and, } \ A \begin{pmatrix} -2\\ 0\\ -1\\ 0\\ -1 \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 0 \end{pmatrix} \end{align*}

So I think that the solution can be to propose a new system of linear equations but I'm not sure how can I do it. How can I find the matrix $A$?

Answer 1

Since $W$ is $3$-dimensional, you need the $2$-dimensional orthogonal space, putting the generating vectors as rows of $A$.

$$0=\begin{pmatrix}1&-1&0&5&1\\1&0&1&0&-2\\-2&0&-1&0&-1\end{pmatrix}\mathbf{x}=\begin{pmatrix}1&0&0&0&3\\0&1&0&-5&2\\0&0&1&0&-5\end{pmatrix}\mathbf{x}$$ so $\mathbf{x}=(-3t,5s-2t,5t,s,t)=t(-3,-2,5,0,1)+s(0,5,0,1,0)$.

Thus $$A=\begin{pmatrix}-3&-2&5&0&1\\0&5&0&1&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{pmatrix}$$ is one possible matrix. (Any equivalent matrix will do.)

Answer 2

Hint: The kernel is the orthogonal complement of the row space. So $A$ needs row space equal to $W^\perp$.