# Common solution to two equations as a subspace

I have a question about an exercise I am working on for a linear algebra class. The exercise is as follows

Let $\vec{v}_1=(1,1,1,1)$ and $\vec{v}_2=(3,5,2,1)$, and let $V$ be the subspace of $\mathbb{R}^4$ spanned by $\vec{v}_1$ and $\vec{v}_2$. Find two equations, each of the form $ax+by+cz+dw=0$, such that the common solution to these equations is the subspace $V$.

So, if $\vec{v}_1=(1,1,1,1)$ and $\vec{v}_2=(3,5,2,1)$ span $V$, then every vector in $V$ can be expressed as a linear combination of $\vec{v}_1=(1,1,1,1)$ and $\vec{v}_2=(3,5,2,1)$. That is, for any two scalars $c$ and $d$,

$$c\begin{bmatrix} 1\\ 1\\ 1\\ 1\\ \end{bmatrix} + d\begin{bmatrix} 3\\ 5\\ 2\\ 1\\ \end{bmatrix} = \begin{bmatrix} x\\ y\\ z\\ w\\ \end{bmatrix} .$$

From here I use an augmented matrix

$$\begin{bmatrix} 1& 3|& x\\ 1& 5|& y\\ 1& 2|& z\\ 1& 1|& w \end{bmatrix}$$

which, after putting into RREF form, yields

$$\begin{bmatrix} 1& 0|& 2w-z\\ 0& 1|& z-w\\ 0& 0|& x-3w-2z\\ 0& 0|& y-5w-4z \end{bmatrix}$$

Does this give me the two equations I am asked to provide? It would be

$$x-3w-2z=0$$ $$y-5w-4z=0$$

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I'm not sure you computed the RREF correctly ($z=w=1$, $x=5$, and $y=9$ satisfy the two equations. This gives $d=0$ and $c=1$; but $cv_1+dv_2\ne[5,9,1,1]$). But your general method is sound. Redo the RREF... – David Mitra Jan 11 '13 at 1:35
You know you can't be right because $x-3w-2z=-4$ when you plug in $(x,y,z,w)=(1,1,1,1)=v_1$. – Alexander Gruber Jan 11 '13 at 1:39
I computed as an echelon form (note, you need not do the "backwards reduction") as: $$\left[\matrix{ 1&3\cr 0&2\cr 0&0\cr0&0} \ \ \ \left|\ \ \matrix{x\cr y-x\cr-3x+2z+y\cr 2x-w-y} \right.\right]$$ – David Mitra Jan 11 '13 at 1:44

You may simply find two nontrivial solutions $(a,b,c,d)$ to the equation $$\begin{pmatrix}1&1&1&1\\3&5&2&1\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\end{pmatrix}=0.$$