Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
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 $$ \begin{pmatrix}1&1&1&1\\3&5&2&1\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\end{pmatrix}=0. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.