Subspace described by solution set of homogenous linear system

Question

Let $U\subset\mathbb{Q}^4$ be the subspace induced by the following vectors: $$u_1=\begin{pmatrix} 2 \\ 2 \\ 6 \\ 0 \end{pmatrix}, \;u_2=\begin{pmatrix} 2 \\ -1 \\ 1 \\ 1 \end{pmatrix}, \;u_3=\begin{pmatrix} -1 \\ 0 \\ 2 \\ -1 \end{pmatrix}$$ Compute the dimension of $U$ and find a homogeneous system of linear equations over $\mathbb{Q}$ with as few equations as possible, such that its solution set is $U$.

I've used Gaussian elimination and found out that $U$ has dimension $3$, hence $u_1,u_2,u_3$ form a basis of $U$.

My problem: I haven't been able to find the according linear system and I have no idea how I should proceed with this.

Answer 1

Since $\dim U=3$ and $\dim\mathbb Q^4=4$, $U$ is a hyperplane, and therefore you can solve your problem with a system which consists of a single equation. One such equation is $-3x+z+5t=0$. Actually, any equation of the type $ax+by+cz+dt=0$ will do as long as$$\left\{\begin{array}{l}2a+2b+6c=0\\2a-b+c+d=0\\-a+2c-d=0\end{array}\right.$$and $(a,b,c,d)\neq(0,0,0,0)$ (that's how I got my solution).

Answer 2

Solve $2a+2b+6c=0\tag1$

$2a-b+c+d=0\tag2$

$-a+2c-d=0.\tag3$.

From (1), $a+b+3c=0$, and from (2)+(3), $a-b+3c=0$; so $b=0$.

Thus $a=-3c$ and $d=5c$.

So $\vec v\cdot \vec u=0,$ where $\vec v$ is any multiple of $(-3,0,1,5)$.