write a system of linear equations that has $U$ (subspace) as the set of solutions

Question

given $U$ a subspace generated by the vectors $v_1=(0,1,2,3),v_2=(1,0,-3,1),v_3=(-3,1,11,0)$

1) find a basis of $U$

$$\begin{bmatrix}0&1&2&3 \\ 1&0&-3&1 \\ -3&1&11&0 \\ \end{bmatrix} \Rightarrow \begin{bmatrix}1&0&-3&1 \\ 0&1&2&3 \\ 0&1&2&3 \\ \end{bmatrix} \Rightarrow \begin{bmatrix}1&0&-3&1 \\ 0 &1&2&3\\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

So two vectors are linear independent.

a base of $U$ is $\{(0,1,2,3),(1,0,-3,1)\}$

2) write a system of linear equations that has $U$ as the set of solutions.

The problem is this question,I don't know where to start. What I can say is that this system of equation cannot be "full rank" because if the matrix associated with the system has max rank that means one unique solution, which is not what is asked. Somehow I need to find infinite solutions such that $v_1,v_2,v_3$ are included.

Answer 1

Consider the system$$\left\{\begin{array}{l}y+2z+3t=0\\x-3z+t=0\end{array}\right.$$Now, get two linearly independent solutions of the system. Say, for instance, that you take $(3,-2,1,0)$ and $(-1,-3,0,1)$. Now, take the system$$\left\{\begin{array}{l}3x-2y+z=0\\-x-3y+t=0\end{array}\right.$$(the coefficients are the coordinates of the solutions). And… it's done!

Answer 2

You can write $$U=\{a(0,1,2,3)+b(1,0,-3,1)\; |\; a,b\in \Bbb K\}=\{(b,a,2a-3b,3a+b)\; |\; a,b\in\Bbb K\}=\{(x,y,z,t)\in\Bbb K^4\; |\; \begin{cases} z=2y-3x\\ t=3y+x\end{cases}\}= \{(x,y,z,t)\in\Bbb K^4\; |\; \begin{cases} 2y-3x-z=0\\ 3y+x-t=0\end{cases}\}$$

Answer 3

Hint:

Write that the matrix $$\begin{bmatrix}0&1&2&3 \\ 1&0&-3&1 \\ -3&1&11&0 \\ x&y&z&t \\ \end{bmatrix}\quad \text{or }\begin{bmatrix}1&0&-3&1 \\ 0 &1&2&3\\ 0 & 0 & 0 & 0 \\ x&y&z&t \end{bmatrix}$$ has rank $2$, in terms of minor subdeterminants