Prove the geometric interpretation?

Question

Let $\vec v_1, \vec v_2, \vec v_3, \vec v_4$ be non-zero vectors in $\mathbb R^4$ such that the solution set of the system $\begin{bmatrix} \vec v_1 & \vec v_2 & \vec v_3 & \vec v_4 \space | \space \vec 0 \end{bmatrix}$ is a hyperplane. What is the geometric interpretation of $Span \left \{\vec v_1, \vec v_2, \vec v_3, \vec v_4\right \}$

Going through an example, it tells me that the geometric interpretation is a line through the origin, but I'm not sure how to prove this more rigorously.

Answer 1

Guide:

By definition, a hyperplane is of dimension $n-1$, hence the solution space has dimension $3$.

By rank-nullity theorem, can you conclude about the rank of the column space?

Answer 2

At least one of the vectors must be non-zero, becaue otherwise you would have more than a hyperplane of solutions. Let $\vec v_1\neq\vec 0$.

The hyperplane of solutions paired with $\vec v_1\neq \vec 0$ means that for any three coefficients $b, c, d\in \Bbb R$, there is an $a\in \Bbb R$ such that $$ a\cdot\vec v_1+b\cdot\vec v_2+c\cdot\vec v_3+d\cdot\vec v_4 = \vec 0 $$ Or, said differently: $$ a\cdot\vec v_1= -b\cdot\vec v_2-c\cdot\vec v_3-d\cdot\vec v_4 $$ In other words, $\operatorname{Span}(\vec v_2, \vec v_3, \vec v_4)\subseteq \operatorname{Span}(\vec v_1)$.