To find a collection of four distinct vectors $V = (v_1, v_2, v_3, v_4)$ in $\mathbb{R}^3$ such that:

$\text{span}(V) = \left\{ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \in \mathbb{R}^3 : x_1 + x_2 + x_3 = 0 \right\},$

we need the following properties:

1. Each vector in $V$ satisfies $x_1 + x_2 + x_3 = 0$.
2. The span of these vectors forms a subspace of dimension 2 (since $x_1 + x_2 + x_3 = 0$ imposes one linear constraint in $\mathbb{R}^3$).

We can construct such a set $V$ by ensuring the vectors are linearly independent and satisfy $x_1 + x_2 + x_3 = 0$. Here is an example:

v_2 = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}, \quad v_3 = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \quad v_4 = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}.$$ ### Verification: 1. **Check the condition $$x_1 + x_2 + x_3 = 0$$:** - For $$v_1: 1 + (-1) + 0 = 0$$, - For $$v_2: 0 + 1 + (-1) = 0$$, - For $$v_3: -1 + 0 + 1 = 0$$, - For $$v_4: 2 + (-3) + 1 = 0$$. All vectors satisfy the condition. 2. **Check if the span of these vectors equals the subspace $$x_1 + x_2 + x_3 = 0$$:** - The subspace defined by $$x_1 + x_2 + x_3 = 0$$ has dimension 2. - Since the first three vectors $$v_1, v_2, v_3$$ are linearly independent, their span already forms a basis for the subspace. Adding $$v_4$$, while linearly dependent on $$v_1, v_2, v_3$$, still ensures all vectors lie in the subspace. Thus, $$V = (v_1, v_2, v_3, v_4)$$ is a valid set of vectors spanning the subspace $$\text{span}(V)$$. --- ### Questions to Explore Further: 1. What is the rank of the matrix formed by $$V = (v_1, v_2, v_3, v_4)$$? 2. Can you construct another set of four distinct vectors satisfying the same condition? 3. What happens if one of the vectors does not satisfy $$x_1 + x_2 + x_3 = 0$$? 4. How does adding more vectors to $$V$$ affect the span of the subspace? 5. How would this solution extend to subspaces in $$\mathbb{R}^n$$? **Tip:** To verify if a set of vectors spans a subspace, use the rank of the matrix formed by placing the vectors as columns.