$a_{1}v_{1}+a_{2}v_{2}+\cdots+a_{i}v_{i}$=$u$, but some of $a_{j}$ is not unique

Question

If $n\times 1$ vector $u$ is spanned by $v_{1},\cdots, v_{i}$, where each $v_{j}$ is $n\times 1$, then we can find scalars $a_{1}, \cdots , a_{i}$ that satisfy $a_{1}v_{1}+a_{2}v_{2}+\cdots+a_{i}v_{i}$=$u$.

Suppose that some of the $a_{j}$ is not unique and infinitely many $a_{j}$ can satisfy the above equation. In that case, can I still say $u$ is spanned by $v_{1}, \cdots, v_{j}$? In other words I'm wondering if uniqueness of those scalars are needed to guarantee that $u$ is spanned by $v_{1}, \cdots, v_{i}$.

Answer 1

Spanning requires independence between the basis vectors. If two such $\{a_n\}$s exist say $a_n$ and $b_n$ we have $$u=a_1v_1+a_2v_2+\cdots+a_nv_n\\u=b_1v_1+b_2v_2+\cdots+b_nv_n$$which yields $$(a_1-b_1)v_1+(a_2-b_2)v_2+\cdots+(a_n-b_n)v_n=0$$where some of $a_n-b_n$ are not zero which contradicts the independence

Answer 2

In other words I'm wondering if uniqueness of those scalars are needed to guarantee that $u$ is spanned by $v_{1}, \cdots, v_{i}$.

If $B = \{v_1,\cdots, v_n\} \in K^n$ is a basis and $x = \{a_1, \cdots, a_n\} \in K^n$ the coordinate tuple. Then $$u = a_1\cdot v_1 + \cdots + a_n\cdot v_n \in Span(B)$$

This means that vectors composing the basis $B$ are linearly independent and every linear combination of scalars will represent a different vector $u$ on the linear subspace spanned by the basis.