Showing that a vector in a subspace formed by linearly dependent vectors can be expressed in infinitely many ways

Question

I'm having trouble with this math problem:

Suppose $a_{1}, \ldots a_{n}$ are linearly dependent vectors that form a subspace $V$ in $\mathbb{R}^{n}$. Let $b \in V$. I want to show that $b$ can be expressed in infinitely many ways as a combination of $a_{1}, \ldots a_{n}$.

---

So I'm pretty sure that I'm going to have to use the fact that $a_{1} \ldots a_{n}$ are linearly dependent, because I tried a few examples with them lineraly independent, and I don't think the statement is true when they are linearly independent. Intuitively, this statement makes sense to me, but I have no idea how to prove it.

EDIT: I think I might have to do something with determinants, but this could be incorrect.

Answer 1

Suppose $b=\sum_i c_i a_i,\,\sum_i d_i a_i= 0$, with the $d_i$ not all zero. Then apply $c_i\mapsto c_i+kd_i$ for any scalar $k$.

Answer 2

As ${a_i}$ span $V$ and are linearly dependent, we can find a basis of $V$ from the $a_i$ say $a_1,..,a_m$ where $m < n$. Then $v = \sum_{i=1}^{m} c_ia_i$ for some $c_i \in \mathcal{F}$. Now we know that $a_{m+1},...,a_n$ are linearly dependent so they can be expressed as linear combination of $\{a_i\}_{i=1}^{m}$. Use this to show that we can write $0$ infinitely many ways for linear combinations of $\{a_i\}_{i=1}^{m}$. Then use $v = \sum_{i=1}^{m} c_ia_i + 0$