I'm stuck on the following problem:
Let $V$ be a vector space and $\mathbf{a_1, a_2, \dots, a_n} \in V$, a set of vectors of which $r$ are linearly independent, where $r \le n$.
Define $\mathbf{b_1, \dots, b_m}$ as below:
$$\begin{aligned}
\mathbf{b_1} &= a_{11} \mathbf{a_1} + \cdots + a_{1n}\mathbf{a_n} \\
\mathbf{b_2} &= a_{21} \mathbf{a_1} + \cdots + a_{2n}\mathbf{a_n} \\
&\vdots \\
\mathbf{b_m} &= a_{m1} \mathbf{a_1} + \cdots + a_{mn}\mathbf{a_n} \\
\end{aligned}$$
Prove that the number of linearly independent vectors in $\mathbf{b_1, \dots, b_m}$ is at most $r$.
I understand that $\mathbf{b_1, \dots, b_m} \in Span(\mathbf{a_1, \dots, a_n})$ and therefore are in a subspace of dimension $r$. Hence, there can only be $r$ linearly independent vectors in $\{\mathbf{b_1, \dots, b_m}\}$. I'm trying not to rely on theorems and taking the following approach to proving this, but am stuck. Here's what I have so far:
Proof.
WLOG, let $\mathbf{a_1, \dots, a_r}$ be linearly independent. Then,
$$\begin{aligned}
\mathbf{a_{r+1}} &= \sum_{i=1}^r \alpha_{r+1, i} \mathbf{a_i} \\
\mathbf{a_{r+2}} &= \sum_{i=1}^r \alpha_{r+2, i} \mathbf{a_i}\\
&\vdots \\
\mathbf{a_n} &= \sum_{i=1}^r \alpha_{n, i}\mathbf{a_i}\\
\end{aligned}$$
from linear dependence.
Then,
$$\begin{aligned}
\mathbf{b_{1}} &= a_{11} \mathbf{a_1} + \cdots + a_{1r}\mathbf{a_r} + a_{1r+1}\sum_{i=1}^r \alpha_{r+1, i} \mathbf{a_i} + \cdots + a_{1n}\sum_{i=1}^r \alpha_{n, i}\mathbf{a_i}:= \lambda_{11}\mathbf{a_1} + \cdots + \lambda_{1r}\mathbf{a_r}\\
&\vdots \\
\mathbf{b_{m}} &= a_{m1} \mathbf{a_1} + \cdots + a_{mr}\mathbf{a_r} + a_{mr+1}\sum_{i=1}^r \alpha_{r+1, i} \mathbf{a_i} + \cdots + a_{mn}\sum_{i=1}^r \alpha_{n, i}\mathbf{a_i}:= \lambda_{m1}\mathbf{a_1} + \cdots + \lambda_{mr}\mathbf{a_r}\\
\end{aligned}$$
When $m \le r$, there can be at most $r$ linearly independent vectors in $\{\mathbf{b_1, \dots, b_m}\}$ since there are at most $r$ vectors to begin with.
When $m > r$, take the first $r$ vectors $\mathbf{b_1, \dots, b_r}$. Let $\begin{cases}\lambda_{ij} = 1 \text{ if } i = j\\ \lambda_{ij} = 0 \text{ if } i \ne j \end{cases}$. Then, $\mathbf{b_1 = a_1, b_2 = a_2, \cdots b_r = a_r}$ and it follows that $\mathbf{b_1, \dots, b_r}$ are linearly independent. Thus, we know that $\{\mathbf{b_1, \dots, b_m}\}$ can have $r$ linearly independent vectors.
From here, assuming that $\mathbf{b_1, \dots, b_r}$ are linearly independent, I want to show that if we introduce $\mathbf{b_{r+1}}$, we can find a nontrivial solution to $\beta_1 \mathbf{b_1} + \cdots \beta_r \mathbf{b_r} + \beta_{r+1} \mathbf{b_{r+1}} = 0$ by using $\mathbf{b_{r+1}} = \lambda_{r+1,1}\mathbf{a_1} + \cdots + \lambda_{r+1,r}\mathbf{a_r}$, but I can't think of a clever way to show it. If anyone could help at all, I would really appreciate it! It would be best if you could give me help to expand my solution, but I would also appreciate pointers in other directions.