The span of a vector space with elements as linear combinations of no more than $r$ vectors has $\dim V \leq r$ If $V=Span \{ \vec{v}_1, \dots, \vec{v}_n \}$ and if every $\{ v_i \}$ is a linear combination of no more than $r$ vectors in $\{ \vec{v}_1, \dots, \vec{v}_r \}$ excluding $\{ v_i \}$, then $\dim V \leq r$
How can I improve the following proof please?
Let's suppose $V=\operatorname{Span} \{ \vec{v}_1, \dots, \vec{v}_n \}$ and every $\{ v_i \}$ is a linear combination of no more than $r$ vectors in $\{ \vec{v}_1, \dots, \vec{v}_r \}$ excluding $\{ v_i \}$ then we can write:
$ \vec{v}_i = \alpha_1 \vec{v}_1 + \dots + \alpha_{i-1} \vec{v}_{i-1} + \alpha_{i+1} \vec{v}_{i+1} + \dots + \alpha_r \vec{v}_r \; (\exists \alpha_i \neq 0)$
If $r<n$ then there are $n-(r+1)$ linearly independent elements with $\vec{v}_i$.  However we can't have this as we could put $\vec{v}_i = \alpha_1 \vec{v}_1$ with both $\vec{v}_i, \vec{v}_1 \in V $which would contradict these vectors spanning V.  Then we must have $r \geq n$  we can think of this intuitively as vectors $\vec{v}_{n+1}, \dots, \vec{v}_r = W$ that are not in the span of $V$.  Then each $\vec{v}_i$ a linear combination of at least 1 element of $W$ and no element in $V$ is a linear combination of the same element or we would get dependence between $\vec{v}_i$

If $V = span \{ v_1, \dots, v_n \}$ and if every $\{ v_i \}$ is a linear combination of no more than $r$ vectors in $\{ v_1, \dots, v_r \}$ excluding $\{ v_i \}$, then $dim V \leq r$
Proof: The hypothesis is the same as, for any $v_i$,
$$v_i = \sum_{j=1}^r b_{ij} v_j$$
with $b_{ii} = 0$, in case $i \leq r$ ("excluding $\{v_i\}$").
Choose $w$ in $V$.
$$w = \sum_{i=1}^n a_i v_i = \sum_{i=1}^n a_i \sum_{j=1}^r b_{ij} v_j$$
Now,
$$w = \sum_{j=1}^r \sum_{i=1}^n a_i b_{ij} v_j = \sum_{j=1}^r c_j v_j$$
and $c_j = \sum_{i=1}^n a_i b_{ij}$.
Since $w$ was arbitrary, any basis in $V$ should have no more than $r$ vectors.
Thus $dim V \leq r$.
$\square$
