Proof for showing linear independence

I think your proof is wrong. From the equality

$$ a_1 v_1 + \dots + a_r v_r + a_{r+1}v_{r+1} = 0 $$

you cannot deduce $a_1 = \dots = a_r = 0$.

But you can carry on in the following way:

If $a_{r+1} = 0$, then you would have $a_1 v_1 + \dots + a_r v_r = 0$, and now you can say that, since $v_1, \dots , v_r$ are linearly independent, $a_1 = \dots = a_r = 0$. So all the vectors $v_1, \dots , v_r, v_{r+1}$ are linearly independent.

On the other hand, if $a_{r+1} \neq 0 $, then you could write $v_{r+1}$ as a linear combination of the rest:

$$ v_{r+1} = -\frac{a_1}{a_{r+1}}v_1 - \dots - \frac{a_r}{a_{r+1}}v_r \ . $$

But $v_{r+1}$ wasn't supposed to be a linear combination of the rest...

There is a problem with your proof: you seem to suppose that if $\;a_1v_1 + a_2v_2 + \dots + a_rv_r + a_{r+1} v_{r+1} = 0$, then necessarily $a_1=a_2=\dots=a_r=0$, and the rest of your proof follows.

Supposing so is not obvious at all: it is not because the only linear relation between $v_1,v_2,\dots, v_r$ is trivial that it is also the case when another vector is involved.

Actually, I would put it this way: considering a linear relation $\;a_1v_1 + a_2v_2 + \dots + a_rv_r + a_{r+1} v_{r+1} = 0$, you have two cases:

• either $a_{r+1}=0$, which means we really have a relation between $v_1,v_2,\dots, v_r$ alone, and by the independence hypothesis, we also have $a_1=a_2=\dots=a_r=0$, which proves the independence of the set of vectors $\{v_1,v_2,\dots, v_r,v_{r+1}\}$ in this case.
• or $a_{r+1}\ne 0$, and we can write $$v_{r+1}=-\tfrac{a_1}{a_{r+1}}v_1 - \tfrac{a_2}{a_{r+1}}v_2 - \dots - \tfrac{a_r}{a_{r+1}}v_r,$$ which shows $a_{r+1}$ is in the span of $v_1,v_2,\dots, v_r$, contrary to the hypothesis.