Proof for showing linear independence

Question

Very recently I started studying proof based linear algebra on my own and I am having a difficulty because there is no one to tell me whether if my proof is right or wrong. Please keep in mind that this is my first time dealing with proofs. The question is the following:

Let a system of vectors $v_1, v_2 … v_r$ be linearly independent but not generating.
Show that it is possible to find a vector $v_{r+1}$ such that the system
$v_1, v_2, \dots, v_r, v_{r+1}$ is linearly independent.

Hint: Take for $v_{r+1}$ any vector that cannot be represented as a linear combination of $v_1,v_2,\dots, v_r$ and show that the system $v_1, v_2, \dots, v_r, v_{r+1}$ is linearly independent.

My attempt:
question says $v_1,\dots,v_r$ is linearly independent so it has trivial coordinates $a_k$.
$a_1v_1 + a_2v_2 + \dots + a_rv_r = 0; a_1=a_2=\dots=a_r=0$

I add $v_{r+1}$ to the set $\{v_1,v_2,\dots,v_r\}$ and see if they can form linear independence.

$a_1v_1 + a_2v_2 + \dots + a_rv_r + a_{r+1} v_{r+1} = 0$
but $a_1=a_2=\dots=a_r=0$
so $a_{r+1} v_{r+1} = 0$ is true if $a_{r+1} = 0$ (which makes it linear independent) as long as $v{r+1}$ is not the $0$ vector.

Therefore, there is a $v_{r+1}$ that makes a system $\{v_1,v_2,\dots,v_{r+1}\}$ linearly independent. My proof ends here.

**Extra question: In the beginning, it says system of vectors given are linearly independent but not generating. Those this mean it has basis but not all of the basis? Like if the space was R^3, then only 2 or 1 out of 3 basis are in the system?

Answer 1

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.