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?

  • 2
    $\begingroup$ Welcome to MSE. You should use MathJax to format math on this site. You'll get a much better response if your posts are easy to read. $\endgroup$
    – saulspatz
    Commented Dec 23, 2019 at 22:14
  • 1
    $\begingroup$ When you write $\sum_{i = 1}^{r} a_i v_i = 0$ then the only solution is $a_i = 0$ for all $i$ is correct. However, when you write $\sum_{i = 1}^{r + 1} a_i v_i = 0$ then why should $a_1 = \dots = a_r = 0$ be true? The coefficients are not the same in both the equations. $\endgroup$
    – sudeep5221
    Commented Dec 23, 2019 at 22:30

3 Answers 3


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.

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...


Your proof does not necessarily work, since if we consider the equation $$ b_1v_1+\dots+b_{n+1}v_{n+1}=0 $$ we do not know if $b_i=a_i$ for all $i\in\{1,\ldots,n\}$ from your initial equation involving just the first $n$ vectors. The proof would go something more like:

Let $V$ be our overlying vector space. Consider $v_1,\ldots,v_n\in V$ such that they are linearly independent but $\operatorname{span}(v_1,\ldots,v_n)\neq V$ (this is what it means by $v_1,\ldots,v_n$ does not "generate" the space, not every vector in $V$ can be written in terms of our set. You may want to investigate the definition of span). Then there exists a vector $v_{n+1}\in V$ such that $v_{n+1}\notin\operatorname{span}(v_1,\ldots,v_n)$ by our assumption that they are not generating. Now, for the sake of contradiction, assume $v_1,\ldots,v_{n+1}$ are linearly dependent. Then there exists scalars $b_1,\ldots,b_{n+1}$ not all 0 such that $$ b_1v_1+\dots+b_{n+1}v_{n+1}=0 $$ We have that the only $b$ that can be non-zero is $b_{n+1}$ (why?). Then we can rearrange our equation and have that $$ v_{n+1}=-\dfrac{b_1}{b_{n+1}}v_1-\dots-\dfrac{b_n}{b_{n+1}}v_n $$ which is a contradiction to our assumption that $v_{n+1}\notin\operatorname{span}(v_1,\ldots,v_n)$. So our set of $n+1$ vectors must be linearly independent.

  • $\begingroup$ Only b_{n+1} can be non zero because v_{1} to v_{n} is linearly independent? $\endgroup$
    – Mardia
    Commented Dec 23, 2019 at 23:10
  • $\begingroup$ @Mardia Precisely. At least one of them must be non-zero. Consider letting $b_{n+1}=0$. Then the resulting equation would be $b_1v_1+\dots+b_nv_n=0$. We still require that one of the $b$'s be 0. However, none of these $b$'s can be 0 since it will instantly contradict our linear independence assumption. $\endgroup$
    – Ken
    Commented Dec 23, 2019 at 23:15
  • $\begingroup$ I suspect that you meant “not every vector in $V$ can...” instead of “every vector in $V$ cannot...” The former is $\forall v\in V \lnot\dots$, while the latter is $\lnot\forall v\in V\dots$, which don’t mean the same thing. $\endgroup$
    – amd
    Commented Dec 23, 2019 at 23:57
  • $\begingroup$ @amd Yes that's what I meant. I was typing in a hurry. Editing now. $\endgroup$
    – Ken
    Commented Dec 24, 2019 at 0:01

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