Suppose that $\{v_1,v_2,v_3\}$ is a linearly independent subset of a vector space $V$, with $\dim(V) = 4$, and that $v_4 \not\in \text{span}\{v_1,v_2,v_3\}$. Prove that $\{v_1,v_2,v_3,v_4\}$ is a basis for $V$.

My current answer would be:

Since $\dim(V) = 4$, it requires another vector (which is not linearly dependent) to form a basis.

Hence $v_4$ is not equal to $k_1 v_1+k_2 v_2+k_3 v_3$ (except if $k1=k2=k3 = 0$)

Since $v_4$ is not in $\text{span}\{v_1,v_2,v_3\}$, it is not linearly dependent and $\{v_1,v_2,v_3,v_4\}$ is a basis for $V$ (since vectors as a basis should be linearly independent).

Just wondering if you can help me whether my answer respond to the question well. Any help will be appreciated.

Thank you very much

  • $\begingroup$ Correct, absolutely fine. $\endgroup$ – Teresa Lisbon Apr 14 '16 at 1:30
  • $\begingroup$ You mean to write "and $v_4$ does not lie in the span of $v_1,v_2,v_3$" Of course a single vector cannot span a three dimensional space. $\endgroup$ – JMoravitz Apr 14 '16 at 1:36
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    $\begingroup$ I think you meant "is not in span..." instead of "does not span..." so I changed the wording when I edited to add typesetting. Please let me know if that is incorrect, or you can edit your post accordingly. $\endgroup$ – Bungo Apr 14 '16 at 1:36
  • $\begingroup$ Here is a Mathjax tutorial for future reference. It's easy to learn and allows you to typeset your posts properly. $\endgroup$ – Bungo Apr 14 '16 at 1:37
  • $\begingroup$ No need for a downvote. Here is an OP who does not blindly ask for an answer but who thinks, shares and wants verification...Hence my upvote $\endgroup$ – imranfat Apr 14 '16 at 1:39

The idea of your proof is correct, but it should be stated more clearly.

First, since $v_4 \not\in \text{span}\{v_1,v_2,v_3\}$, this means that there are no scalars $k_1,k_2,k_3$ for which $v_4 = k_1 v_1 + k_2 v_2 + k_3 v_3$, and this is remains true if $k_1 = k_2 = k_3 = 0$. You don't need to treat that as a special case. It is only when discussing linear independence that we need to exclude that case.

Then, to argue that $\{v_1,v_2,v_3,v_4\}$ is linearly independent, suppose for a contradiction that this is false. This means that there are scalars $k_1,k_2,k_3,k_4$, not all zero, such that $$k_1 v_1 + k_2 v_2 + k_3 v_3 + k_4 v_4 = 0$$ Now $k_4$ can't be zero, since otherwise this equation becomes $$k_1 v_1 + k_2 v_2 + k_3 v_3 = 0$$ which is only true if $k_1 = k_2 = k_3 = 0$ since $\{v_1,v_2,v_3\}$ is a linearly independent set. Therefore, since $k_4 \neq 0$, we can divide by $k_4$ and rearrange to obtain $$v_4 = -\frac{k_1}{k_4} v_1 - \frac{k_2}{k_4}v_2 - \frac{k_3}{k_4}v_3$$ But this expresses $v_4$ as a linear combination of $v_1,v_2,v_3$, so $v_4 \in \text{span}\{v_1,v_2,v_3\}$, which is a contradiction.

We conclude that $\{v_1,v_2,v_3,v_4\}$ must be linearly independent. Moreover, it is a linearly independent set of $4$ vectors, in a vector space of dimension $4$, hence it is a basis for $V$.

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  • $\begingroup$ Thank you very much or your detailed answer. I'd like to have a question, please. if k4 = nonzero, can we still say that the set (v1,v2,v3,v4) is linearly independent? Since for linear independence should there only be the trivial solution for case AX=0? $\endgroup$ – Lucky Boy Apr 14 '16 at 2:06
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    $\begingroup$ @LuckyBoy If $k_1 v_1 + k_2 v_2 + k_3 v_3 + k_4 v_4$ where any of the $k_n$'s (including $k_4$) is nonzero, then this mean that $\{v_1,v_2,v_3,v_4\}$ is a linearly dependent set. $\endgroup$ – Bungo Apr 14 '16 at 2:14
  • $\begingroup$ I see, thanks a lot for your explanation $\endgroup$ – Lucky Boy Apr 14 '16 at 2:27

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