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Let $V$ be a vector space over a field $F$ and $v_1,v_2,v_3,v_4$ four vectors that are different from zero then:

a) if $\{v_1,v_2,v_3,v_4\} $ are linearly dependent and $\{v_1 + v_2 + v_3 + v_4, v_2 + 2v_3, v_3 - v_4, v_2 + v_3+kv_4 \}$ are linearly dependent then $k=1$.

My thoughts: I used the theorem that if we get zero rows while reducing a matrix to it's row echelon form then the rows are linearly independent. Here's my matrix after subtracting row $2$ from row $4$:
$$\begin{bmatrix}1&1&1&1\\0&1&2&0\\0&0&1&-1\\0&0&-1&k\end{bmatrix}$$

Now I answered that in order to get a zero row we need $k=1$.

But well that was wrong, I guess I'm missing some fundamental knowledge about linear dependence but I can't get what part is my mistake, is it using the matrix? or I shouldn't even bother since the vectors are linearly dependent?

b) if $\{ v_1, v_2, v_3\}$ , $\{v_3,v_4\}$ are linearly independent sets and $v_4 \in sp \{v_1, v_3\}$ then $\{ v_2, v_3, v_4 \}$ are linearly independent.

This statement is true, I sat for too long trying to come out with a counter example and I failed, and while doing it I tried to prove, but I had no idea where to start the proof and how to construct it.

I would really appreciate any help and feedback, thanks in advance!

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    $\begingroup$ "But well that was wrong..." Says who? $\endgroup$
    – JMoravitz
    Mar 5, 2021 at 15:41
  • $\begingroup$ @JMoravitz The final answers.. the answer was that it is false, and $k$ can be anything. $\endgroup$
    – Pwaol
    Mar 5, 2021 at 15:43
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    $\begingroup$ Ah... I think I see the issue. The hypothesis was "if $\{v_1,\dots,v_4\}$ were linearly dependent and...then..." I misread that as $v_1,\dots,v_4$ being independent. If it were read that way then your answer and approach would have been correct. Since $v_1,\dots,v_4$ are dependent however... $\endgroup$
    – JMoravitz
    Mar 5, 2021 at 15:48
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    $\begingroup$ Hmm... well, in general it might be difficult to say... if you have linearly dependent $v_1,v_2,\dots,v_n$ and you were asked if some set of $m$ vectors, each of whom are linear combinations of $v_1,\dots,v_n$ with $m<n$ were linearly dependent or not there wouldn't be enough information. We'd need to know in what way the original vectors were dependent. $\endgroup$
    – JMoravitz
    Mar 5, 2021 at 16:02
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    $\begingroup$ For this however... we know that $v_1,v_2,v_3,v_4$ being linearly dependent span at most a $3$-dimensional space (since they can't have spanned a 4 dimensional space, else they would have been independent). Now, your four vectors $v_1+v_2+v_3+v_4, v_2+2v_3,\dots$ these are all elements of that same at-most-$3$-dimensional space spanned by $v_1,v_2,v_3,v_4$. Since there are more of these $v_1+v_2+v_3+v_4,v_2+2v_3,\dots$ than the dimension of the space in which they reside, they must be dependent. $\endgroup$
    – JMoravitz
    Mar 5, 2021 at 16:04

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For part (b): I suspect the condition is supposed to be $\{v_3,v_4\}$ being linearly independent (you have $\{v_3,v_3\}$); it's the only pair that makes sense, since any other pair's independence is subsumed in the assumption that $\{v_1,v_2,v_3\}$ is linearly independent. And if we don't have that second condition, we get a counterexample by letting $v_4=v_3$. So, assuming that:

Because $v_4\in\mathrm{span}(v_1,v_3)$, then we can write $v_4=\alpha v_1+\beta v_3$ for some scalars $\alpha$ and $\beta$. We also know $v_4\neq0$, so $\alpha$ and $\beta$ are not both equal to zero. And since $\{v_3,v_4\}$ is linearly independent, $\alpha$ must be nonzero (otherwise, $v_4=\beta v_3$ and $\{v_3,v_4\}$ would be linearly dependent).

Now let's take a linear combination of $v_2$, $v_3$, and $v_4$ that is equal to $0$: $$r_2v_2 + r_3v_3 + r_4v_4=0.$$ We want to prove that $r_2=r_3=r_4=0$.

Plugging in the value of $v_4$, we get $$\alpha r_4v_1 + r_2v_2 + (r_3+\beta r_4)v_3 = 0.$$ Since we know that $\{v_1,v_2,v_3\}$ is linearly independent, this means that $\alpha r_4=0$, $r_2=0$, and $r_3+\beta r_4=0$.

Since $\alpha r_4=0$, either $\alpha=0$ or $r_4=0$. But we know from the fact that $\{v_3,v_4\}$ is linearly independent that $\alpha\neq 0$, hence $r_4=0$. And then the last equation becomes $r_3=0$, proving that $r_2=r_3=r_4=0$, which is what we wanted to prove.

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  • $\begingroup$ Thanks alot for the amazing proof and explanation!! and as you've said I had a typo I fixed it now, tysm. $\endgroup$
    – Pwaol
    Mar 5, 2021 at 20:06

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