# Prove or disprove these two statements

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!

• "But well that was wrong..." Says who? Mar 5, 2021 at 15:41
• @JMoravitz The final answers.. the answer was that it is false, and $k$ can be anything. Mar 5, 2021 at 15:43
• 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... Mar 5, 2021 at 15:48
• 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. Mar 5, 2021 at 16:02
• 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. Mar 5, 2021 at 16:04

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.