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I have that the set of vectors:

$$\vec u, \vec v, \vec w$$ is L.I.

This means that:

$$a_1\vec u + a_2\vec v + a_3\vec w = \vec 0 \implies a_1 = a_2 = a_3 = 0$$

I need to prove that the set:

$$(\vec u + \vec v + \vec w, \vec u - \vec v, 3\vec v)$$

Is also L.I.

What I did was to make a linear combination with the vectors, so:

$$k_1(\vec u + \vec v + \vec w) + k_2(\vec u - \vec v) + k_3(3\vec v) = \vec 0 \tag{1}$$

If this set is L.I., then $k_1 = k_2 = k_3 = 0$.

Expanding the sum:

$$k_1\vec u + k_1\vec v + k_1\vec w + k_2\vec u - k_2\vec v + 3k_3\vec v = \vec 0$$

So I have:

$$(k_1+k_2)\vec u + (k_1 -k_2 + 3k_3)\vec v + k_1 \vec w = \vec 0$$

By $(1)$ I have that this is a L.I. set iff $$(k_1+k_2) = (k_1 -k_2 + 3k_3) = k_1 = 0$$

Am I correct? How do I prove, then, that $(k_1+k_2) = (k_1 -k_2 + 3k_3) = k_1 = 0$?

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Your argument is correct. Since you have by assumption that $v, u,w$ are linearly independent and you are assuming $(k_1 + k_2)u + (k_1 -k_2 +3k_3)v + k_1 w=0$, then by definition of linearly independence you have

$k_1+k_2 = k_1 - k_2 +3k_3 = k_1 =0$

Now you can simply use substitution. You know $k_1=0$, then you put it in $k_1 + k_2 =0$ and get $k_2 =0$. Then you put $k_1=0$ and $k_2=0$ in $k_1-k_2+3k_3=0$ and you have $k_3=0$. And now you are done.

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