Proving a Given Set Spans a Vector Space

Question

Show that if $\{u, v, w\}$ is a basis for a vector space $V$, then $\{2u-v-w,3u-v,2w\}$ is a basis for $V$.

Ok, so it was relatively easy to prove the set of vectors $\{2u-v-w,3u-v,2w\}$ were linearly independent. However, my thought process for proving their $\operatorname{span} = V$ was the following:

Since $\{u, v, w\}$, a set containing 3 vectors, is a basis for $V$, then all bases of $V$ must contain 3 vectors.
This means that given a set, a necessary condition for it to be a basis and hence for its span to equal $V$ would be that it contains 3 elements.
But $\{2u-v-w,3u-v,2w\}$ contains three vectors and since it is linearly independent, then it spans $V$.

I am convinced with this, however when looking here, it seems no one mentioned it and they are all doing a pretty long proof.

Would anyone enlighten me as to what or where I am going wrong?

A basis for a vector space is by definition a set of linearly independent vectors that spans $V$... — Yes, Oct 24 '15 at 14:11
Exactly, so why are they going into a long-winded proof in the linked question? Instead of just directly proving they span V because they contain 3 elements? — Kevin Zakka, Oct 24 '15 at 14:13
the first answer to the question you linked proves the independence of those vectors and concludes that they must be a basis since they are as many vectors as those in the given basis, isn't that exactly what you're doing here? — Alessandro Codenotti, Oct 24 '15 at 14:15
Gudson! I figured it out. I am using the theorem that says if the dimension of v is n then proving S is linearly independent is enough for it to be a basis. In your example, they are not linearly independent, that's why the span part doesn't hold — Kevin Zakka, Oct 24 '15 at 14:26

score 1 · Accepted Answer · edited Oct 28 '23 at 20:30

I was able to figure this out and can now answer it a few weeks later.

Basically, since $\{u, v, w\}$ is a basis for V, then $\dim(V) = 3$.

This means that for a set $S$ containing 3 vectors, it is enough to prove one of the following:

The vectors in $S$ are linearly independent $\implies \operatorname{span}(S) = V$ and $S$ is a basis.
$\operatorname{span}(S) = V \implies$ $S$ is linearly independent and $S$ is a basis.

So let’s show that $\{2u-v-w,3u-v,2w\}$ are linearly independent by examining the following equation:

$$c_1(2u-v-w) = c_2(3u-v) + c_3(2w) = 0.$$

Regrouping $u$, $v$ and $w$:

$$u(2c_1 + 3c_2) + v(-c_1 -c_2)+w(2c_3 - c1) = 0.$$

But $u$, $v$ and $w$ are linearly independent. Thus,

\begin{align} 2c_1 +3c_2 &= 0 \\ -c_1 -c_2 &= 0 \\ 2c_3 -c_1 &= 0. \end{align}

From this system, we get $c_1 = c_2 = c_3 = 0$, thus $\{2u-v-w,3u-v,2w\}$ are linearly independent and constitute a basis for $V$.

Proving a Given Set Spans a Vector Space

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