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 $span = V$ 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 contain 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?

  • $\begingroup$ A basis for a vector space is by definition a set of linearly independent vectors that spans $V$... $\endgroup$
    – Megadeth
    Oct 24, 2015 at 14:11
  • $\begingroup$ 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? $\endgroup$ Oct 24, 2015 at 14:13
  • $\begingroup$ $(1,1,1), (1,1,1), (1,0,0)$ do not span $\mathbb{R}^{3}$. $\endgroup$
    – Megadeth
    Oct 24, 2015 at 14:15
  • $\begingroup$ 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? $\endgroup$ Oct 24, 2015 at 14:15
  • $\begingroup$ 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 $\endgroup$ Oct 24, 2015 at 14:26

1 Answer 1


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 $\implies$ $span(S) = V$ and S is a basis.
  • $span(S) = V \implies$ $S$ 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 $u,v,w$ are linearly independent thus,

$$2c_1 + 3c_2 = 0$$ $$-c_1 -c_2 = 0$$ $$2c_3 -c_1 = 0$$

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.


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