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 '15 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? $\endgroup$ – Kevin Zakka Oct 24 '15 at 14:13
  • $\begingroup$ $(1,1,1), (1,1,1), (1,0,0)$ do not span $\mathbb{R}^{3}$. $\endgroup$ – Megadeth Oct 24 '15 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? $\endgroup$ – Alessandro Codenotti Oct 24 '15 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 $\endgroup$ – Kevin Zakka Oct 24 '15 at 14:26

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.