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?
Best Answer
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:
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.