Thread: [SOLVED] Basis for the subpace V= Span S of R^3?Any ideas?

1. [SOLVED] Basis for the subpace V= Span S of R^3?Any ideas?

This is a two part question:

a) Let S= (v1,v2,v3) , where:
v1 = (1,2,3) v2 = (2,1,4) v3 = (1,1,2)
Find a basis for the subspace V= span S of R^3.

I thought I understood how to do this but now I think I'm weong because the second part doesn't apply to my answer. I expressed the vectors as the columns of a matrix and found the basis of the column space to be all 3 vectors which are then the basis for span S. But the second part states:

b) Express each vector not in the basis as a linear combination of the found basis vector.

With my answer all 3 vectors are in the basis so where did I go wrong?

Any help would be appreciated.

2. Originally Posted by chocaholic
This is a two part question:

a) Let S= (v1,v2,v3) , where:
v1 = (1,2,3) v2 = (2,1,4) v3 = (1,1,2)
Find a basis for the subspace V= span S of R^3.

I thought I understood how to do this but now I think I'm weong because the second part doesn't apply to my answer. I expressed the vectors as the columns of a matrix and found the basis of the column space to be all 3 vectors which are then the basis for span S. But the second part states:

b) Express each vector not in the basis as a linear combination of the found basis vector.

With my answer all 3 vectors are in the basis so where did I go wrong?

Any help would be appreciated.

$\displaystyle \left|\begin{pmatrix}1&2&3\\2&1&4\\1&1&2\end{pmatr ix}\right|$ $\displaystyle =2+8+6-4-8-3=1\neq 0\Longrightarrow$ the set of vectors $\displaystyle \{(1,2,3)\,,\,(2,1,4)\,,\,(1,1,2)\}$ is linearly independent and is thus a basis for $\displaystyle \mathbb{R}^3$ ,which
of course means that in fact $\displaystyle S=\mathbb{R}^3$ and thus you went wrong nowhere.
I suppose part (b)'s answer could be: there's no vector from the original given three ones that is no element of the basis above for $\displaystyle S=\mathbb{R}^3$ so there's no more to be done.