# Dimension of a span

1. May 3, 2015

### hitemup

1. The problem statement, all variables and given/known data

A. Let {t,u,v,w} be a basis for a vector space V. Find dim(U) where

U = span{t+2u+v+w, t+3u+v+2w, 3t+4u+2v, 3t+5u+2v+w}

B. Compute the dimension of the vector subspace V= span{(-1,2,3,0),(5,4,3,0),(3,1,1,0)} of R^4

2. Relevant equations

3. The attempt at a solution

I know that the dimension is the number of vectors in a basis. Since we're given a span, all we need to do is determine linearly independent vectors. But there's something confuses me in these two questions. I've got the solutions for them and in the first one, it forms a matrix whose columns are the vectors of the span, and in the second one, it constructs a matrix whose rows are the given vectors of the span to get the linearly independent vectors. So my question is simply, when to form the matrix as columns, and as rows?

Last edited: May 3, 2015
2. May 3, 2015

### Zondrina

The answer to part A is correct.

If you were given $U = \text{span} \{ v_1, v_2, v_3, v_4 \}$, and were asked to find $\text{dim}(U)$, then forming a matrix and row reducing would yield a linearly independent basis for $U$. Call this basis $B$, then $\text{dim}(B) = \text{dim}(U)$.

As for part B, the same logic applies. I don't think it matters how you reduce the vectors, i.e. it doesn't matter if you reduce the row vectors or the column vectors. The dimension of the basis will still be the same because the vectors forming it are linearly independent.

3. May 3, 2015

### hitemup

So there is not any difference between forming the matrix as whether rows or columns and then row reduce, is there?
EDIT:
It turns out we can do it in both ways as you suggest, thank you.

Last edited: May 3, 2015