# How many dimensions is this vector space?

1. Feb 28, 2013

### hello95

So I just got out of my linear algebra midterm, and this question is confusing the hell out of me. Basically, it's a subspace of R^4, such that the coordinates satisfy the following qualifications:

(a, b - a, b, 2(b - a))

So basically, a and b can range over the xz plane, and y and w sort of follow suit. I said that this space had the basis given by the three vectors:

(a,0,0,0)
(0,0,b,0)
(0,b-a,0,2(b-a))

But now that I look back at it, the more it seems like this is a 2-dimensional subspace, since you're essentially mapping from the xz plane to a line in yw, which means that it's basically isomorphic to a plane in three space.

Any thoughts?

2. Feb 28, 2013

### jbunniii

Note that you can write $(a, b-a, b, 2(b-a))$ as the linear combination of two vectors: $$(a, b-a, b, 2(b-a)) = a(1, -1, 0, -2) + b(0, 1, 1, 2)$$
What does this tell you about the dimension of the subspace?

3. Feb 28, 2013

### pasmith

It is indeed 2-dimensional: you can re-write it as $a(1, -1, 0, -2) + b(0,1,1,2)$.

None of the vectors you give as basis vectors are actually in that subspace for arbitrary a and b: to get the second component to vanish you must take a = b, so either the first and third components are both non-zero or they are both zero, in which case all components are zero. To make the first and third components vanish you must take a = b = 0, in which case all components are zero.