# Really quick question on linear spans

1. Mar 2, 2009

### NeonVomitt

If I want to find if

span ([4, 0, -3], [2,2,1]) = span ([2,-2,-4], [0,1,5]) do I first find their reduced row echelon form, and then see if they match? For instance, if I found both matrices to reduce to:
[ 1 0]
[ 0 1]
[ 0 0]

does that mean that they equal each other? Or do I have to do something else?

Also what is the vector space? Is it R2 in this case?

2. Mar 2, 2009

### Werg22

Essentially, if two matrices are equivalent, then augmenting them with the same vector will give the same solution set. Another way of saying this is that if A ~ B, then in Ax = b and Bx = b, where b can be anything, if x satisfies the first equality, then it satisfies the second, and vice-versa. Also note that when there are no solutions in one, there are no solutions in the other.

If the rref of A is C and the rref of B is also C, then A ~ C and B ~ C. So we have that Ax = b, Cx = b, Bx = b, have the same solutions for x given any vector b (again if there are no solutions for particular b, none of the equations will have a solution). More elaborately, equivalence is transitive and reflexive, so A ~ C and B ~ C implies A ~ B.

That tells us that a vector b that has a solution in Ax = b, also has a solution in Bx = b. And vice-versa. Finding a linear combination of a set {v1, v2, ..., vn} of vectors equal to a vector b, amounts to solving [v1 v2 v3 ... vn]x = b. If {w1, w2, ..., wn} is another set of vectors, we would solve [w1 w2 ... wn]x = b. If [v1 v2 v3 ... vn] ~ C and [w1 w2 ... wn] ~ C then, by the above, one matrix has a solution when the other does. This means when there's a linear combination in {v1, v2, ..., vn} equal to a vector b, there's also one in {w1, w2, ... , wn} and vice-versa. This shows you that {v1, v2, ..., vn} and {w1, w2, ... , wn} have the same span.

3. Mar 2, 2009

### bpet

Remember that row operations transform the row vectors to an equivalent basis, so you look at the rref of

[4, 0, -3]
[2, 2, 1]

etc.