# Spanning set R^3

1. Feb 23, 2010

### Dustinsfl

Determine the spanning set of {(1,0,0)^T, (0,1,1)^T, (1,0,1)^T, (1,2,3)^T}

v=a*(1,0,0)^T+b*(0,1,1)^T+c*(1,0,1)^T+d*(1,2,3)^T

Augmented Matrix(3, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 1, (1, 4) = 1, (1, 5) = v1, (2, 1) = 0, (2, 2) = 1, (2, 3) = 0, (2, 4) = 2, (2, 5) = v2, (3, 1) = 0, (3, 2) = 1, (3, 3) = 1, (3, 4) = 3, (3, 5) = v3})

In reduced row echelon, Matrix(3, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = v1-v3+v2, (2, 1) = 0, (2, 2) = 1, (2, 3) = 0, (2, 4) = 2, (2, 5) = v2, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1, (3, 4) = 1, (3, 5) = v3-v2})

I am not sure how to finish this off since there is an extra column.

2. Feb 23, 2010

### Staff: Mentor

Most of the stuff from "Augmented Matrix ..." on down is gobbledygook that I will ignore.

You have a set of vectors in R3 that includes four vectors, so obviously there are too many for a basis. What you want to do is to eliminate any of the four vectors that is a linear combination of the others. To save myself some typing I'm going to call your four vectors v1, v2, v3, and v4. If you put your four vectors as columns in a 4 x 3 (not augmented) matrix and row-reduce it, you are finding the solutions to the equation c1v1 + c2v2 + c3v3 + c4v4 = 0.

From that you should be able to see which vector is a linear combination of the others.