Spanning Set of R^3: (1,0,0)^T,(0,1,1)^T,(1,0,1)^T,(1,2,3)^T

[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.

 

[Mark44]

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.

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.