Proving Linear Independence and Span of B={(1,-1,-1),(1,0,1),(0,-1,1)} in R^3

[DanielFaraday]

Homework Statement

Prove that B={(1,-1,-1),(1,0,1),(0,-1,1)} spans R^3.

(Actually, the problem asks to show that B is a basis for R^3. This would require that I prove linear independence AND that it spans R^3. The first is easy, but I'm not sure about the second.)

Homework Equations

None

The Attempt at a Solution

I know how to prove that these vectors are linearly independent, but does that prove that B spans real space?

 

[g_edgar]

It has to do with the notion of "dimension" ... If R^3 has dimension 3, and you have a linearly independent set of 3 vectors, then that set spans R^3.

 

[DanielFaraday]

Oh, I see. So the fact that they are 3 linearly independent vectors means that they do span R^3. Got it. Thanks!

 

[HallsofIvy]

If you have that theorem:
"A basis for vector space has three properties:
1) It spans the space
2) It is a set of independent vectors
3) The number of vectors in the basis is equal to the dimension of the vector space"
and you know that R³ has dimension 3, then you can say "this is a set of 3 independent vectors so it spans R³."

If you do not have all of those facts, you could do it directly:
Show how to find, for any real numbers, x, y, z, real numbers, a, b, c, such that
a(1,-1,-1)+ b(1,0,1)+ c(0, -1, 1)= (x,y,z). That is the same as showing that the three equations, a+ b= x, -a- c= y, -a+ b+ c= z, has a solution (not necessarily unique) no matter what x, y, and z are.