Find a basis for R3 that contains the vectors 123 and 321Sol

Find a basis for R^3 that contains the vectors (1,2,3) and (3,2,1)

Solution

A basis for R^3 must consist of 3 linearly independent vectors. You are given two linearly independent vectors (lets call these A and B) so you need to find a third vector that is linearly independent of A and B i.e. the third vector must not be in the plane spanned by A and B.

One way to pick a third vector would be to find a vector that is perpendicular to both A and B - this is what you get if you take the cross-product of A and B. But there are other ways ...

One simple way is that you can just pick a third vector at random, and then test it to see if it is linearly independent of A and B. If it is, we are done ... if it is not, pick another random vector and try again.

For example, let\'s try C=(1,0,0) as out third vector. We know that the plane spanned by A and B contains all vectors of the form rA+sB where r and s are real numbers. So all the vectors in this plane will have co-ordinates which look like

(m+3n, 2m+2n, 3m+n) for some m and n

so (1,0,0) only lies in this plane if we can find values for m and n which simultaneously solve :-

m + 3n = 1
2m + 2n = 0
3m +n = 0

Is this possible ? If not, then C does not lie in the plane spanned by A and B ... so C is linearly independent of A and B ... so A, B and C are a basis for R^3.