Linear Algebra question!!!???

Question

Suppose the vectors v,w are in R^n and {v,w} is linearly independent. Prove that { v-w , 2v+w } is linearly independent as well.

Answer 1

Suppose the vectors are not linearly independent. Then there exists a nontrivial relation of the form c(v-w) + k(2v+w) = 0 for some scalars c and k. Rearranging terms, this implies that (c+2k)v + (k-c)w=0. Since v and w are linearly independent, this implies that c+2k=0 and k-c=0. This system of equations has the unique solution c=0, k=0. Since this solution is unique, there are no solutions where either c or k is not equal to zero. However, earlier we said that this was a nontrivial relation, contradicting this result. Therefore our initial assumption, that v-w and 2v+w are linearly dependent, is false. Q.E.D.

Answer 2

There are a few ways to do this. By thought, or by using determinants.

Recall that vectors are linearly dependant if and only if the determinant a matrix composed of the two vectors is 0.

Thus,

we make a matrix

A = [v -w ; 2v w]

The determinant of this matrix is

v*w - (-w*2v) = vw + 2vw = 3vw =! 0
(=! means does not equal)

Thus, since the determinant is non-zero it means the vectors can't be linearly dependant, and thus they are linearly independant.

Answer 3

You need to show that if there are L and M such that L(v-w)+M(2v+w) = 0, then both L and M are 0 as well. Simplify this expression and gather together two terms, one involving u and the other involving v. This gives you a linear combination of u and v with coefficients that are polynomials in L and M. From the independence of u and v, conclude that the coefficients are 0. Setting the coefficients equal to 0 gives you two equations in two unknowns L and M. If this yields L = M = 0, you will have proved your proposition. Try it.

Answer 4

you said they were independent