linear algebra?

Question

if u v and w are vectors that are linearly independent will u-v v-w and u-w be linearly independent?

justify your answer

Answer 1

First poster is very wrong. Consider that (u-v) + (v-w) - (u-w) = 0, which is a nontrivial linear dependency in (u-v), (v-w), and (u-w), so those three vectors are linearly dependent no matter what u, v, and w are.

In order for his "proof" to work, he would have needed to show that the system of equations a₁+a₃ = 0, a₂-a₁ = 0, -a₂-a₃ = 0 has no solutions other than a₁=0, a₂=0, a₃=0, and of course, it has several other solutions, among them a₁=1, a₂=1, a₃=-1.

Answer 2

No.

Here is a linear dependence relation among u - v, v - w, and u - w:

1(u - v) + 1(v - w) - 1(u - w) = 0

(In fact, it does not matter whether u, v, and w are linearly independent or not; The vectors u - v, v - w, and u - w are linearly independent for *any* vectors u, v, w.)

Answer 3

just to expand on the previous answer. the final step implies that u, v, and w are linearly independent, but since the assumption was that they are not this is a contradiction and thus proves, by contradiction that the differences are linearly independent.

Answer 4

The matrix version of the difficulty is AX = b the place A is a 2x2 matrix [4, ok; ok, a million] X is a vector variable [x; y] and b is a vector consistent [7; 0] there'll be a special answer as long as A is non-singular. there is not any longer a special answer if A *is* singular. For A to be singular det(A) = 0 the place det() is the "determinant" of a sq. matrix for a 2x2 matrix [a, b; c, d] det([a, b; c, d]) = advert - bc for this reason on your case 4*a million - ok*ok = 0 ok^2 = 4 ok = +/- 2

Answer 5

Proof by contradiction:

Assume that they are not linearly independent. Therefore there exists a1, a2 and a3 such that

a1*(u-v) + a2*(v-w) + a3*(u-w) = 0

or a1*u -a1*v + a2*v - a2*w + a3*u - a3*w = 0

so (a1+a3)*u +(a2-a1) * v + (-a2-a3) * w = 0

However, this would imply that u,v,w are not linearly independent. Therefore u-v, v-w, and u-w are linearly independent.

QED