# Linear Algebra: Linear Independence

## Homework Statement

Let u,v,w be three linearly independent vectors in ℝ7. Determine a value of k,

k= , so that the set S={u-3v,v-5w,w-ku} is linearly dependent.

## The Attempt at a Solution

I don't really know why knowing that we're in ℝ7 will help. I know a couple of ways to solve this, but I don't think that they really apply. If we go:
A(u-3v)+B(v-5w)+C(w-ku)=0
[u-3v,v-5w,w-ku][A;B;C]=0

If it was square we could set the determinant equal to zero and solve for k. Otherwise, we have the matrix,

[u-3v,v-5w,w-ku,0]

If the matrix was little friendlier we could do some G-J elimination. Er... actually I'm not sure if that would help.

Any hints?
Thanks!

Related Calculus and Beyond Homework Help News on Phys.org
The first equation you listed is the right way to go. Just solve for A,B,C, and k using:

A(u-3v)+B(v-5w)+C(w-ku)=0

You don't need any matrices, since u, v, and w are linearly independent, you can just set all the coefficients in front of the individual u, v, and w to 0 and solve for the A,B,C, and k in the coefficients.

For example, (-3A+B) = 0, for the v term in that above equation.

Wow. Okay so I did it and got 1/15 for k and got the right answer, but am still a bit clueless as to why that works. Are we saying that u,v, and w equal 1 and saying that each term equals zero? From my understanding of linear (in)dependence a set of vectors is linearly independent if the only way that their sum can equal zero is if the linear combination where A, B, and C equal zero. er.. I don't know if I said that right. What I meant was that vectors x,y, z are linearly independent provided that

Ax+By+Cz=0 only if A,B,and C = 0.

How can we get to the method used to solve this problem knowing only that?

You know:

Ru+Sv+Tw = 0

is impossible since u,v,w are linearly independent unless R=S=T=0. That is the only way a linear combination of linearly independent vectors can sum to 0.

But you also wrote out:

A(u-3v)+B(v-5w)+C(w-ku)=0

which can be reorganized as:

(A-Ck)u + (-3A+B)v + (-5B+C)w = 0

which looks like Ru+Sv+Tw = 0

Therefore each of the coefficients must be 0, since u,v,w are linearly independent.