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Linear Algebra: Linear Independence

  • Thread starter Breedlove
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  • #1
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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.


Homework Equations





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!
 

Answers and Replies

  • #2
674
2
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.
 
  • #3
27
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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?
 
  • #4
674
2
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.
 

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