Linearly Independent Sets After Subtraction

  1. 1. The problem statement, all variables and given/known data
    Here is a really simple lin.alg problem that for some reason I'm having trouble doing.

    Assume that [itex]\left\{ v_i \right\} [/itex] is a set of linearly independent vectors. Take w to be a non-zero vector that can be written as a linear combination of the [itex] v_i [/itex]. Show that [itex] \left\{ v_i - w \right\} [/itex] is still linearly independent.

    3. The attempt at a solution
    For some reason I'm quite stuck on this. My first goal was to let [itex] b_i [/itex] be such that we can write
    [tex] w = \sum_j b_j v_j [/tex]
    and then consider the sum
    [tex] \sum_i a_i (v_i-w) = 0 [/itex]
    and show that each [itex] a_i [/itex] must necessarily be zero. Substituting the first equation into the other yields
    [tex] \begin{align*}\sum_i a_i (v_i - \sum_j b_j v_j ) &= \sum_i a_i - \sum_{i,j} a_i b_j v_j \\
    &= \sum_i \left( a_i - \sum_j a_j b_i \right) v_i
    where in the last step I've switched the indices in the double summation. By linear independence of the [itex] v_i [/itex] it follows that
    [tex] a_i = \sum_j a_j b_i [/tex]
    but that's where I'm stuck.

    It's possible that I'm doing this the wrong way also. Any help would be appreciated.
  2. jcsd
  3. Note that it may be necessary to add that [itex] w \neq v_i [/itex] for any i.
  4. Mark44

    Staff: Mentor

    You are given that the set {v1, v2, ..., vn} is linearly independent, which means that the equation
    c1v1 + c2v2 + ... + cnvn = 0 has only the trivial solution.

    Now look at the equation a1(v1 - b) + a2(v2 - b) + ... + an(vn - b) = 0, where b != 0, and b != vi, and show that this equation has only the trivial solution.
  5. Hey Mark,

    Thanks for the reply. This is exactly what I did in "My attempt at the solution," though I got stuck. Do you have any advice on whether I should take a different approach, or how to resolve where I got stuck?
  6. Dick

    Dick 25,910
    Science Advisor
    Homework Helper

    Take the case of two vectors {v1,v2}. Let w=(v1+v2)/2.
  7. Okay, so in this case we get
    a_1(v_1 -\frac12 v_1 -\frac12 v_2) + a_2(v_2 - \frac12 v_1 -\frac12 v_2 ) &= \frac{a_1}2 (v_1 -v_2) + \frac{a_2}2 (v_2 - v_1) \\
    &= \frac12(a_1-a_2) v_1 + \frac12(a_2-a_1) v_2

    By linear independence of [itex] v_1,v_2[/itex] we get that [itex] a_1 = a_2[/itex]. Why does this imply that either is zero?
  8. Dick

    Dick 25,910
    Science Advisor
    Homework Helper

    You are staring too hard at the ai's. v1-w=(v1-v2)/2, v2-w=(v2-v1)/2. (v1-w)=(-1)*(v2-w). They aren't linearly independent. I'm saying your proposed theorem is false. That's why you are having a hard time proving it.
  9. Okay, so then that hypothesis goes out the window.

    Tell me, does it then make sense to instead say that if [itex] \left\{ v_i \right\}_{i=1}^d [/itex] span a d dimensional space, then for w a linear combo of the [itex] v_i [/itex] it follows that [itex] \left\{ v_i - w \right\} [/itex] span a d-1 dimensional space?
  10. Dick

    Dick 25,910
    Science Advisor
    Homework Helper

    No. The {vi-w} are 'usually' independent, if you pick some random w. But for a family of special values of w it will fail.
  11. See, here's the issue. I've tried to make it simple so that it doesn't complicate things, but maybe putting it in a proper framework will make it better.

    Consider a finite dimensional complex Hilbert space [itex] \mathcal H [/itex] of dimension d, and fix an orthonormal basis [itex] \left\{ e_i \right\} [/itex]. Let [itex] P_i:\mathcal H \to \mathcal H [/itex] represent projection operators taking each element of [itex] \mathcal H [/itex] to the one dimensional space spanned by [itex] e_i [/itex]. In particular then, the set of [itex] P_i [/itex] span a d-dimensional subspace of [itex] \mathcal B(\mathcal H) [/itex], the set of bounded linear operators on [itex] \mathcal H[/itex]. Furthermore since the [itex] P_i [/itex] decompose [itex] \mathcal H [/itex] into a direct sum of the orthgonal subspaces, it follows that
    [tex] \sum_{i=1}^d P_i = \text{id} [/tex]
    where id is the identity operator in [itex] \mathcal B(\mathcal H) [/itex]. I now have a paper in front of me saying that the set [itex] \left\{ P_i - \text{id} \right\} [/itex] spans a d-1 dimensional subspace of [itex]\mathcal B(\mathcal H) [/itex] and I am not certain why this is true.
  12. Clearly, I was wrong to cast this into a simplified framework. So I'm wondering, do you see why these new projectors span 1-less dimensional space?
  13. Dick

    Dick 25,910
    Science Advisor
    Homework Helper

    No I don't. Take the case d=2. Then the set is {P1-id,P2-id}={-P2,-P1}. Looks to me like the span is the same as the span of {P1,P2}. If it were {Pi-id/d} then I can see where the dimension of the span would drop. That's the example I just gave you.
  14. Thanks Dick, you've been quite helpful.

    I checked this too, and in the case when [itex] \mathcal H = \mathbb C^d [/itex] and the orthogonal basis is taken to be the standard basis, this always seems to be the case.

    I think then that maybe the author is stating that [itex] d-1 [/itex] is all that is necessary to do their work, even though the span is a d dimensional space. So alternatively, with this specific set-up can we say that the [itex] \left\{ P_i - \text{id} \right\} [/itex] are linearly independent and hence span a d-dimensional space?

    This is where the original problem came from and combined with checking the simple cases I assumed this is why it had to be true. Any thoughts on this one?
  15. Dick

    Dick 25,910
    Science Advisor
    Homework Helper

    Sure, sum {Pi-id} for i=1 to d, you get id-d*(id)=(1-d)*id, right? So id is in the span (except for the silly case d=1). Pi-id is in the span. So (Pi-id)+id=Pi is in. So all of the Pi are also in the span. Hence span has dimension d. Perhaps they are just trying to make a statement that includes d=1??
  16. Very nice, thank you.
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