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 \end{align*}[/tex] 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.
You are given that the set {v_{1}, v_{2}, ..., v_{n}} is linearly independent, which means that the equation c_{1}v_{1} + c_{2}v_{2} + ... + c_{n}v_{n} = 0 has only the trivial solution. Now look at the equation a_{1}(v_{1} - b) + a_{2}(v_{2} - b) + ... + a_{n}(v_{n} - b) = 0, where b != 0, and b != v_{i}, and show that this equation has only the trivial solution.
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?
Okay, so in this case we get [tex]\begin{align*} 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 \end{align*} [/tex] 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?
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.
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?
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.
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.
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?
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.
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?
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??