Linearly Independent Sets After Subtraction
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 nonzero 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_iw) = 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. 
Re: Linearly Independent Sets After Subtraction
Note that it may be necessary to add that [itex] w \neq v_i [/itex] for any i.

Re: Linearly Independent Sets After Subtraction
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. 
Re: Linearly Independent Sets After Subtraction
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? 
Re: Linearly Independent Sets After Subtraction
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Re: Linearly Independent Sets After Subtraction
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_1a_2) v_1 + \frac12(a_2a_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? 
Re: Linearly Independent Sets After Subtraction
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Re: Linearly Independent Sets After Subtraction
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 d1 dimensional space? 
Re: Linearly Independent Sets After Subtraction
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Re: Linearly Independent Sets After Subtraction
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 ddimensional 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 d1 dimensional subspace of [itex]\mathcal B(\mathcal H) [/itex] and I am not certain why this is true. 
Re: Linearly Independent Sets After Subtraction
Clearly, I was wrong to cast this into a simplified framework. So I'm wondering, do you see why these new projectors span 1less dimensional space?

Re: Linearly Independent Sets After Subtraction
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Re: Linearly Independent Sets After Subtraction
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] d1 [/itex] is all that is necessary to do their work, even though the span is a d dimensional space. So alternatively, with this specific setup can we say that the [itex] \left\{ P_i  \text{id} \right\} [/itex] are linearly independent and hence span a ddimensional 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? 
Re: Linearly Independent Sets After Subtraction
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Re: Linearly Independent Sets After Subtraction
Very nice, thank you.

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