# Linearly Independent Sets After Subtraction

1. Mar 10, 2011

### Kreizhn

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 $\left\{ v_i \right\}$ 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 $v_i$. Show that $\left\{ v_i - w \right\}$ 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 $b_i$ be such that we can write
$$w = \sum_j b_j v_j$$
and then consider the sum
\sum_i a_i (v_i-w) = 0 [/itex] and show that each $a_i$ 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*}
where in the last step I've switched the indices in the double summation. By linear independence of the $v_i$ it follows that
$$a_i = \sum_j a_j b_i$$
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. Mar 10, 2011

### Kreizhn

Note that it may be necessary to add that $w \neq v_i$ for any i.

3. Mar 10, 2011

### 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.

4. Mar 10, 2011

### Kreizhn

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?

5. Mar 10, 2011

### Dick

Take the case of two vectors {v1,v2}. Let w=(v1+v2)/2.

6. Mar 10, 2011

### Kreizhn

Okay, so in this case we get
\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*}

By linear independence of $v_1,v_2$ we get that $a_1 = a_2$. Why does this imply that either is zero?

7. Mar 10, 2011

### Dick

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.

8. Mar 10, 2011

### Kreizhn

Okay, so then that hypothesis goes out the window.

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

9. Mar 10, 2011

### Dick

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.

10. Mar 10, 2011

### Kreizhn

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 $\mathcal H$ of dimension d, and fix an orthonormal basis $\left\{ e_i \right\}$. Let $P_i:\mathcal H \to \mathcal H$ represent projection operators taking each element of $\mathcal H$ to the one dimensional space spanned by $e_i$. In particular then, the set of $P_i$ span a d-dimensional subspace of $\mathcal B(\mathcal H)$, the set of bounded linear operators on $\mathcal H$. Furthermore since the $P_i$ decompose $\mathcal H$ into a direct sum of the orthgonal subspaces, it follows that
$$\sum_{i=1}^d P_i = \text{id}$$
where id is the identity operator in $\mathcal B(\mathcal H)$. I now have a paper in front of me saying that the set $\left\{ P_i - \text{id} \right\}$ spans a d-1 dimensional subspace of $\mathcal B(\mathcal H)$ and I am not certain why this is true.

11. Mar 10, 2011

### Kreizhn

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?

12. Mar 10, 2011

### Dick

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.

13. Mar 10, 2011

### Kreizhn

Thanks Dick, you've been quite helpful.

I checked this too, and in the case when $\mathcal H = \mathbb C^d$ 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 $d-1$ 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 $\left\{ P_i - \text{id} \right\}$ 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?

14. Mar 10, 2011

### Dick

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??

15. Mar 10, 2011

### Kreizhn

Very nice, thank you.