Linearly Independent Sets After Subtraction

[Kreizhn]

Homework Statement

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.

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]<br /> and show that each a_i must necessarily be zero. Substituting the first equation into the other yields<br /> \begin{align*}\sum_i a_i (v_i - \sum_j b_j v_j ) &amp;amp;= \sum_i a_i - \sum_{i,j} a_i b_j v_j \\&lt;br /&gt; &amp;amp;= \sum_i \left( a_i - \sum_j a_j b_i \right) v_i &lt;br /&gt; \end{align*}<br /> where in the last step I&#039;ve switched the indices in the double summation. By linear independence of the v_i it follows that <br /> a_i = \sum_j a_j b_i<br /> but that&#039;s where I&#039;m stuck.<br /> <br /> It&#039;s possible that I&#039;m doing this the wrong way also. Any help would be appreciated.

 

[Kreizhn]

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

 

[Mark44]

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

Now look at the equation a₁(v₁ - b) + a₂(v₂ - b) + ... + a_n(v_n - b) = 0, where b != 0, and b != v_i, and show that this equation has only the trivial solution.

 

[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?

 

[Dick]

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?

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

 

[Kreizhn]

Okay, so in this case we get
\begin{align*}<br /> a_1(v_1 -\frac12 v_1 -\frac12 v_2) + a_2(v_2 - \frac12 v_1 -\frac12 v_2 ) &amp;= \frac{a_1}2 (v_1 -v_2) + \frac{a_2}2 (v_2 - v_1) \\<br /> &amp;= \frac12(a_1-a_2) v_1 + \frac12(a_2-a_1) v_2<br /> \end{align*}<br />

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

 

[Dick]

Okay, so in this case we get
\begin{align*}<br /> a_1(v_1 -\frac12 v_1 -\frac12 v_2) + a_2(v_2 - \frac12 v_1 -\frac12 v_2 ) &amp;= \frac{a_1}2 (v_1 -v_2) + \frac{a_2}2 (v_2 - v_1) \\<br /> &amp;= \frac12(a_1-a_2) v_1 + \frac12(a_2-a_1) v_2<br /> \end{align*}<br />

By linear independence of v_1,v_2 we get that a_1 = a_2. 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.

 

[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?

 

[Dick]

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?

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.

 

[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.

 

[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?

 

[Dick]

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.

 

[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?

 

[Dick]

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?

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??

 

[Kreizhn]

Very nice, thank you.