Proving Orthogonal Bases Homework Statement

[Mark44]

Let's try it again -

x $\cdot$ (c₁b₁ + c₂b₂ + ... + c_kb_k) = ?

Please distinguish between vectors and scalars. You're writing them all the same, so they are probably becoming an amorphous mush in your mind.

Don't use parentheses.

 

[LosTacos]

It would be the sum of (c_ix)⋅b_i

 

[Mark44]

It would be the sum of (c_ix)⋅b_i

Which things in this expression are vectors? Aside from this, this is OK.

Now what about this?
So v₁ $\cdot$ v₂ = (a₁b₁ + a₂b₂ + ... + a_kb_k) $\cdot$ (c₁b₁ + c₂b₂ + ... + c_kb_k)

Again, note that I am making clear which things are vectors and which things are scalars. Please try to follow this pattern. And we're looking for the intermediate steps, not necessarily the final result.

 

[Fredrik]

If it makes things easier for you, you can start by using the definition

x=a₁b₁ + a₂b₂ + ... + a_kb_k

to continue this:

(c_ix)·b_i =c_i (x·b_i) = ...

As Mark said, we're more interested in the intermediate steps than in the final result.

 

[LosTacos]

v₁ ⋅ v₂ = (a₁b₁ + a₂b₂ + ... + a_k b_k) ⋅ (c₁b₁ + c₂b₂ + ... + c_k b_k)
= (a₁b₁ ⋅ c₁b₁) + (a₂b₂⋅c₂b₂) + ... + (a_kb_k ⋅ c_kb_k)

= (a₁c[/B]₁ b₁b₁) + (a₂c[/B]₂ b₂b₂) + ... + (a_kc[/B]_k b_kb_k)

= sum of (a_ic_i)b_i

= sum of (a_i)b_i + sum of (c_i)b_i

= [v₁]_B ⋅ [v₂]_B

 

[Fredrik]

v₁ ⋅ v₂ = (a₁b₁ + a₂b₂ + ... + a_k b_k) ⋅ (c₁b₁ + c₂b₂ + ... + c_k b_k)
= (a₁b₁ ⋅ c₁b₁) + (a₂b₂⋅c₂b₂) + ... + (a_kb_k ⋅ c_kb_k)

You are close to a complete solution now, but we are still interested in how you're getting from the first line to the second line. In particular, why are there only k terms in the second line?

 

[LosTacos]

I don't understand. Back on #16 u ⋅ v = (u₁)(u₁) + (u₂)(u₂) + (u₃)(u₃)[/QUOTE]

And there are k terms because B is the orthonormal basis for a k-dimensional subspace

 

[Fredrik]

#16 is just the definition of the dot product on $\mathbb R^3$. It doesn't have a lot to do with this.

Let's consider a simpler problem. Suppose that $\{u,v\}$ is an orthonormal basis for $\mathbb R^2$. Then what is $u\cdot(u+v)$? Obivously, we have
$$u\cdot (u+v)=u\cdot u+u\cdot v=1+0=1.$$ Our objection is that since your calculation doesn't have any terms like $u\cdot v$, we can't see if you used the properties of the dot product correctly.

What you're doing is like writing the above as
$$u\cdot (u+v)=u\cdot u.$$ This looks strange, because if we haven't yet used (any part of) the orthonormality of the set {u,v}, the result is $u\cdot u+u\cdot v$, and if we have used the orthonormality, we can write the result as 1+0.

It looks a lot better if you use only the properties of the dot product in the first step, and then use the orthonormality in the second. What you did in the first step was (I hope) to use the properties of the dot product, and that B is an orthogonal set (while not using the other aspect of orthonormality). This isn't wrong, but it hides the fact that you understand what you're doing.

It would have been OK if you had added a comment like this: "In step 1, I'm using the properties of the dot product, and that B is an orthogonal set. In step 2, I'm using that the members of B are unit vectors."

 

[LosTacos]

What do you mean that in step 2 you are using the members of B as unit vectors? And what was wrong with k elements?

 

[Fredrik]

I'm not sure I can explain that any clearer than in my previous post. Was I using terms that you don't know the definition of? In that case, please ask about the terms you don't know, and then read my previous post again.

A unit vector is a vector with norm 1.

Do you understand that for all vectors u and v, we have (u+v)·(u+v) = u·u + u·v + v·u + v·v? This holds even if {u,v} is an orthonormal basis for $\mathbb R^2$. Note that there are 4 terms, not 2 terms.

 

[LosTacos]

Okay I think I understand.

v₁ ⋅ v₂ = (a₁b₁ + ... + a_k b_k) ⋅ (c₁b₁ + ... + c_k b_k)

= (a₁b₁ ⋅ c₁b₁) + (a₁b₁ ⋅ c_kb_k) + (a_kb_k ⋅ c₁b₁) + (a_kb_k ⋅ c_kb_k)

=
(a₁c₁ b₁b₁) +(a₁c_k b₁b_k) +(a_kc₁ b_kb₁) +
(a_kc_k b_kb_k)

=
(a₁c₁ b₁b₁) + (a₁c_k⋅ 0) +
(a_kc₁⋅ 0) +
(a_kc_k b_kb_k)

=
(a₁c₁ b₁b₁) +
(a_kc_k b_kb_k)
= sum of (a_ic_i)b_i

= sum of (a_i)b_i + sum of (c_i)b_i

= [v₁]_B ⋅ [v₂]_B

 

[Fredrik]

Can you please edit those B tags? This is too hard to read. [noparse]Start with or , end with or .[/noparse] You can edit your post for 11 hours and 40 minutes.

 

[Mark44]

Okay I think I understand.

v₁ ⋅ v₂ = (a₁b₁ + ... + a_k b_k) ⋅ (c₁b₁ + ... + c_k b_k)

= (a₁b₁ ⋅ c₁b₁) + (a₁b₁ ⋅ c_kb_k) + (a_kb_k ⋅ c₁b₁) + (a_kb_k ⋅ c_kb_k)

The B tags come in pairs - a B tag at the start of what you want to bold, and an /B tag at the end of what you want to bold. The easiest way is to use the B icon on the menu you get when you click Go Advanced (if the menu isn't already showing). A tag by itself doesn't do anything.

Your equation above isn't right, as it is missing most of the terms.

You're getting warmer on the work below, but it is not correct.

=
(a₁c₁ b₁b₁) +(a₁c_k b₁b_k) +(a_kc₁ b_kb₁) +
(a_kc_k b_kb_k)

=
(a₁c₁ b₁b₁) + (a₁c_k⋅ 0) +
(a_kc₁⋅ 0) +
(a_kc_k b_kb_k)

=
(a₁c₁ b₁b₁) +
(a_kc_k b_kb_k)



= sum of (a_ic_i)b_i

= sum of (a_i)b_i + sum of (c_i)b_i

= [v₁]_B ⋅ [v₂]_B

 

[Mark44]

Let's make a simpler problem that will help you with this one, since there are some important concepts you aren't understanding.

Suppose u and v are in R³, and we have an orthonormal basis {x₁, {x₂, {x₃}.

Then we can write u and v in this way:
u = u₁x₁ + u₂x₂ + u₃x₃
v = v₁x₁ + v₂x₂ + v₃x₃

The dot product of u and v is:
u $\cdot$ v = (u₁x₁ + u₂x₂ + u₃x₃) $\cdot$ (v₁x₁ + v₂x₂ + v₃x₃)

Please do the calculation for me, showing all nine terms in this product, each of which will include a dot product. Also, please use a dot ($\cdot$) in all nine intermediate dot products.

 

[LosTacos]

Would I create an i where 1 < i < k, and then all the terms would cancel out bc they would equal zero except the term (a_ic_i )⋅ (b_ib_i)

 

[LosTacos]

(u₁x₁)⋅(v₁x₁) +
(u₁x₁)⋅(v₂x₂) +
(u₁x₁)⋅(v₃x₃) +
(u₂x₂)⋅(v₁x₁) +
(u₂x₂)⋅(v₂x₂) +
(u₂x₂)⋅(v₃x₃) +
(u₃x₃)⋅(v₁x₁) +
(u₃x₃)⋅(v₂x₂) +
(u₃x₃)⋅(v₃x₃)

 

[Mark44]

(u₁x₁)⋅(v₁x₁) +
(u₁x₁)⋅(v₂x₂) +
(u₁x₁)⋅(v₃x₃) +
(u₂x₂)⋅(v₁x₁) +
(u₂x₂)⋅(v₂x₂) +
(u₂x₂)⋅(v₃x₃) +
(u₃x₃)⋅(v₁x₁) +
(u₃x₃)⋅(v₂x₂) +
(u₃x₃)⋅(v₃x₃)

Good - now keep going, but let's go at it in baby steps. To help you along, the first product above can be written as
u₁v₁x₁⋅x₁
This is NOT u₁v₁x₁!
What's the simplest way to write this product?

What does the second product in the list above, in simplest form?

 

[LosTacos]

u₁v₁x₁⋅x₁ = (u₁v₁)(x₁)^2

u₁v₂x₁⋅x₂ = (u₁v₂)(0)

 

[micromass]

u₁v₁x₁⋅x₁ = (u₁v₁)(x₁)^2

u₁v₂x₁⋅x₂ = (u₁v₂)(0)

Why is $\mathbf{x}_1 \cdot \mathbf{x}_1 = 0$?

 

[LosTacos]

Because it is an orthogonal set.

 

[micromass]

Because it is an orthogonal set.

I think you mean orthonormal. What is the definition of an orthonormal set?

 

[Mark44]

u₁v₁x₁⋅x₁ = (u₁v₁)(x₁)^2

No, you can't square a vector. What is x₁ ⋅ x₁? This question gets to the heart of the problems you're having.

u₁v₂x₁⋅x₂ = (u₁v₂)(0)

Yes.

 

[Fredrik]

No, you can't square a vector.

I think the notation $\mathbf x^2=\mathbf x\cdot\mathbf x$ is not uncommon. My objection to it here is that it's just the same thing in a different notation, so it doesn't bring us closer to the final answer.

 

[LosTacos]

Orthonormal set is set of vectors where all are unit vectors.

 

[Mark44]

Orthonormal set is set of vectors where all are unit vectors.

There's more.

 

[LosTacos]

x₁ ⋅ x₁ = 1

 

[Mark44]

I think the notation $\mathbf x^2=\mathbf x\cdot\mathbf x$ is not uncommon.

I don't think this shortcut is helpful to the OP's understanding. I am not convinced that the OP would be able to distinguish between, say, c₁² and x²; i.e., that there are different kinds of multiplication occurring.

My objection to it here is that it's just the same thing in a different notation, so it doesn't bring us closer to the final answer.

 

[Mark44]

So what does this simplify to?
(u₁x₁)⋅(v₁x₁) +
(u₁x₁)⋅(v₂x₂) +
(u₁x₁)⋅(v₃x₃) +
(u₂x₂)⋅(v₁x₁) +
(u₂x₂)⋅(v₂x₂) +
(u₂x₂)⋅(v₃x₃) +
(u₃x₃)⋅(v₁x₁) +
(u₃x₃)⋅(v₂x₂) +
(u₃x₃)⋅(v₃x₃)

 

[LosTacos]

(u₁v₁) + (u₂v₂) + (u₃v₃)

 

[Mark44]

(u₁v₁) + (u₂v₂) + (u₃v₃)

Right.

Now back to the original problem:

So v₁ $\cdot$ v₂ = (a₁b₁ + a₂b₂ + ... + a_kb_k) $\cdot$ (c₁b₁ + c₂b₂ + ... + c_kb_k)

When the dot product on the right is calculated, how many terms will there be, before simplification?
How many terms will there be after simplification.

 

[LosTacos]

v₁ ⋅ v₂ = (a₁b₁ + a₂b₂ +... + a_k b_k) ⋅ (c₁b₁ +c₂b₂ ... + c_k b_k)

= (a₁c₁b₁b₁) + (a₂c₂b₂b₂) + ... +
(a_kc_kb_kb_k)

= (a₁c₁ + a₂c₂ + ... +
a_kc_k

The only terms that simplify are the ones where bi = bi. If bi does not equal bi, then it will be multiplied by 0 and the dot product will cancel.

 

[Mark44]

OK, good. We've been struggling to get you to realize that b_i $\cdot$ b_i = 1 and b_i $\cdot$ b_j = 0 if i ≠ j.

Now you're in a better position to tackle the problem you posted in #1. It would be nice if you restated the problem.

 

[LosTacos]

Problem: Let B be an ordered orthonormal basis for a k-dimensional subspace V of ℝn. Prove that for all v1,v2 ∈ V, v1·v2 = [v1]B · [v2]B, where the first dot product takes place in ℝn and the second takes place in ℝk.

Okay so:

v₁ ⋅ v₂ = (a₁b₁ + a₂b₂ +... + a_k b_k) ⋅ (c₁b₁ +c₂b₂ ... + c_k b_k)

= (a₁c₁b₁b₁) + (a₂c₂b₂b₂) + ... +
(a_kc_kb_kb_k)

= (a₁c₁ + a₂c₂ + ... +
a_kc_k

= [v₁]_B ⋅ [v₂]_B

 

[LosTacos]

Now for the opposite direction:

Let v₁ = a₁b₁ + ... + a_kb_k

Then,
[v₁]_B = [a₁, a₂, ... , a_k]

Let v₂ = c₁b₁ + ... + c_kb_k

Then,
[v₂]_B = [c₁, c₂, ... , c_k]

So,

[v₁]_B ⋅ [v₂]_B = [a₁c₁, a₂c₂, ... , a_kc_k]

Since each v is expressed as the coordinatization with respect to basis B, these can just be expanded to the linear combination of each and therefore = v₁⋅[v₂

 

[Fredrik]

Problem: Let B be an ordered orthonormal basis for a k-dimensional subspace V of ℝn. Prove that for all v1,v2 ∈ V, v1·v2 = [v1]B · [v2]B, where the first dot product takes place in ℝn and the second takes place in ℝk.

Okay so:

v₁ ⋅ v₂ = (a₁b₁ + a₂b₂ +... + a_k b_k) ⋅ (c₁b₁ +c₂b₂ ... + c_k b_k)

= (a₁c₁b₁b₁) + (a₂c₂b₂b₂) + ... +
(a_kc_kb_kb_k)

= (a₁c₁ + a₂c₂ + ... +
a_kc_k

= [v₁]_B ⋅ [v₂]_B

The calculation is correct, but the second equality (the step where you go from the first line to the second) looks very strange. Can you please make another attempt to understand my post #43? I think I explained it there.