Proving Orthogonal Bases Homework Statement

[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.

 

[Mark44]

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

What you have here is the right side of the equation you're trying to prove. IOW, what you have shown is that [v₁]B · [v₂]B = (a₁c₁ + a₂c₂ + ... +
a_kc_k.

What you need to do is work with the other side of the equation - v₁ ⋅ v₂ - taking into consideration that, although v₁ and v₂ are vectors in the k-dimensional subspace V, they are also vectors in the n-dimensional vector space Rⁿ.

 

[Fredrik]

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

This part would require more explanation. (Edit: Actually, it's wrong. See below). Of course if you do the calculation one step at a time, then all you have to do to "go in the other direction" is to read the string of equalities from right to left.

 

[Mark44]

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

This part would require more explanation. Of course if you do the calculation one step at a time, then all you have to do to "go in the other direction" is to read the string of equalities from right to left.

The equation above doesn't make sense to me. I'm reading the right side as a vector, which doesn't make sense as the output of a dot product.

 

[Fredrik]

The equation above doesn't make sense to me. I'm reading the right side as a vector, which doesn't make sense as the output of a dot product.

Ah, yes you're right. I was too fast on the trigger there.

 

[Mark44]

Ah, yes you're right. I was too fast on the trigger there.

I can say from personal experience, it happens.

 

[LosTacos]

I am confused as to what is correct for the reverse direction.

From the definition of coordinatization,
Let B = (b₁, b₂, ..., b_k) be an ordered basis. Suppose v₁= a₁b₁ + a₂b₂ + ... + a_nb_n. Then, [v₁]_B, the coordinatization of v₁ with respect to B is the n-vector [a₁, a₂, ..., a_n]

So if this follows true for [v₂]_B as well, the dot product will give me

[v₁]_B⋅[v₁]_B = [a₁, a₂, ..., a_n] ⋅ [c₁, c₂, ..., c_n].

So, how when doing the dot product, why do the non-identical terms cancel out. OR is this not the right approach.

 

[Fredrik]

I am confused as to what is correct for the reverse direction.

From the definition of coordinatization,
Let B = (b₁, b₂, ..., b_k) be an ordered basis. Suppose v₁= a₁b₁ + a₂b₂ + ... + a_nb_n. Then, [v₁]_B, the coordinatization of v₁ with respect to B is the n-vector [a₁, a₂, ..., a_n]

So if this follows true for [v₂]_B as well, the dot product will give me

[v₁]_B⋅[v₁]_B = [a₁, a₂, ..., a_n] ⋅ [c₁, c₂, ..., c_n].

So, how when doing the dot product, why do the non-identical terms cancel out. OR is this not the right approach.

This is fine, but it's much harder to see what the next step is when you start at this end. If you want to see what the next step is, all you have to do is to write out all the steps of the previous calculation, the one that started with

v₁ ⋅ v₂ = ...

and ended with

= [v₁]_B ⋅ [v₂]_B. Now you can just read this string of equalities from right to left, and you should see it. (It might still be somewhat hard to see it, because of the strange way you have continued to write the calculation, in spite of what I said in #43).

 

[LosTacos]

Well to prove that each side is equal to one another, I have to prove that each is a subset of the other. In essence, prove it both ways. So when doing the dot product of [a₁, a₂, ..., a_n] ⋅ [c₁, c₂, ..., c_n] are you saying that i need to create 9 different products for this example. And then bc it is orthonormal, 6 of them cancel out and the other 3 are left which then show that this equals v₁ ⋅ v₂

 

[Fredrik]

Well to prove that each side is equal to one another, I have to prove that each is a subset of the other. In essence, prove it both ways.

I was wondering why you were talking about "both ways". You don't have to do anything like that here. It's true that every equality in mathematics (at least in the branch of mathematics defined by ZFC set theory) is an equality between sets, and that this means that the equality

v₁ ⋅ v₂ = [v₁]_B ⋅ [v₂]_B​

holds if and only if every element of the left-hand side is an element of the right-hand side, and vice versa.

This doesn't mean that you haven't already proved the equality above. You have. You did it by proving a string of equalities

v₁ ⋅ v₂ = X = Y = Z = [v₁]_B ⋅ [v₂]_B.​

This is sufficient since equality is a transitive operation. (If x=y and y=z, then x=z).

Another way of looking at it is that the first of these equalities means that every element of v₁ ⋅ v₂ is an element of X and that every element of X is an element of v₁ ⋅ v₂. That's just what the equality sign means. The other equalities can be interpreted similarly. Because of this, there's no need to do anything more than to prove this string of equalities.

I thought that what you wanted to do was just to start with [v₁]_B ⋅ [v₂]_B, and here use the definitions of [v₁]_B, [v₂]_B and the dot product, to discover each step in the string of equalities in the opposite order compared to before. This is why I said is that all you need to do to see the steps is to read the string of equalities from right to left.

 

[Fredrik]

Did you understand that explanation? Is everything clear now? Do you understand why we found the way you wrote the solution confusing?

Since you solved the problem, I'm going to show you how I would have done it. I typed this up a week ago. I just didn't want to post it until you had worked out the solution for yourself.

Define $v_1,\cdots,v_k\in V$ by $B=(v_1,\cdots,v_n)$. Let $x,y\in V$ be arbitrary. Let $x_1,\dots,x_k$ and $y_1,\dots,y_k$ be the unique real numbers such that $x=\sum_{i=1}^k x_i v_i$ and $y=\sum_{i=1}^k y_i v_i$. We have
$$x\cdot y=\bigg(\sum_{i=1}^k x_i v_i\bigg)\cdot\bigg(\sum_{j=1}^k y_j v_j\bigg)=\sum_{i=1}^k\sum_{j=1}^k x_i y_j \underbrace{v_i\cdot v_j}_{=\delta_{ij}}=\sum_{i=1}^k x_i y_i =[x]_B\cdot[y]_B.$$
In case you're not familiar with the Kronecker delta, it's defined by
$$\delta_{ij}=\begin{cases}1 &\text{if }i=j\\0 &\text{if }i\neq j.\end{cases}$$