Proving Orthogonal Bases Homework Statement

[LosTacos]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

Let B = (b1,...,bk)
Express v1 and v2 as linear combinations of the vectors in B:
v1 = a1v1 + a2v2 + ... + akvk
v2 = b1v1 + b2v2 + ... + bkvk
I am confused as to where to go from here.

 

[Fredrik]

It seems like a terrible idea to use the same notation for the first two basis vectors as for the two arbitrary vectors. Other than that, good start. What happens if you use what you wrote down to evaluate $v_1\cdot v_2$?
$$v_1\cdot v_2=\cdots $$

 

[LosTacos]

Okay so let the first basis be x = x1v1 + x2v2 + ... + xkvk and the second basis be y = y1v1 + y2v2 ... + ykvk. Then, the dot product of xy = (x1y1)v1 + (x2y2)v2 + ... + (xkyk)vk

 

[Fredrik]

Okay so let the first basis be x = x1v1 + x2v2 + ... + xkvk and the second basis be y = y1v1 + y2v2 ... + ykvk.

Uhm...I assume that you meant to say that you now intend to use the notation x and y for the two arbitrary vectors in V.

Then, the dot product of xy = (x1y1)v1 + (x2y2)v2 + ... + (xkyk)vk

The result of a dot product is a number, but what you wrote here is a vector.

Could you please use sub tags for the indices? Like this: x₂. Hit the quote button to see how I typed this. And here's a dot that you can copy and paste: ·

 

[LosTacos]

Okay, well since B is an orthonormal basis, then the set of all its vecors are unit vectors. So when taking the dot product of xy, I must turn them into unit vectors:
xy = [(x1y1)v1]/IIv1II + [(x2y2)v2]/IIv2II + ... + [(xkyk)vk]/IIvkII.
But, since B is orthnormal:
IIvkII = 1

 

[Fredrik]

Your notation is pretty hard to read, and that's still a vector.

 

[LosTacos]

Sorry, xy = [x₁y₁]v₁ / (Unit vector v1) + [x₂y₂]v₂ / (Unit vector v2) + ...+ [x_ky_k]v_k (Unit vector vk) = 1

 

[Fredrik]

That's easier to read, but it's still wrong. As I've been saying, the result of a dot product isn't a vector, and what you're writing is. Also, you could write ||v₁|| for the norm of v₁. It's wrong to call it "unit vector v₁".

 

[LosTacos]

I am confused as to how the dot product works here. So i am going to try it from the other direction. Since B is an orthonormal basis, [v1]_b = [v₁·v₁, + v₁·v₂, + ...+ v₁·v_k

Then, [v2]_b = [v₂·v₁, + v₂·v₂, + ...+ v₂·v_k

So how could I represent teh dot product of both of these?

 

[Mark44]

Let B = (b1,...,bk)
Express v1 and v2 as linear combinations of the vectors in B:
v1 = a1v1 + a2v2 + ... + akvk
v2 = b1v1 + b2v2 + ... + bkvk

As Fredrik already noted, your notation is very confusing. Here's my attempt to unravel it.
v₁ = a₁b₁ + a₂b₂ + ... + a_kb_k
Here, v₁ is written as a linear combination of the basis vectors b₁, b₂, ... , b_k.
v₂ = c₁b₁ + c₂b₂ + ... + c_kb_k
Here, v₂ is written as a different linear combination of the same basis vectors b₁, b₂, ... , b_k.

Now, what is v₁ $\cdot$ v₂?

 

[Fredrik]

I am confused as to how the dot product works here. So i am going to try it from the other direction. Since B is an orthonormal basis, [v1]_b = [v₁·v₁, + v₁·v₂, + ...+ v₁·v_k

Then, [v2]_b = [v₂·v₁, + v₂·v₂, + ...+ v₂·v_k

So how could I represent teh dot product of both of these?

I can't imagine a more confusing notation than this

You started at the right end before, with an OK notation. You just need to use what you know about of the dot product to evaluate what you get when you replace the x and the y in x·y with the corresponding linear combinations of elements of B.

 

[LosTacos]

v₁ ⋅ v₂ = a₁c₁(b₁) + a₂c₂(b₂) + ... + a_kc_k(b_k)

Since B = (b₁, b₂, ... , b_k) is the orthonormal basis,

v₁ ⋅ v₂ = [v₁]_B ⋅ [v₂]_B

 

[Fredrik]

Still wrong, in the same way as before. What you're writing there is a vector, but $v_1\cdot v_2$ is not a vector, so what you wrote can't possibly be right. Think about what you know about the dot product, and write out the intermediate steps in your calculation.

 

[LosTacos]

I understand v₁ ⋅ v₂ is not a vector. I don't know how to represent it. I understand that v₁ ⋅ v₂ = the sum of (1st element of v₁)x(1st element of v₂) + (2nd element of v₁) x (2nd element of v₂) + ... + (kth element of v₁)x(kth element of v₂)

 

[Mark44]

Suppose u = u₁i + u₂j + u₃k, and
v = u₁i + u₂j + u₃k

What is u $\cdot$ v? In the above, i, j, and k are the usual unit vectors in R³. Note that they are an orthonormal set.

 

[LosTacos]

u ⋅ v = (u₁)(u₁) + (u₂)(u₂) + (u₃)(u₃)

 

[Fredrik]

Do you know any theorems about the properties of the dot product? For example, if x,y,z are vectors and k is a number, what can you tell me about these expressions?
\begin{align}x\cdot (y+z)=?\\
(kx)\cdot y=?
\end{align}

 

[LosTacos]

x⋅(y+z) = x⋅y + x⋅z

(kx)⋅y = k(x⋅y)

 

[Mark44]

u ⋅ v = (u₁)(u₁) + (u₂)(u₂) + (u₃)(u₃)

Yes, that's correct. Do you understand why there are no i$\cdot$ j or i$\cdot$ k (etc.) terms? Or why you don't need to include the i$\cdot$ i and j$\cdot$ j (etc.) terms?

Think about this when you're calculating v₁ $\cdot$ v₂. As Fredrik already said, this dot product represents a number, so should not involve any vectors.

 

[Fredrik]

x⋅(y+z) = x⋅y + x⋅z

(kx)⋅y = k(x⋅y)

Exactly right. So what do you get if you use these rules to evaluate $v_1\cdot v_2$? (First express $v_1$ and $v_2$ as linear combinations of the basis vectors. Then use these rules).

 

[LosTacos]

v₁ = a₁b₁ + a₂b₂ + ... + a_kb_k

v₂ = c₁b₁ + c₂b₂ + ... + c_kb_k

v1⋅v2 = a₁c₁ + a₂c₂ + ... + a_kc_k

 

[Fredrik]

This is correct, but I would like to see an intermediate step that shows that you know why that last equality holds.

 

[LosTacos]

v1⋅v2 = a₁c₁(b₁) + a₂c₂(b₂) + ... + a_kc_k(b_k)

= a₁c₁(1) + a₂c₂(1) + ... + a_kc_k(1)

 

[Fredrik]

No, that first equality is wrong. Each of those terms is a vector, so their sum is too. This is assuming that (b₁) means b₁. I don't know what else it could mean.

In the first step, you should just use the first two equalities from post #21, nothing else. In the second step, you should use the rules I mentioned.

 

[Mark44]

You didn't answer the questions I asked earlier.

Do you understand why there are no i$\cdot$ j or i$\cdot$ k (etc.) terms? Or why you don't need to include the i$\cdot$ i and j$\cdot$ j (etc.) terms?

v1⋅v2 = a₁c₁(b₁) + a₂c₂(b₂) + ... + a_kc_k(b_k)

= a₁c₁(1) + a₂c₂(1) + ... + a_kc_k(1)

 

[LosTacos]

v1⋅v2 = a₁b₁ + a₂b₂ + ... + a_kb_k
v2 = b₁c₁ + b₂c₂ + ... + b_kc_k

v1⋅v2 = (a₁b₁)c₁ + (a₂b₂)c₂ + ... + (a_kb_k)c_k

v1⋅v2 = (a₁c₁)b₁ + (a₂c₂)b₂ + ... + (a_kc_k)b_k

 

[Mark44]

Now you're just guessing...

 

[LosTacos]

I used the fact that (kx)⋅y = k(x⋅y). Since the basis is orthonormal, all the vectors are unit vectors. Therefore, they are each = 1. So, I took the fact that 1 times xy = xy

 

[Mark44]

v1⋅v2 = a₁b₁ + a₂b₂ + ... + a_kb_k
v2 = b₁c₁ + b₂c₂ + ... + b_kc_k

If you're following Fredrik's suggestion, the first equation above should be
v₁ = a₁b₁ + a₂b₂ + ... + a_kb_k

The second equation would be better written as
v₂ = c₁b₁ + c₂b₂ + ... + c_kb_k

(I added bolding to make the vectors more obvious.)

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

Now what do you get, showing the intermediate steps?

v1⋅v2 = (a₁b₁)c₁ + (a₂b₂)c₂ + ... + (a_kb_k)c_k

v1⋅v2 = (a₁c₁)b₁ + (a₂c₂)b₂ + ... + (a_kc_k)b_k

I used the fact that (kx)⋅y = k(x⋅y). Since the basis is orthonormal, all the vectors are unit vectors. Therefore, they are each = 1. So, I took the fact that 1 times xy = xy

No, their magnitudes are 1. They are all still vectors, so they can't be equal to the number 1. For example, |b₂| = 1, but b₂ ≠ 1.

 

[LosTacos]

v1⋅v2 = (a₁b₁ + a₂b₂ + ... + a_kb_k)⋅ (b₁c₁ + b₂c₂ + ... + b_kc_k)
= ((a₁b₁)(b₁c₁) + (a₂b₂)(b₂c₂) + ... + (a_kb_k)(b_kc_k)

 

[Mark44]

v1⋅v2 = (a₁b₁ + a₂b₂ + ... + a_kb_k)⋅ (b₁c₁ + b₂c₂ + ... + b_kc_k)
= ((a₁b₁)(b₁c₁) + (a₂b₂)(b₂c₂) + ... + (a_kb_k)(b_kc_k)

Since each vector has k terms, there should be k² products. I see only k of them. You're missing most of them. Also, which things are vectors in what you wrote. I made an extra effort to indicate the things that were vectors in my last post.

 

[LosTacos]

The b's would be squared because they are in each vector.

 

[Fredrik]

v1⋅v2 = (a₁b₁ + a₂b₂ + ... + a_kb_k)⋅ (b₁c₁ + b₂c₂ + ... + b_kc_k)
=

Would you know how to handle this if we (temporarily) denote the first factor by x?

v1⋅v2 = (a₁b₁ + a₂b₂ + ... + a_kb_k)⋅ (b₁c₁ + b₂c₂ + ... + b_kc_k)
= x·(b₁c₁ + b₂c₂ + ... + b_kc_k) =

Edit: I just noticed something that Mark44 has already made a comment about. The order of the bs and the cs there is pretty odd. We usually write the number first, and the vector second. So it would be better to write $c_1b_1$ instead of $b_1c_1$, and it would be even better to use a notation like $c_1\mathbf b_1$.

 

[LosTacos]

Yes, it would be (b₁x)c₁ + (b₂x)c₂ + ... +(b₁x)c_k

 

[Fredrik]

Yes, but it would make more sense to write it as $c_1 x\cdot b_1+\cdots+c_k x\cdot b_k$. Now consider an arbitrary term from this sum, $c_i x\cdot b_i$. What will this term look like if you use the definition of x?

 

[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)