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 the 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 the 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)