Orthogonal Bases

1. May 8, 2013

LosTacos

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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.

2. May 8, 2013

Fredrik

Staff Emeritus
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 $v_1\cdot v_2$?
$$v_1\cdot v_2=\cdots$$

3. May 8, 2013

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

4. May 8, 2013

Fredrik

Staff Emeritus
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.

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: ·

5. May 8, 2013

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

6. May 8, 2013

Fredrik

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

7. May 8, 2013

LosTacos

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

8. May 8, 2013

Fredrik

Staff Emeritus
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".

9. May 8, 2013

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·vk

Then, [v2]b = [v2·v1, + v2·v2, + ......+ v2·vk

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

10. May 8, 2013

Staff: Mentor

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

Now, what is v1 $\cdot$ v2?

11. May 8, 2013

Fredrik

Staff Emeritus
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.

12. May 8, 2013

LosTacos

v1 ⋅ v2 = a1c1(b1) + a2c2(b2) + .... + akck(bk)

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

v1 ⋅ v2 = [v1]B ⋅ [v2]B

13. May 8, 2013

Fredrik

Staff Emeritus
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.

14. May 8, 2013

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)

15. May 8, 2013

Staff: Mentor

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.

16. May 8, 2013

LosTacos

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

17. May 8, 2013

Fredrik

Staff Emeritus
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}

18. May 8, 2013

LosTacos

x⋅(y+z) = x⋅y + x⋅z

(kx)⋅y = k(x⋅y)

19. May 8, 2013

Staff: Mentor

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.

20. May 8, 2013

Fredrik

Staff Emeritus
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).