Cross Product

[DanielFaraday]

Homework Statement

Which of the following functions is not an inner product in $$\mathbb{R}^2$$, where u=(a, b), v=(c,d)?

a. <u,v> = ac+bd
b. <u,v> = 4ad+7bc
c. <u,v> = ac/2+bd/3
d. <u,v> = 2ac+9bd

Homework Equations

None

The Attempt at a Solution

u x v = det(u v) = ad-bc

I don't see how ANY of these answers could be right. Any ideas?

 

[Office_Shredder]

Inner product, not cross product. Look up the axioms for what a function has to satisfy to be an inner product

 

[DanielFaraday]

Inner product, not cross product. Look up the axioms for what a function has to satisfy to be an inner product

Oops! I forgot the definition of inner product - for some reason I was thinking of cross products.

Okay, so <u,v> = ac+bd, which means b is the correct answer because it doesn't have this form (the others have this form but are simply multiplied by constants).

Thanks!

 

[Office_Shredder]

For your final answer you'll actually need to prove either all three parts of the definition are satisfied, or demonstrate that one of them isn't for each one

For example

ac-bd is not an inner product since <(1,1), (1,1)> = 0 even though it's of the 'form' ac+bd with constants

 

[DanielFaraday]

For your final answer you'll actually need to prove either all three parts of the definition are satisfied, or demonstrate that one of them isn't for each one

For example

ac-bd is not an inner product since c even though it's of the 'form' ac+bd with constants

I don't quite follow you. Where does the <(1,1), (1,1)> = 0 come from? And doesn't <(1,1), (1,1)> = 2?

 

[DanielFaraday]

Isn't it enough just to say that 4ad+7bc is impossible because there is no way that a and b could be multiplied by d and c (respectively)?

 

[cipher42]

You have the standard inner product stuck in your head. It's true that in Euclidean-space (the usual space we're used to working with), the inner product is defined as:
$$<(x_1,x_2),(y_1,y_2)>=x_1y_1+x_2y_2$$
but this is just one possible definition of the inner product. It turns out that you can define lots of different inner products (in terms of the vector's components) as long as you follow three rules:
1) Symmetry: $<\mathbf{x},\mathbf{y}>=<\mathbf{y},\mathbf{x}>$
2) Linearity: $<a\mathbf{x}+b\mathbf{y},\mathbf{z}>=a<\mathbf{x},\mathbf{z}>+b<\mathbf{y},\mathbf{z}>$
3) Positive-Definiteness: $<\mathbf{x},\mathbf{x}>\geq 0, \mathbf{x}\neq 0$

You need to check to see if those different inner products satisfy these conditions.

 

[DanielFaraday]

You have the standard inner product stuck in your head. It's true that in Euclidean-space (the usual space we're used to working with), the inner product is defined as:
$$<(x_1,x_2),(y_1,y_2)>=x_1y_1+x_2y_2$$
but this is just one possible definition of the inner product. It turns out that you can define lots of different inner products (in terms of the vector's components) as long as you follow three rules:
1) Symmetry: $<\mathbf{x},\mathbf{y}>=<\mathbf{y},\mathbf{x}>$
2) Linearity: $<a\mathbf{x}+b\mathbf{y},\mathbf{z}>=a<\mathbf{x},\mathbf{z}>+b<\mathbf{y},\mathbf{z}>$
3) Positive-Definiteness: $<\mathbf{x},\mathbf{x}>\geq 0, \mathbf{x}\neq 0$

You need to check to see if those different inner products satisfy these conditions.

Thanks for your input. It was very helpful. Tell me if my logic is correct:

The correct answer is b since it fails the symmetry test below.

$$\langle u,v\rangle = 4ad+7bc = 4da+7cb = \langle (d,c),(b,a)\rangle \neq \langle v,u\rangle$$

Is this correct?

 

[Office_Shredder]

Thanks for your input. It was very helpful. Tell me if my logic is correct:

The correct answer is b since it fails the symmetry test below.

$$\langle u,v\rangle = 4ad+7bc = 4da+7cb = \langle (d,c),(b,a)\rangle \neq \langle v,u\rangle$$

Is this correct?

You didn't actually show that symmetry fails, you just said it did. You have to find two vectors u and v, and show that <u,v> is not equal to <v,u>

 

[DanielFaraday]

You didn't actually show that symmetry fails, you just said it did. You have to find two vectors u and v, and show that <u,v> is not equal to <v,u>

What about this. My two vectors are u=(a,b) and v=(c,d), assuming a, b, c, and d are unique real numbers.

<u,v> = 4ad+7bc
<v,u> = 4cb+7da

Since 4ad+7bc is not equal to 4cb+7da, symmetry fails.

P.S. I still think I was right in the first place. The only reason why this fails is because it doesn't follow the form (first*first)+(last*last).

 

[Office_Shredder]

You still haven't shown anything. How do you know 4ad+7bc=/=4cb+7da? Pick two ACTUAL vectors, with ACTUAL numbers. It doesn't need to be that complicated.

And I demonstrated how something of the form (first*first) + (last*last) can not be an inner product: ac-bd. Then the inner product of (1,1) with itself comes out to

1*1-1*1 = 1-1=0. So while your intuition was correct, you need a more formal proof because the way you described was, in its entirety, incorrect

 

[DanielFaraday]

Let u=(1,2) and v=(3,4):

$$<u,v>=4 (1*4)+7 (2*3)=58$$

$$<v,u>=4 (3*2)+7 (4*1)=52$$

$$58\neq 52$$

 

[cipher42]

There you go! Now that looks like a compelling reason why that is not a well-defined inner product.

 

[DanielFaraday]

Okay, I finally get it. Thanks for all your help!

 

[Mark44]

Let u=(1,2) and v=(3,4):

$$<u,v>=4 (1*4)+7 (2*3)=58$$

$$<v,u>=4 (3*2)+7 (4*1)=52$$

$$58\neq 52$$

Your conclusion shouldn't be that $58\neq 52$, which is obvious, but that $<u,v> \neq <v,u>$