Homework Help: Hilbert, Inner Product

1. Mar 30, 2012

bugatti79

I have just realized that I accidently put it in wrong sub forum. This should be in 'calculus and beyond'.

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

Prove the function <x,y>=x_1y_1+x_2y_2+x_3y_3 defines an inner product space on the real vector space R^3 where x=(x1,x2,x3) and y=(y1,y2,y3)

3. The attempt at a solution

Axiom 1 <x,y> >=0 since we have that x_n and y_n for n=1,2,3 are in R

Axiom 2a <x,y> =x1y1+x2y2+x3y3, then <x,y>=0 iff x_n and y_n for n=1,2,3 both =0

Axiom 2b <ax,y>=a<x,y>

<ax,y> = ax1y1+ax2y2+ax3y3
= a(x1y1+x2y2+x3y3)
=a<x,y>

Axiom 3 <y,x>= complex of <x,y>

<y,x>=(y1x1+y2x2+y3x3) but y complex =y and x complex=x in R therefore
= (y1x1 complex+y2x2 complex +x3y3 complex)
=<y,x> complex
=<x,y> complex

Axiom 4 <x+y,z>=<x,z>+<y,z>, let z=(z1,z2,z3) in R^3

<x+y,z>=(x1+y1+x2+y2+x3+y3)(z1+z2+z3)
=x1z1 +x2z2+x3z3+y1z1+y2zy3z3
=<x,z>+<y,z>
...?

Last edited: Mar 30, 2012
2. Mar 30, 2012

Staff: Mentor

This isn't Axiom 1. It has to do with <x, x>.
Don't you mean "conjugate transpose"?
Since the underlying vector space is R3, all you need to show is that <x, y> = <y, x>.

3. Mar 30, 2012

bugatti79

Oh I see,

Axiom 1 <x,x> >=0 since we have that x_n for n=1,2,3 are in R

Yes, the bar on top of them.
Anyway,

<y,x>=y1x1+y2x2+y3x3
=x1y1+x2y2+x3y3
=<x,y>

Is axiom 4 ok?

Thanks

4. Mar 30, 2012

Ray Vickson

Your statement "Axiom 1 <x,y> >=0" is NOT a property of an inner product; we can have <x,y> > 0, = 0 or < 0. However, we do have <x,x> >= 0, with <x,x> = 0 iff x is the zero vector. Can you prove that?

Your statement "Axiom 2a <x,y> =x1y1+x2y2+x3y3, then <x,y>=0 iff x_n and y_n for n=1,2,3 both =0" is FALSE: <x,y> = 0 means only that the vectors x and y are perpendicular to one another.

Your statement "Axiom 3 <y,x>= complex of <x,y>" is meaningless; perhaps you mean "complex conjugate". Anyway, in real space, the components of x and y are all real and <x,y> is real; the property you state is trivially true, because x1*y1 + x2*y2 + x3*y3 = y1*x1 + y2*x2 + y3*x3.

Your statement of Axiom 4 is correct, but your proof is absolutely incorrect! Go back and read what you did.

RGV

5. Mar 30, 2012

Staff: Mentor

You're waving your arms here. Expand <x, x> using the definition of this inner product and it should be obvious why <x, x> >= 0.
Yes, #4 is fine.

6. Mar 30, 2012

bugatti79

<x,x>=x1x1+x2x2+x3x3>=0 where x_n>=0 for n=1,2,3 since x_n is in R

<x,x>=0 iff x1x1+x2x2+x3x3=0 ie iff x_nx_n=0 for n=1,2,3, iff x_n=0 for n=1,2,3, ie x=x_n=0 the 0 vector.

7. Mar 30, 2012

Staff: Mentor

No, you're still not getting it. Any or all of the coordinates of a given vector can be negative, such as x = [1, -2, 5]. Here x2 < 0, but it's easy to show that <x, x> > 0.
If the sum of three numbers is zero, why in this case is it necessarily true that all three numbers have to be zero?

8. Mar 30, 2012

bugatti79

The product of a negative number times a negative number gives a positive therefore <x,x> will always be greater than 0..?

Because if at least one of them is non 0 then <x,x> is non 0..?

9. Mar 30, 2012

Staff: Mentor

That's closer. For any real number x, x2 ≥ 0, and x2 = 0 iff x = 0.
Now, why is x12 + x22 + x32 ≥ 0? This is what you need to establish in order to verify the <x, x> ≥ 0

10. Mar 31, 2012

bugatti79

because we have that

|x1*x1|+|x2*x2|+|x3*x3|>=0 and |x1*x1|+|x2*x2|+|x3*x3|=0 iff each |x_n*x_n|=0 for n=1,2,3

11. Mar 31, 2012

Staff: Mentor

Why do you have the absolute values? Your inner product is defined this way:
<x, y> = x1y1 + x2y2 + x3y3

So again, why is x12 + x22 + x32 ≥ 0?

Take a closer look at what I said in post # 9.

12. Apr 3, 2012

bugatti79

Because we have that x=(x1,x2,x3) in R, therefore x^2>=0 in R^3..?

13. Apr 3, 2012

Staff: Mentor

This makes no sense. x is a vector in R3, so x2 is not defined. Therefore you can't say that x2 ≥ 0.

Start with <x, x> and expand it, using the definition in post #1. It is really very simple to show that <x, x> ≥ 0.

14. Apr 3, 2012

bugatti79

<x,x>=x1x1+x2x2+x3x3
=x1^2+x2^2+x3^3
but x1x1>=0, x2x2>=0 and x3x3>=0 since x1,x2,x3 are in R
implies <x,x>>=0

15. Apr 3, 2012

Staff: Mentor

Yes.

16. Apr 3, 2012

bugatti79

Thank you.

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