## Hilbert, Inner Product

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>
...?

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Mentor
 Quote by bugatti79 1. The problem statement, all variables and given/known data Prove the function =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 >=0 since we have that x_n and y_n for n=1,2,3 are in R
This isn't Axiom 1. It has to do with <x, x>.
 Quote by bugatti79 Axiom 2a =x1y1+x2y2+x3y3, then =0 iff x_n and y_n for n=1,2,3 both =0 Axiom 2b =a = ax1y1+ax2y2+ax3y3 = a(x1y1+x2y2+x3y3) =a Axiom 3 = complex of
Don't you mean "conjugate transpose"?
Since the underlying vector space is R3, all you need to show is that <x, y> = <y, x>.
 Quote by bugatti79 =(y1x1+y2x2+y3x3) but y complex =y and x complex=x in R therefore = (y1x1 complex+y2x2 complex +x3y3 complex) = complex = complex Axiom 4 =+, let z=(z1,z2,z3) in R^3 =(x1+y1+x2+y2+x3+y3)(z1+z2+z3) =x1z1 +x2z2+x3z3+y1z1+y2zy3z3 =+ ...?

 Quote by Mark44 This isn't Axiom 1. It has to do with .
Oh I see,

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

 Quote by Mark44 Don't you mean "conjugate transpose"? Since the underlying vector space is R3, all you need to show is that = .
Yes, the bar on top of them.
Anyway,

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

Is axiom 4 ok?

Thanks

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Homework Help

## Hilbert, Inner Product

 Quote by 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_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 >=0 since we have that x_n and y_n for n=1,2,3 are in R Axiom 2a =x1y1+x2y2+x3y3, then =0 iff x_n and y_n for n=1,2,3 both =0 Axiom 2b =a = ax1y1+ax2y2+ax3y3 = a(x1y1+x2y2+x3y3) =a Axiom 3 = complex of =(y1x1+y2x2+y3x3) but y complex =y and x complex=x in R therefore = (y1x1 complex+y2x2 complex +x3y3 complex) = complex = complex Axiom 4 =+, let z=(z1,z2,z3) in R^3 =(x1+y1+x2+y2+x3+y3)(z1+z2+z3) =x1z1 +x2z2+x3z3+y1z1+y2zy3z3 =+ ...?
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

Mentor
 Quote by bugatti79 Oh I see, Axiom 1 >=0 since we have that x_n for n=1,2,3 are in R
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.
 Quote by bugatti79 Yes, the bar on top of them. Anyway, =y1x1+y2x2+y3x3 =x1y1+x2y2+x3y3 = Is axiom 4 ok? Thanks
Yes, #4 is fine.

 Quote by Mark44 You're waving your arms here. Expand using the definition of this inner product and it should be obvious why >= 0.
 Quote by Ray Vickson Your statement "Axiom 1 >=0" is NOT a property of an inner product; we can have > 0, = 0 or < 0. However, we do have >= 0, with = 0 iff x is the zero vector. Can you prove that?
<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.

Mentor
 Quote by bugatti79 =x1x1+x2x2+x3x3>=0 where x_n>=0 for n=1,2,3 since x_n is in R
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.
 Quote by bugatti79 =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.
If the sum of three numbers is zero, why in this case is it necessarily true that all three numbers have to be zero?

 Quote by Mark44 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 > 0.
The product of a negative number times a negative number gives a positive therefore <x,x> will always be greater than 0..?

 Quote by Mark44 If the sum of three numbers is zero, why in this case is it necessarily true that all three numbers have to be zero?
Because if at least one of them is non 0 then <x,x> is non 0..?

Mentor
 Quote by bugatti79 The product of a negative number times a negative number gives a positive therefore will always be greater than 0..?
That's closer. For any real number x, x2 ≥ 0, and x2 = 0 iff x = 0.
 Quote by bugatti79 Because if at least one of them is non 0 then is non 0..?
Now, why is x12 + x22 + x32 ≥ 0? This is what you need to establish in order to verify the <x, x> ≥ 0

 Quote by Mark44 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 ≥ 0
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

 Mentor Why do you have the absolute values? Your inner product is defined this way: = x1y1 + x2y2 + x3y3 So again, why is x12 + x22 + x32 ≥ 0? Take a closer look at what I said in post # 9.

 Quote by Mark44 Why do you have the absolute values? Your inner product is defined this way: = x1y1 + x2y2 + x3y3 So again, why is x12 + x22 + x32 ≥ 0? Take a closer look at what I said in post # 9.
Because we have that x=(x1,x2,x3) in R, therefore x^2>=0 in R^3..?

Mentor
 Quote by bugatti79 Because we have that x=(x1,x2,x3) in R, therefore x^2>=0 in R^3..?
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.

 Quote by Mark44 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 and expand it, using the definition in post #1. It is really very simple to show that ≥ 0.
<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

 Mentor Yes.

 Quote by Mark44 Yes.
Thank you.