Hilbert Spaces: Inner Product on R^3

[bugatti79]

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

Homework Statement

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)

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>
...?

 

[Mark44]

Homework Statement

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)

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

This isn't Axiom 1. It has to do with <x, x>.

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>

Don't you mean "conjugate transpose"?
Since the underlying vector space is R³, all you need to show is that <x, y> = <y, x>.

<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>
...?

 

[bugatti79]

This isn't Axiom 1. It has to do with <x, x>.

Oh I see,

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

Don't you mean "conjugate transpose"?

Since the underlying vector space is R³, all you need to show is that <x, y> = <y, x>.

Yes, the bar on top of them.
Anyway,

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

Is axiom 4 ok?
Thanks

 

[Ray Vickson]

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

Homework Statement

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)

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>
...?

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

 

[Mark44]

Oh I see,

Axiom 1 <x,x> >=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.

Yes, the bar on top of them.
Anyway,

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

Is axiom 4 ok?



Thanks

Yes, #4 is fine.

 

[bugatti79]

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.

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?

<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.

 

[Mark44]

<x,x>=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.

<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.

If the sum of three numbers is zero, why in this case is it necessarily true that all three numbers have to be zero?

 

[bugatti79]

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.

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

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

 

[Mark44]

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

That's closer. For any real number x, x² ≥ 0, and x² = 0 iff x = 0.

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

 

[bugatti79]

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

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

 

[Mark44]

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.

 

[bugatti79]

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.

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

 

[Mark44]

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.

 

[bugatti79]

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.

<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

 

[Mark44]

Yes.

 

[bugatti79]

Yes.

Thank you.