Hilbert Spaces: Inner Product on R^3

In summary: 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 SolutionAxiom 1 <x,y> >=0 since we have that x_n for n=1,2,3 are in RAxiom 2a <x,y> =x1y1+x2y2+x3y3Axiom 2b <ax,y>=a<x,y>Axiom 3 <y,x>= complex of <x
  • #1
bugatti79
794
1
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>
...?
 
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  • #2
bugatti79 said:

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>.
bugatti79 said:
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>.
bugatti79 said:
<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>
...?
 
  • #3
Mark44 said:
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
Mark44 said:
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
 
  • #4
bugatti79 said:
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
 
  • #5
bugatti79 said:
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.
bugatti79 said:
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.
 
  • #6
Mark44 said:
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.

Ray Vickson said:
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.
 
  • #7
bugatti79 said:
<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.
bugatti79 said:
<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?
 
  • #8
Mark44 said:
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..?


Mark44 said:
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..?
 
  • #9
bugatti79 said:
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.
bugatti79 said:
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
 
  • #10
Mark44 said:
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
 
  • #11
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
Mark44 said:
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..?
 
  • #13
bugatti79 said:
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.
 
  • #14
Mark44 said:
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
 
  • #16
Mark44 said:
Yes.

Thank you.
 

1. What is Hilbert space?

Hilbert space is a mathematical concept that refers to a complete vector space with an inner product defined on it. It is named after the German mathematician David Hilbert and has various applications in physics, engineering, and mathematics.

2. What is an inner product?

An inner product is a mathematical operation that takes two vectors and returns a scalar value. It is similar to the dot product, but it is defined on more general vector spaces, such as Hilbert spaces. It satisfies certain properties such as linearity, symmetry, and positive definiteness.

3. How is an inner product used in Hilbert space?

In Hilbert space, the inner product is used to define the norm of a vector, which measures its length or magnitude. It is also used to define the angle between two vectors and to generalize the concept of orthogonality. Additionally, inner products are used to define important concepts like projections, orthogonal complements, and adjoint operators.

4. What are the applications of Hilbert space and inner product?

Hilbert space and inner product have numerous applications in various fields of mathematics, physics, and engineering. Some examples include quantum mechanics, signal processing, control theory, and functional analysis. They are also used in machine learning and data science for tasks such as dimensionality reduction and feature extraction.

5. What is the difference between Hilbert space and Euclidean space?

While both Hilbert space and Euclidean space are vector spaces, the main difference lies in the type of inner product defined on each space. In Euclidean space, the inner product is the dot product, which is geometrically interpreted as the projection of one vector onto another. In Hilbert space, the inner product is more general and can take various forms depending on the specific space and application. Additionally, Hilbert space is infinite-dimensional, while Euclidean space is finite-dimensional.

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