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Scalar product

FilipVz

New member
Oct 21, 2013
8
Hi,

can somebody help me with the problem:

Suppose that in a vector space over field of real numbers a positive defined norm is defined for each vector which satisfies the triangle inequality and ||aU||=|a|*||u||. Show that a real valued scalar product can de defined as follows:

(U,V)=1/2*(||U+V||^2 - ||U||^2- ||V||^2)

which satisfied the postulates, if following identity is satisfied by the norms:

||U+V||^2 + ||U-V||^2 = 2*||U||^2 + 2*||V||^2
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Hi,

can somebody help me with the problem:

Suppose that in a vector space over field of real numbers a positive defined norm is defined for each vector which satisfies the triangle inequality and ||aU||=|a|*||u||. Show that a real valued scalar product can de defined as follows:

(U,V)=1/2*(||U+V||^2 - ||U||^2- ||V||^2)

which satisfied the postulates, if following identity is satisfied by the norms:

||U+V||^2 + ||U-V||^2 = 2*||U||^2 + 2*||V||^2
Welcome to MHB, FilipVz! :)

How far do you get with the postulates?

That is:
  1. (Conjugate) symmetry
  2. Linearity in the first argument
  3. Positive definiteness
See e.g. wiki.

For starters, is it symmetric?
That is, does for all u and v hold that (u,v) = (v,u)?
 

FilipVz

New member
Oct 21, 2013
8
Welcome to MHB, FilipVz! :)

How far do you get with the postulates?

That is:
  1. (Conjugate) symmetry
  2. Linearity in the first argument
  3. Positive definiteness
See e.g. wiki.

For starters, is it symmetric?
That is, does for all u and v hold that (u,v) = (v,u)?

Hi, I like Serena,

Scalar product is symmetric.

Could you please explain to me what is "Linearity in the first argument"?

Thanks,

Filip
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Hi, I like Serena,

Scalar product is symmetric.

Could you please explain to me what is "Linearity in the first argument"?

Thanks,

Filip
From the wiki reference I gave you can see that an inner product over the field of the real numbers must satisfy the following axioms:

1. Symmetry:
$$\langle x,y\rangle =\langle y,x\rangle.$$
2. Linearity in the first argument:
$$\langle ax,y\rangle= a \langle x,y\rangle.$$
$$\langle x+y,z\rangle= \langle x,z\rangle+ \langle y,z\rangle.$$
3. Positive-definiteness:
$$\langle x,x\rangle \geq 0 \text{ with equality only for }x = 0.$$


Yes, your inner product is symmetric, but you did not mention why.
The reason is in the expression for your inner product you can swap u and v around, and you'll end up with the same expression.


Next step is linearity for a factor in the first argument.
To find if it is, you need to write your inner product with $a \mathbf u$ and $\mathbf v$ instead of $\mathbf u$ and $\mathbf v$.
And then figure out if the resulting expression is the same as it was for $\mathbf u$ and $\mathbf v$, except for a factor $a$.
 

FilipVz

New member
Oct 21, 2013
8
From the wiki reference I gave you can see that an inner product over the field of the real numbers must satisfy the following axioms:

1. Symmetry:
$$\langle x,y\rangle =\langle y,x\rangle.$$
2. Linearity in the first argument:
$$\langle ax,y\rangle= a \langle x,y\rangle.$$
$$\langle x+y,z\rangle= \langle x,z\rangle+ \langle y,z\rangle.$$
3. Positive-definiteness:
$$\langle x,x\rangle \geq 0 \text{ with equality only for }x = 0.$$


Yes, your inner product is symmetric, but you did not mention why.
The reason is in the expression for your inner product you can swap u and v around, and you'll end up with the same expression.


Next step is linearity for a factor in the first argument.
To find if it is, you need to write your inner product with $a \mathbf u$ and $\mathbf v$ instead of $\mathbf u$ and $\mathbf v$.
And then figure out if the resulting expression is the same as it was for $\mathbf u$ and $\mathbf v$, except for a factor $a$.
Linearity for a factor in the firs arugment: (aU,V)=a(U,V)

Positive-definiteness: (U,U)>=0
(U,U)=0, iff U=0

What is the next step?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Linearity for a factor in the firs arugment: (aU,V)=a(U,V)

Positive-definiteness: (U,U)>=0
(U,U)=0, iff U=0

What is the next step?
You skipped the step for linearity of addition in the first argument.

After that, there is no next step.
If you can prove that they are satisfied, you are done - then it is an inner product.
You didn't really verify or prove them yet though.
 

FilipVz

New member
Oct 21, 2013
8
You skipped the step for linearity of addition in the first argument.

After that, there is no next step.
If you can prove that they are satisfied, you are done - then it is an inner product.
You didn't really verify or prove them yet though.



So, all i need to do is to prove the postulates of Scalar product?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Hi,

can somebody help me with the problem:

Suppose that in a vector space over field of real numbers a positive defined norm is defined for each vector which satisfies the triangle inequality and ||aU||=|a|*||u||. Show that a real valued scalar product can de defined as follows:

(U,V)=1/2*(||U+V||^2 - ||U||^2- ||V||^2)

which satisfied the postulates, if following identity is satisfied by the norms:

||U+V||^2 + ||U-V||^2 = 2*||U||^2 + 2*||V||^2
This is the Jordan–von Neumann theorem. As I like Serena has pointed out, you need to show that the inner product satisfies the additivity property $\langle x+y,z\rangle = \langle x,z\rangle + \langle y,z\rangle$ and the scalar multiplication property $\langle ax,y\rangle = a\langle x,y\rangle$. Neither of those is at all easy. Unless you are a budding Jordan or von Neumann, you are unlikely to be able to find a proof unaided.

The proof is usually given (as in the above link) for a vector space over the complex numbers. The proof for the real case follows the same route and is basically no easier.