# Polarization Formulae for Inner-Product Spaces ....

• MHB
• Math Amateur
In summary, the conversation discusses the polarization formula for the complex case in Garling's book on Metric and Topological Spaces, specifically in Chapter 11. The formula is given as: $\langle x,y \rangle = \frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$. The conversation then asks for a demonstration of how to prove this formula, which involves expanding the terms and using the properties of inner products in complex vector spaces.
Math Amateur
Gold Member
MHB
I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help with the polarization formula for the complex case ...

Garling's statement of the polarization formulae reads as follows:https://www.physicsforums.com/attachments/7914In the above text from Garling we read the following:" ... ... in the complex case we have the polarization formula $$\displaystyle \langle x,y \rangle = \frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$$ ... ... "
Can someone please demonstrate how to prove that $$\displaystyle \langle x,y \rangle = \frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$$ ...?Help will be appreciated ...

Peter
==========================================================================================***NOTE***

It may help readers of the above post to know Garling's notation and approach to inner-product spaces ... ... so I am providing the same ... as follows:
https://www.physicsforums.com/attachments/7915
https://www.physicsforums.com/attachments/7916
https://www.physicsforums.com/attachments/7917
https://www.physicsforums.com/attachments/7918Hope that helps ...

Peter

Peter said:
Can someone please demonstrate how to prove that $$\displaystyle \langle x,y \rangle = \frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$$ ...?
You just need to expand the four terms in that sum, remembering that (i) the square of the norm of a vector is equal to the inner product of the vector with itself, and (ii) in a complex vector space, the inner product is linear in the first variable and conjugate-linear in the second variable.

So for example the second term in the above sum (the term given by $j=1$) is $i\|x+iy\|^2 = i\langle x+iy,x+iy \rangle$. But $$\langle x+iy,x+iy \rangle = \langle x,x \rangle + \langle iy,x \rangle + \langle x,iy \rangle + \langle iy,iy \rangle = \langle x,x \rangle + i\langle y,x \rangle - i\langle x,y \rangle + \langle y,y \rangle.$$ So $$i\|x+iy\|^2 = i\langle x,x \rangle - \langle y,x \rangle + \langle x,y \rangle + i\langle y,y \rangle.$$ When you do that for all four terms in the sum $$\displaystyle \frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$$, you will find that the coefficients of $\langle x,x \rangle$, $\langle y,x\rangle$ and $\langle y,y \rangle$ all cancel out, and you are just left with $\langle x,y \rangle.$

Opalg said:
You just need to expand the four terms in that sum, remembering that (i) the square of the norm of a vector is equal to the inner product of the vector with itself, and (ii) in a complex vector space, the inner product is linear in the first variable and conjugate-linear in the second variable.

So for example the second term in the above sum (the term given by $j=1$) is $i\|x+iy\|^2 = i\langle x+iy,x+iy \rangle$. But $$\langle x+iy,x+iy \rangle = \langle x,x \rangle + \langle iy,x \rangle + \langle x,iy \rangle + \langle iy,iy \rangle = \langle x,x \rangle + i\langle y,x \rangle - i\langle x,y \rangle + \langle y,y \rangle.$$ So $$i\|x+iy\|^2 = i\langle x,x \rangle - \langle y,x \rangle + \langle x,y \rangle + i\langle y,y \rangle.$$ When you do that for all four terms in the sum $$\displaystyle \frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$$, you will find that the coefficients of $\langle x,x \rangle$, $\langle y,x\rangle$ and $\langle y,y \rangle$ all cancel out, and you are just left with $\langle x,y \rangle.$

Thanks Opalg ...

Just working through your post now ...

Thanks again,

Peter

## 1. What is the polarization formula for inner-product spaces?

The polarization formula for inner-product spaces is a mathematical expression that relates the inner product of two vectors to their norm. It is written as ⟨x,y⟩ = (||x+y||^2 - ||x-y||^2)/4, where x and y are vectors in the inner-product space.

## 2. How is the polarization formula derived?

The polarization formula is derived using the properties of inner products and the Pythagorean theorem. It can also be derived using the Cauchy-Schwarz inequality and the parallelogram law.

## 3. What are the applications of the polarization formula?

The polarization formula has various applications in mathematics and other fields such as physics and engineering. It is used to simplify calculations involving inner products and to prove theorems in functional analysis and geometry.

## 4. Can the polarization formula be extended to other types of spaces?

Yes, the polarization formula can be extended to other types of spaces such as Banach spaces and Hilbert spaces. However, the formula may have different forms and properties in these spaces.

## 5. Are there any limitations to the polarization formula?

The polarization formula is only applicable to inner-product spaces and cannot be used in other types of vector spaces. It also assumes that the inner product is positive definite, meaning that the inner product of a vector with itself is always positive. If this is not the case, the formula may not hold.

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