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

[Math Amateur]

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:

In the above text from Garling we read the following:" ... ... in the complex case we have the polarization formula$\langle x,y \rangle = \frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$ ... ... "
Can someone please demonstrate how to prove that $\langle x,y \rangle = \frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$ ...?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:

Hope that helps ...

Peter

 

[andrewkirk]

We don't need any of the attached support material. The result follows directly from the axioms of a complex inner product:

\begin{align*}
\sum_{j=0}^3 i^j\|x+i^j y\|^2
% 1
&= \sum_{j=0}^3 i^j \langle x+i^j y ,x+i^j y\rangle
\\ % 2
&= \sum_{j=0}^3 i^j
\left(
\langle x ,x\rangle +
i^j \langle y,x\rangle +
\left( i^j\right)^*\langle x , y \rangle +
i^j \left( i^j\right)^* \langle y , y\rangle
\right)
\\ % 3
&= \sum_{j=0}^3 i^j
\left(
\langle x ,x\rangle +
i^j \langle y,x\rangle +
\left(i^j \langle y,x\rangle \right)^* +
\left( i\ i^*\right)^j \langle y , y\rangle
\right)
\\ % 4
&= \left(\langle x,x\rangle + \langle y,y\rangle\right)
\sum_{j=0}^3 i^j
+
\sum_{j=0}^3 i^j\cdot
2\ \Re \left( i^j \langle y,x\rangle\right)
\\ % 5
&=0
+
2
\left(
\Re \langle y,x\rangle +
i\cdot \Re \left( i \langle y,x\rangle\right)
- \Re \left( - \langle y,x\rangle\right)
-i\cdot \Re \left(-i \langle y,x\rangle \right)
\right)
\\ % 6
&=
2
\left(
\Re \langle y,x\rangle
- i\cdot \Im \langle y,x\rangle
+\Re \langle y,x\rangle
-i\cdot \Im \langle y,x\rangle
\right)
\\ % 7
&=
2
\left(
\langle y,x\rangle^*
+
\langle y,x\rangle^*
\right)
\\ % 8
&=
4
\langle x,y\rangle
\end{align*}

 

[StoneTemplePython]

A lot of times greater generality isn't necessarily helpful but because $m$ evenly spaced points on the unit circle comes up so often, it seems worth pointing out that your complex polarization identity works for any natural number $m \geq 3$, not just the $m = 4$ case that the book stated.

- - - -
where we recall that $\sum_{k=0}^{m-1} \omega^k = 0$, and again with the restriction that natural number $m \geq 3$ we recall $\sum_{k=0}^{m-1} \omega^{2k} = 0$
(why?)
- - - -
start by expanding the squared 2 norm
$\langle \mathbf x, \mathbf y \rangle = \frac{1}{m} \sum_{k=0}^{m-1} \big \Vert \mathbf x + \omega^k \mathbf y \big \Vert_2^2 \omega^k = \frac{1}{m} \sum_{k=0}^{m-1} \Big(\big \Vert \mathbf x \big \Vert_2^2 \omega^k + \big \Vert \mathbf y \big \Vert_2^2 \omega^k + \langle \mathbf x, \omega^k \mathbf y \rangle \omega^k + \langle \omega^k \mathbf y, \mathbf x \rangle \omega^k \Big)$

split the summation, and make use of conjugate linearity by removing the $\omega^k$ from the $\mathbf y$ terms of the inner product

$\langle \mathbf x, \mathbf y \rangle = \frac{1}{m} \sum_{k=0}^{m-1} \Big(\big(\big \Vert \mathbf x \big \Vert_2^2 + \big \Vert \mathbf y \big \Vert_2^2\big)\omega^k \Big) + \frac{1}{m} \sum_{k=0}^{m-1} \Big(\langle \mathbf x, \mathbf y \rangle \bar{\omega}^k\omega^k\Big) + \frac{1}{m} \sum_{k=0}^{m-1} \Big(\langle \mathbf y, \mathbf x \rangle \omega^{k} \omega^k\Big)$

finally:

$\langle \mathbf x, \mathbf y \rangle = \Big(\big( \big \Vert \mathbf x \big \Vert_2^2 + \big \Vert \mathbf y \big \Vert_2^2 \big) \frac{1}{m} \big(\sum_{k=0}^{m-1} \omega^k\big) \Big) + \langle \mathbf x, \mathbf y \rangle \Big(\frac{1}{m} \sum_{k=0}^{m-1} 1 \big)+ \langle \mathbf y, \mathbf x \rangle\Big(\frac{1}{m} \sum_{k=0}^{m-1} \omega^{2k} \Big) = \big(0\big) + \langle \mathbf x, \mathbf y \rangle \big(1\big)+ \big(0\big) = \langle \mathbf x, \mathbf y \rangle$

 

[Math Amateur]

THanks Andrew, StoneTemplePython ...

Appreciate the help and insights ...

Just working through your posts now ...

Thanks again ...

Peter