Polarization identity in Hilbert space

The polarization identity in Hilbert space is given as $$\langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2)$$ Why is this called "polarization"?

Also, the book that I am reading says that the polarization identity shows that the inner product is uniquely determined by its values on the diagonal, that is, by its values when the first and second arguments are equal. But how does the following $$\langle x,x\rangle=\frac{1}{4}(\|x+x\|^2+i\|x+ix\|^2-i\|x-ix\|^2)$$ determines uniquely the inner product?

• It might be helpful to state the textbook that you are referring to. – Troy Dec 23 '16 at 8:52
• You can see that the RHS of the identity involves lengths only. – velut luna Dec 23 '16 at 9:22
• Look at the first equation you wrote: the scalar product between any two vectors $x,y$ can be obtained as a sum of just squared norms, which are scalar products between a vector and itself, i.e. they are values of $\langle \cdot,\cdot \rangle$ on the diagonal – Del Dec 23 '16 at 10:39
• @Del Makes perfect sense! Thank you – yumiko Dec 23 '16 at 20:55
• @Del I think you should put your comment into an answer – Surb Mar 6 '17 at 13:23

Regarding Polarisation:
To each symmetric bilinear form or sesquilinear form there's an associated quadratic form, determined by evaluation on the diagonal.
A Polarisation identity expresses (the values of) the bi- or sesquilinear form in terms of its associated quadratic form; it thus works in the less obvious direction.

Strictly speaking, this does not answer your question "Why it's called polarisation?"
but $\exists$ mathSEtymology question dealing with it.

You wrote down the polarisation identity for a complex Hilbert space where the squared norm $\|\cdot\|^2$ is the quadratic form associated to the inner product $\langle\,\cdot\,,\cdot\,\rangle$ being a sesquilinear form. Slightly altered it reads $$\langle x,y\rangle\: =\: \frac{1}{4}\sum_{k=0}^3\:i^k\,\langle x+i^ky\,,\, x+i^ky\rangle\,,$$ this might clarify the meaning "when [on the rhs] the first and second arguments are equal".

This context is inviting to mention the (pre-)Hilbert space criterion for a normed space $(X,\|\cdot\|)$,

attributed to Fréchet, von Neumann, and P. Jordan: $$\text{The parallelogram law holds for }\|\cdot\| \quad\Longleftrightarrow\quad \exists\text{ Inner product}\bigm\lvert \langle x,x\rangle=\|x\|^2\;\forall x\in X$$

When strolling from left to right—the harder way—then one may get hold of the inner product via polarisation.

My understanding of this (and I admit I do not have a very thorough background in linear algebra- I'm learning, but thought I might add my two cents in case it is helpful!) is as follows. Clearly an inner product is uniquely determined by a norm, since the inner product can be written exclusively as a fucntion of norms as in the polarisation identity (note here that the polarisation identity takes the norm as being the inner product of a vector with itself; so this particular norm that arises from a given inner product- which happens because the requirements for a norm are automatically satisfies by inner products- determines the inner product).

It uniquely determines the inner product, which can be seen by contradiction argument: suppose we have two different inner products but they give rise to the same norm. The using the polarisation identity, same norms $\implies$ same inner product, which contradicts our starting condition of different inner products!

As for the second part, every norm in the polarization identity is the inner product of a vector with itself. We could relabel

$\textbf{z}_1=\textbf{x}+\textbf{y}$

$\textbf{z}_2=\textbf{x}-\textbf{y}$

etc.

Then we get

$\langle x,y\rangle=\frac{1}{4}(||\textbf{z}_1||^2-\|\textbf{z}_2||^2+i\|\textbf{z}_3||^2-i\|\textbf{z}_4||^2)$

which is indeed the values on the diagonal (all norms)!