I am new to this branch of math, so bear with me.


This question started when reading Kevin McCrimmon's "A Taste of Jordan Algebras"
It talks about polarization and gives a general description.



> 
> the general process of linearization
> (often called polarization, espe-
> cially in analysis in dealing with
> quadratic mappings on a complex
> space). This is an important technique
> in nonassociative algebras ... Given a homogeneous
> polynomial p(x) of degree n, the
> process of linearization is designed
> to create a symmetric multilinear
> polynomial p'(x1,... ,xn) in n
> variables such that the original
> polynomial arises by specializing all
> the variables to the same value x :
> p'(x,... ,x) = p(x). For example, the
> full linearization of the square x^2=
> x*x is 1/2 (x1*x2+x2*x1), and the full
> linearization of the cube x3= x*x\*x of
> degree 3 is1/6(x1x2x3+ x1x3x2+ x2x1x3+
> x2x3x1+ x3x1x2+ x3x2x1).
> 
> 
> 


I understand the first part



> 
> Full linearization is usually achieved
> one step at a time by a series of
> partial linearizations, in which a
> polynomial homogeneous of degree n in
> a particular variable x is replaced by
> one of degree n − 1 in x and linear in
> a new variable y.
> Intuitively, in an expression with n
> occurrences of x we simply replace
> each occurrence of x, one at a time,
> by a y, add up the results
> 
> 
> 


Ok to here...But then he says



> 
> we often have no very
> explicit expression for p, and must
> describe linearization in a more
> intrinsic way. The clearest
> formulation is to take p(x+ λy) for an
> indeterminate scalar λ and expand this
> out as a polynomial in λ: p(x + λy) =
> p(x) + λp1(x;y) + λ2p2(x;y) + ··· +
> λnp(y). Here pi(x;y) is homogeneous of
> degree n − i in x and i in y
> (intuitively, we obtain it by
> replacing i of the x’s in p(x) by y’s
> in all possible ways)
> 
> 
> 


This is where i almost get it, but not quite. where does the y come from here. Is it the same y as above, and so one... and wish I could look at another description of it...
How would I symmetrize/linearize/polarize something like
3x^2yz + 2y^2x^2 + z^2x + z^3y


Also later he suggests this method for cases where p is quadratic ( ===q)



> 
> we take the value [of q[x]] on the sum x + y of
> two elements, and then subtract the
> pure x and y terms to obtain q(x,y) :=
> q(x + y) − q(x) − q(y).
> 
> 
> 


This only works then for squares of one variable then? ( ax^2?). not very useful...


I am looking for some more details about this, but links such as the following are confusing me as much as they are enlightening me because the notation and context are different. Is there a good textbook that explains this clearly?


* <http://planetmath.org/polarizationbydifferentialoperators>
* <https://en.wikipedia.org/wiki/Polarization_of_an_algebraic_form>
* [polarization formula for homogeneous polynomials](https://mathoverflow.net/questions/61884/polarization-formula-for-homogeneous-polynomials)



polarization/linearization as in jordan forms
Alternatively, you can polarise right away as follows: if $p(x)$ is homogeneous of degree $n$ (here $x$ may be a variable with values in $\mathbb{R}^k$, e.g. $p(x)=\mathop{\mathrm{tr}}(x^4)$, where $x$ is a matrix), then you can look at 
 $$
p(\lambda\_1x\_1+\lambda\_2x\_2+\cdots+\lambda\_nx\_n),
 $$
where $\lambda\_i$ are elements of the ground field (e.g. real numbers), and $x\_i$ are variables of the same type as $x$ (e.g. matrices like in my example above). In that expression, extract the coefficient of $\lambda\_1\lambda\_2\cdots\lambda\_n$ there. That coefficient is the result of polarisation (multilinearisation).


