linearization
Linearization is the process of reducing a homogeneous polynomial^{} into a multilinear map over a commutative ring. There are in general two ways of doing this:

•
Method 1. Given any homogeneous polynomial $f$ of degree $n$ in $m$ indeterminates over a commutative^{} scalar ring $R$ (scalar simply means that the elements of $R$ commute with the indeterminates).

Step 1
If all indeterminates are linear in $f$, then we are done.

Step 2
Otherwise, pick an indeterminate $x$ such that $x$ is not linear in $f$. Without loss of generality, write $f=f(x,X)$, where $X$ is the set of indeterminates in $f$ excluding $x$. Define $g({x}_{1},{x}_{2},X):=f({x}_{1}+{x}_{2},X)f({x}_{1},X)f({x}_{2},X)$. Then $g$ is a homogeneous polynomial of degree $n$ in $m+1$ indeterminates. However, the highest degree of ${x}_{1},{x}_{2}$ is $n1$, one less that of $x$.

Step 3
Repeat the process, starting with Step 1, for the homogeneous polynomial $g$. Continue until the set $X$ of indeterminates is enlarged to one ${X}^{{}^{\prime}}$ such that each $x\in {X}^{{}^{\prime}}$ is linear.

Step 1

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Method 2. This method applies only to homogeneous polynomials that are also homogeneous^{} in each indeterminate, when the other indeterminates are held constant, i.e., $f(tx,X)={t}^{n}f(x,X)$ for some $n\in \mathbb{N}$ and any $t\in R$. Note that if all of the indeterminates in $f$ commute with each other, then $f$ is essentially a monomial^{}. So this method is particularly useful when indeterminates are noncommuting. If this is the case, then we use the following algorithm^{}:

Step 1
If $x$ is not linear in $f$ and that $f(tx,X)={t}^{n}f(x,X)$, replace $x$ with a formal linear combination^{} of $n$ indeterminates over $R$:
$${r}_{1}{x}_{1}+\mathrm{\cdots}+{r}_{n}{x}_{n}\text{, where}{r}_{i}\in R.$$ 
Step 2
Define a polynomial^{} $g\in R\u27e8{x}_{1},\mathrm{\dots},{x}_{n}\u27e9$, the noncommuting free algebra^{} over $R$ (generated by the noncommuting indeterminates ${x}_{i}$) by:
$$g({x}_{1},\mathrm{\dots},{x}_{n}):=f({r}_{1}{x}_{1}+\mathrm{\cdots}+{r}_{n}{x}_{n}).$$ 
Step 3
Expand $g$ and take the sum of the monomials in $g$ whose coefficent is ${r}_{1}\mathrm{\cdots}{r}_{n}$. The result is a linearization of $f$ for the indeterminate $x$.

Step 4
Take the next nonlinear indeterminate and start over (with Step 1). Repeat the process until $f$ is completely linearized.

Step 1
Remarks.

1.
The method of linearization is used often in the studies of Lie algebras, Jordan algebras^{}, PIalgebras and quadratic forms^{}.

2.
If the characteristic of scalar ring $R$ is 0 and $f$ is a monomial in one indeterminate, we can recover $f$ back from its linearization by setting all of its indeterminates to a single indeterminate $x$ and dividing the resulting polynomial by $n!$:
$$f(x)=\frac{1}{n!}\mathrm{linearization}(f)(x,\mathrm{\dots},x).$$ Please see the first example below.

3.
If $f$ is a homogeneous polynomial of degree $n$, then the linearized $f$ is a multilinear map in $n$ indeterminates.
Examples.

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$f(x)={x}^{2}$. Then $f({x}_{1}+{x}_{2})f({x}_{1})f({x}_{2})={x}_{1}{x}_{2}+{x}_{2}{x}_{1}$ is a linearization of ${x}^{2}$. In general, if $f(x)={x}^{n}$, then the linearization of $f$ is
$$\sum _{\sigma \in {S}_{n}}{x}_{\sigma (1)}\mathrm{\cdots}{x}_{\sigma (n)}=\sum _{\sigma \in {S}_{n}}\prod _{i=1}^{n}{x}_{\sigma (i)},$$ where ${S}_{n}$ is the symmetric group^{} on $\{1,\mathrm{\dots},n\}$. If in addition all the indeterminates commute with each other and $n!\ne 0$ in the ground ring, then the linearization becomes
$$n!{x}_{1}\mathrm{\cdots}{x}_{n}=\prod _{i=1}^{n}i{x}_{i}.$$ 
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$f(x,y)={x}^{3}{y}^{2}+xyxyx$. Since $f(tx,y)={t}^{3}f(x,y)$ and $f(x,ty)={t}^{2}f(x,y)$, $f$ is homogeneous over $x$ and $y$ separately, and thus we can linearize $f$. First, collect all the monomials having coefficient $abc$ in $(a{x}_{1}+b{x}_{2}+c{x}_{3},y)$, we get
$$g({x}_{1},{x}_{2},{x}_{3},y):=\sum {x}_{i}{x}_{j}{x}_{k}{y}^{2}+{x}_{i}y{x}_{j}y{x}_{k},$$ where $i,j,k\in 1,2,3$ and $(ij)(jk)(ki)\ne 0$. Repeat this for $y$ and we have
$$h({x}_{1},{x}_{2},{x}_{3},{y}_{1},{y}_{2}):=\sum {x}_{i}{x}_{j}{x}_{k}({y}_{1}{y}_{2}+{y}_{2}{y}_{1})+({x}_{i}{y}_{1}{x}_{j}{y}_{2}{x}_{k}+{x}_{i}{y}_{2}{x}_{j}{y}_{1}{x}_{k}).$$
Title  linearization 

Canonical name  Linearization 
Date of creation  20130322 14:53:52 
Last modified on  20130322 14:53:52 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  5 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 15A63 
Classification  msc 15A69 
Classification  msc 16R99 
Classification  msc 17A99 
Synonym  polarization 
Defines  linearized 