linearization (original) (raw)

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  Method 1. Given any homogeneous polynomial f of degree n in m indeterminates over a commutativescalar ring R (scalar simply means that the elements of R commute with the indeterminates).
  1. Step 1
     If all indeterminates are linear in f, then we are done.
  2. 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. Defineg⁢(x1,x2,X):=f⁢(x1+x2,X)-f⁢(x1,X)-f⁢(x2,X). Then g is a homogeneous polynomial of degree n in m+1indeterminates. However, the highest degree of x1,x2 is n-1, one less that of x.
  3. Step 3
     Repeat the process, starting with Step 1, for the homogeneous polynomial g. Continue until the set Xof indeterminates is enlarged to one X′ such that each x∈X′ is linear.
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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⁢(t⁢x,X)=tn⁢f⁢(x,X) for some n∈ℕ and any t∈R. Note that if all of the indeterminates in f commute with each other, then f is essentially amonomial. So this method is particularly useful when indeterminates are non-commuting. If this is the case, then we use the following algorithm:
  1. Step 1
     If x is not linear in f and that f⁢(t⁢x,X)=tn⁢f⁢(x,X), replace x with a formal linear combination ofn indeterminates over R:

     r1⁢x1+⋯+rn⁢xn⁢, where ⁢ri∈R.

  2. Step 2
  3. Step 3
     Expand g and take the sum of the monomials in g whose coefficent is r1⁢⋯⁢rn. The result is a linearization of f for the indeterminate x.
  4. Step 4
     Take the next non-linear indeterminate and start over (with Step 1). Repeat the process until f is completely linearized.

Remarks.

2. 1. 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)=1n!⁢linearization⁡(f)⁢(x,…,x).
      Please see the first example below.

3. 1. If f is a homogeneous polynomial of degree n, then the linearized f is a multilinear map in nindeterminates.

Examples.

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  f⁢(x)=x2. Then f⁢(x1+x2)-f⁢(x1)-f⁢(x2)=x1⁢x2+x2⁢x1 is a linearization of x2. In general, if f⁢(x)=xn, then the linearization of f is

  ∑σ∈Snxσ⁢(1)⁢⋯⁢xσ⁢(n)=∑σ∈Sn∏i=1nxσ⁢(i),
  where Sn is the symmetric group on {1,…,n}. If in addition all the indeterminates commute with each other and n!≠0 in the ground ring, then the linearization becomes

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  f⁢(x,y)=x3⁢y2+x⁢y⁢x⁢y⁢x. Since f⁢(t⁢x,y)=t3⁢f⁢(x,y) and f⁢(x,t⁢y)=t2⁢f⁢(x,y), f is homogeneous over x and yseparately, and thus we can linearize f. First, collect all the monomials having coefficient a⁢b⁢c in(a⁢x1+b⁢x2+c⁢x3,y), we get

  g⁢(x1,x2,x3,y):=∑xi⁢xj⁢xk⁢y2+xi⁢y⁢xj⁢y⁢xk,
  where i,j,k∈1,2,3 and (i-j)⁢(j-k)⁢(k-i)≠0. Repeat this for y and we have
  h⁢(x1,x2,x3,y1,y2):=∑xi⁢xj⁢xk⁢(y1⁢y2+y2⁢y1)+(xi⁢y1⁢xj⁢y2⁢xk+xi⁢y2⁢xj⁢y1⁢xk).
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