288 lines
7.2 KiB
Plaintext
288 lines
7.2 KiB
Plaintext
---
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title: "Exploring Finite Fields, Part 2 Appendix"
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description: |
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Additional notes about polynomial evaluation.
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format:
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html:
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html-math-method: katex
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date: "2024-01-15"
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date-modified: "2025-07-16"
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categories:
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- algebra
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- finite field
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- haskell
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---
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In the [second post in this series](../), we briefly discussed alternate means
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of evaluating polynomials by "plugging in" different structures.
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Different Kinds of Polynomials
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------------------------------
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Rather than redefining evaluation for each of these cases,
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we can map the polynomial into a structure compatible with it should be evaluated.
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Essentially, this means that from a polynomial in the base structure,
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we can derive polynomials in these other structures.
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In particular, there is a distinction between a matrix of polynomials or a polynomial in matrices:
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:::: {layout-ncol="2"}
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::: {}
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$$
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\begin{align*}
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p &: K[x]
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\\
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p(x) &= x^n + p_{n-1}x^{n-1} + ...
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\\
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\phantom{= p} & + p_1 x + p_0
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\end{align*}
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$$
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:::
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::: {}
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$x$ is a scalar indeterminate
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```haskell
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p :: Polynomial k
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```
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:::
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::::
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:::: {layout-ncol="2"}
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::: {}
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$$
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\begin{align*}
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P &: (K[x])^{m \times m}
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\\
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P(x I) &= (x I)^n + (p_{n-1})(x I)^{n-1} + ...
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\\
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& + p_1(x I)+ p_0 I
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\end{align*}
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$$
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:::
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::: {}
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$x$ is a scalar indeterminate, $P(x I)= p(x) I$ is a matrix of polynomials in $x$
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```haskell
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asPolynomialMatrix
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:: Polynomial k -> Matrix (Polynomial k)
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pMat :: Matrix (Polynomial k)
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pMat = asPolynomialMatrix p
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```
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:::
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::::
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:::: {layout-ncol="2"}
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::: {}
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$$
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\begin{align*}
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\hat P &: K^{m \times m}[X]
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\\
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\hat P(X) &= X^n + (p_{n-1}I)X^{n-1} + ...
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\\
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& + (p_1 I) X + (p_0 I)
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\end{align*}
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$$
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:::
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::: {}
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$X$ is a matrix indeterminate, $\hat P(X)$ is a polynomial over matrices
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```haskell
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asMatrixPolynomial
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:: Polynomial k -> Polynomial (Matrix k)
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pHat :: Polynomial (Matrix k)
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pHat = asMatrixPolynomial p
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```
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:::
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::::
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It's easy to confuse the latter two, but the Haskell makes the difference in types clearer.
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There exists a natural isomorphism between the two, which is discussed further
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in the [fourth post in this series](../../4/).
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Cayley-Hamilton Theorem, Revisited
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----------------------------------
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As a reminder, the
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[Cayley-Hamilton theorem](https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem)
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says that a matrix satisfies its own characteristic polynomial.
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In a type-stricter sense, it says the following relationship holds:
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```haskell
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evalPoly :: a -> Polynomial a -> a
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mA :: Matrix a
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charpolyA :: Polynomial a
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charpolyA = charpoly mA
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charpolyA :: Polynomial (Matrix a)
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matCharpolyA = asMatrixPolynomial charPolyA
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evalPoly mA matCharpolyA == (0 :: Matrix a)
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```
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Due to the aformentioned isomorphism, factoring a polynomial "inside" a matrix turns
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out to give the same answer as factoring a polynomial over matrices.
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:::: {layout-ncol="2"}
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::: {}
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$$
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\begin{gather*}
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P(xI) = \left( \begin{matrix}
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x^2 + x + 1 & 0 \\
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0 & x^2 + x + 1
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\end{matrix}\right)
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\\ \\
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= (xI - C)(xI - C')
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\\ \\
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= \left( \begin{matrix}
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x & -1 \\
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1 & x + 1
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\end{matrix} \right)
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\left( \begin{matrix}
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x - a & -b \\
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-c & x - d
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\end{matrix} \right)
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\\ \\
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\begin{align*}
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x(x - a) + c &= x^2 + x + 1
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\\
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\textcolor{green}{x(-b) - (x - d)} &\textcolor{green}{= 0}
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\\
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\textcolor{blue}{(x - a) + (x + 1)(-c)} &\textcolor{blue}{= 0}
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\\
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(-b) + (x + 1)(x - d) &= x^2 + x + 1
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\end{align*}
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\\ \\
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\textcolor{green}{(-b -1)x +d = 0} \implies b = -1, ~ d = 0 \\
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\textcolor{blue}{(1 - c)x - a - c = 0} \implies c = 1, ~ a = -1
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\\ \\
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C' =
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\left( \begin{matrix}
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-1 & -1 \\
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1 & 0
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\end{matrix} \right)
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\end{gather*}
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$$
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:::
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::: {}
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$$
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\begin{gather*}
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\hat P(X) = X^2 + X + 1I
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\\[10pt]
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= (X - C)(X - C')
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\\[10pt]
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= X^2 - (C + C')X + CC'
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\\[10pt]
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\implies
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\\[10pt]
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C + C' = -I, ~ C' = -I - C
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\\[10pt]
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CC' = I, ~ C^{-1} = C'
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\\[10pt]
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C' = \left( \begin{matrix}
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-1 & -1 \\
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1 & 0
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\end{matrix} \right)
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\end{gather*}
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$$
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:::
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::::
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It's important to not that a matrix factorization is not unique.
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*Any* matrix with a given characteristic polynomial can be used as a root of that polynomial.
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Of course, choosing one root affects the other matrix roots.
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### Moving Roots
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All matrices commute with the identity and zero matrices.
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A less obvious fact is that for a matrix factorization, all roots *also* commute with one another.
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By the Fundamental Theorem of Algebra,
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[Vieta's formulas](https://en.wikipedia.org/wiki/Vieta%27s_formulas) state:
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$$
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\begin{gather*}
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\hat P(X)
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= \prod_{[i]_n} (X - \Xi_i)
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= (X - \Xi_0) (X - \Xi_1)...(X - \Xi_{n-1})
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\\
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= \left\{ \begin{align*}
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& \phantom{+} X^n
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\\
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& - (\Xi_0 + \Xi_1 + ... + \Xi_{n-1}) X^{n-1}
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\\
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& + (\Xi_0 \Xi_1+ \Xi_0 \Xi_2 + ... + \Xi_0 \Xi_{n-1} + \Xi_1 \Xi_2 + ... + \Xi_{n-2} \Xi_{n-1})X^{n-2}
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\\
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& \qquad \vdots
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\\
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& + (-1)^n \Xi_0 \Xi_1 \Xi_2...\Xi_n
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\end{align*} \right.
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\\
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= X^n -\sigma_1([\Xi]_n)X^{n-1} + \sigma_2([\Xi]_n)X^{n-2} + ... + (-1)^n \sigma_n([\Xi]_n)
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\end{gather*}
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$$
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The product range \[*i*\]~*n*~ means that the terms are ordered from 0 to *n* - 1 over the index given.
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On the bottom line, *σ* are
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[elementary symmetric polynomials](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial)
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and \[*Ξ*\]~*n*~ is the list of root matrices from *Ξ*~*0*~ to Ξ~*n-1*~.
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By factoring the matrix with the roots in a different order, we get another factorization.
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It suffices to only focus on *σ*~2~, which has all pairwise products.
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$$
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\begin{gather*}
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\pi \in S_n
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\\
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\qquad
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\pi \circ \hat P(X) = \prod_{\pi ([i]_n)} (X - \Xi_i)
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\\ \\
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= X^n
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- \sigma_1 \left(\pi ([\Xi]_n) \vphantom{^{1}} \right)X^{n-1}
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+ \sigma_2 \left(\pi ([\Xi]_n) \vphantom{^{1}} \right)X^{n-2} + ...
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+ (-1)^n \sigma_n \left(\pi ([\Xi]_n) \vphantom{^{1}} \right)
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\\[10pt]
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\pi_{(0 ~ 1)} \circ \hat P(X) = (X - \Xi_{1}) (X - \Xi_0)(X - \Xi_2)...(X - \Xi_{n-1})
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\\
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= X^n + ... + \sigma_2(\Xi_1, \Xi_0, \Xi_2, ...,\Xi_{n-1})X^{n-2} + ...
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\\[10pt]
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\begin{array}{}
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e & (0 ~ 1) & (1 ~ 2) & ... & (n-2 ~~ n-1)
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\\ \hline
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\textcolor{red}{\Xi_0 \Xi_1} & \textcolor{red}{\Xi_1 \Xi_0} & \Xi_0 \Xi_1 & & \Xi_0 \Xi_1
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\\
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\Xi_0 \Xi_2 & \Xi_0 \Xi_2 & \Xi_0 \Xi_2 & & \Xi_0 \Xi_2
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\\
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\Xi_0 \Xi_3 & \Xi_0 \Xi_3 & \Xi_0 \Xi_3 & & \Xi_0 \Xi_3
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\\
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\vdots & \vdots & \vdots & & \vdots
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\\
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\Xi_0 \Xi_{n-1} & \Xi_0 \Xi_{n-1} & \Xi_{0} \Xi_{n-1} & & \Xi_{n-1} \Xi_0
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\\
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\textcolor{green}{\Xi_1 \Xi_2} & \Xi_1 \Xi_2 & \textcolor{green}{\Xi_2 \Xi_1} & & \Xi_1 \Xi_2
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\\
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\vdots & \vdots & \vdots & & \vdots
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\\
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\textcolor{blue}{\Xi_{n-2} \Xi_{n-1}} & \Xi_{n-2} \Xi_{n-1} & \Xi_{n-2} \Xi_{n-1} & & \textcolor{blue}{\Xi_{n-1} \Xi_{n-2}}
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\end{array}
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\end{gather*}
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$$
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The "[path swaps](/posts/math/permutations/1/)" shown commute only the adjacent elements.
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By contrast, the permutation $(0 ~ 2)$ commutes *Ξ*~0~ past both *Ξ*~1~ and *Ξ*~2~.
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But since we already know *Ξ*~0~ and *Ξ*~1~ commute by the above list,
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we learn at this step that *Ξ*~0~ and *Ξ*~2~ commute.
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This can be repeated until we reach the permutation $(0 ~ n-1)$ to prove commutativity between all pairs.
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