A circulant matrix and its transpose It is well-known that a circulant matrix $A$ of size $n \times n$ is isomorphic to a polynomial $$p(x) \bmod x^n - 1.$$  If we consider the transpose $A^T$, what is the corresponding polynomial called? Is there something such as transpose of a polynomial?
 A: The answer is yes, there is indeed something such as the transpose of a polynomial. Being somewhat esoteric, it rarely [to my knowledge] occurs outside coding theory. The transpose is $$A^T \sim a(x) ^T  = \sum^{n-1}_{i=0} a_{n-i} x^i.$$
Easily verified, it also holds that
$$ a(x)^T = a(x^{n-1}) \bmod (x^n-1). $$
A: Using the equipment provided by @CarlLöndahl, another way of presenting that polynomial's transpose is
$a(x)^T = a(x^{-1}) \mod (x^n - 1)$
Furthermore, the product
$a(x)^T a(x) \sim A^TA$
is also an $n \times n$ circulant, one which is of some interest in autocorrelation modelling, and even musical Pitch Class Theory.
It allows you to evaluate the difference distribution of a set. For example where $a(x) = 1 + x + x^3 + x^4$ represents a set of integers, $\{0, 1, 3, 4 \}$ from $\mathbb{Z}_6$.
The multiset of differences is
$\{ 1-0, 3-0, 4-0, 3-1, 4-1, 4-3 \} \cup \{ 0-1, 0-3, 0-4, 1-3, 1-4, 3-4 \}$
$= \{ 1, 3, 4, 2, 3, 1 \} \cup \{ -1, -3, -4, -2, -3, -1 \}$
$\equiv_6 \{ 1, 3, 4, 2, 3, 1 \} \cup \{ 5, 3, 2, 4, 3, 5 \}$
$= \{ 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5 \}$
From which we may account 2 of 1, 2 of 2, 4 of 3, 2 of 4 and 2 of 5. Present also is 4 of 0 if you include differences between identical elements, to take the total to the expected 16.
But instead of counting them 'by hand' we can multiply $a(x)$ by its transpose, $a(x^5)$ or, more conveniently in this case by $a(x^{-1})$ - it saves a little time with the exponents:
$(1 + x + x^3 + x^4)(1 + x^{-1} + x^{-3} + x^{-4})$
$= (1 + x^{-1} + x^{-3} + x^{-4}) + (x + 1 + x^{-2} + x^{-3}) + (x^3 + x^2 + 1 + x^{-4}) + (x^4 + x^3 + x + 1)$
$= 4 + 2x + x^2 + x^{-4} + 2x^3 + 2x^{-3} + x^4 + x^{-2} + 2x^{-1}$
$\equiv_6 4 + 2x + 2x^2 + 4x^3 + 2x^4 + 2x^5$
whereupon we recover the frequency of differences of size $d$ directly by reading off the coefficients of $x^d$ in $a^T(x)a(x)$.
One may of course recover the same result by multiplication of the two $6 \times 6$ circulant matrices (in either order - it's commutative) derived from row shifts of (1, 1, 0, 1, 1, 0) with its transpose - the circulant matrix derived similarly from (1, 0, 1, 1, 1, 0). The result will be the (perforce symmetric) circulant matrix (4, 2, 2, 4, 2, 2).
