# Finding the characteristic polynomial of a matrix modulus n

Given a square matrix, is it possible to calculate its characteristic polynomial modulo n? Unfortunately, this function CharacteristicPolynomial doesn't have the Modulus option that many other functions have.

There is no need for the Modulus option in CharacteristicPolynomial, since PolynomialMod serves that purpose. Assume we have a matrix m e.g. :

m = RandomInteger[10, {5, 5}]
m // MatrixForm

{{10, 1, 4, 10, 9}, {1, 9, 6, 1, 5}, {9, 7, 9, 1, 0}, {1, 10, 8, 0, 4}, {4, 0, 4, 7, 10}} then

CharacteristicPolynomial[m, x]

2310 - 4008 x + 1739 x^2 - 370 x^3 + 38 x^4 - x^5


while e.g. a characteristic polynomial modulo 5 is :

PolynomialMod[ CharacteristicPolynomial[m, x], 5]

2 x + 4 x^2 + 3 x^4 + 4 x^5


Edit

If there are specific reasons for a characteristic polynomial different than knowing its PolynomialMod, one can use directly Modulus in functions like Solve, Factor or other with that option.

Let us factor PolynomialMod[ CharacteristicPolynomial[m, x], 5] as well as CharacteristicPolynomial[m, x] over $\mathbb{Z}_5$ :

Factor[ CharacteristicPolynomial[ m, x], Modulus -> 5]

Factor[ PolynomialMod[ CharacteristicPolynomial[ m, x], 5], Modulus -> 5] ===
Factor[ CharacteristicPolynomial[ m, x], Modulus -> 5]

4 x (3 + x + 2 x^3 + x^4)

True


analogically with Solve, e.g. :

Solve[ CharacteristicPolynomial[ m, x] == 0, x, Modulus -> 5]

Solve[ PolynomialMod[ CharacteristicPolynomial[ m, x], 5] == 0, x, Modulus -> 5] ==
Solve[ CharacteristicPolynomial[ m, x] == 0, x, Modulus -> 5]

{{x -> 0}}
True


Therefore the option Modulus -> n in CharacteristicPolynomial would be superfluous.

Consider another simple polynomial :

p2 = PolynomialMod[ -1 + 4 x^2, 5]
p1 = Factor[        -1 + 4 x^2, Modulus -> 5]

4 + 4 x^2
4 (2 + x) (3 + x)


apparently they are different, however they are certainly the same modulo 5, i.e. :

PolynomialMod[ p1 - p2, 5]

0