Question

Show that if a square matrix K over Z_p ( p prime) is involutory ( or self-inverse), then det K=+-1

(An nxn matrix K is called involutory if K is invertible and K^-1 = K)

from Applied algebra

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Answer 1

Note that Z_p where p is prime is a field

As involutary matrix A is defined as

A = A^-1

i.e. A² = I

Thus characterstic polynomial is given by x² -1 = (x + 1)(x-1)

Minimal polynomial can be (x+1) or (x-1) or (x+1)(x-1)

If mimimal polynomial is

a) x + 1

i.e. matrix A must satisfy it

A + I = 0

A = -I

thus, det(A) = det(-I) = -1

b) If minimal polynomial is x -1

then,    Matrix A satisfies it

A - I = 0

A = I

det(A) = det(I) = 1

c) Minimal polynomial is (x-1)(x+1)

Implies that eigen values are a mix of +1 or -1

We know det A = product of all eigen values = 1^m(-1)ⁿ

if n is even the det is +1 and if n is odd then   -1

thus det A = +-1