Question

Prove if A is skew-Hermitian, then |A| is either real or is a pure imaginary number.

@dan815 @ikram002p @ganeshie8

Answer 1

prove that all the eigenvalues are purely imaginary or zero.

then use |A|=product of its eigenvalues.

Answer 2

The overline means complex conjugate, T is transpose, and - is what defines a matrix being skew hermitian. \[\LARGE A=- \overline {A^T}\]

So we take the determinate of both sides to get: \[\LARGE |A|=|- \overline {A^T}|\] When we distribute the -1 to all the entries, we factor it out of each row of the determinant and note that transposing doesn't change the determinant \[\LARGE |A|=(-1)^n| \overline A|\] so now let's represent the determinant of |A|=a+bi so we have: \[\LARGE a+bi=(-1)^n ( a-bi)\]

So when the number of rows is even it's real and when the number of rows is odd then it's imaginary.