__Problem statement:__
Let

be skew-Hermitian, i.e,

.

Show that

is nonsingular.

__Attempt:__
Well, I know that an arbitrary square matrix can be written as the sum of a Hermitian matrix and a skew-Hermitian matrix.

This tells me that

is Hermitian, and since the determinant of a Hermitian matrix is real it is

is not singular.

First, is not hermitian in general, as you can easily check by using . Second, even if it were then its determinant is real so...so what? It could be zero as zero is a real number! In fact, the zero matrix is Hermitian, and thus your argument doesn't work. Tonio
Now, I would like to express this in an more elegant manner using actual math

Hope someone can give me a hint or two, thanks.