Prove that the characteristic polynmial of a skew mapping satisfies $c(-\lambda) = (-1)^n c(\lambda)$

Question

In the book of Linear Algebra by Greub at page 233, it is asked that

Prove that the characteristic polynomial of a skew mapping satisfies the equation $$c(-\lambda) = (-1)^n c(\lambda).$$

I have tried showing this directly from the definition of determinant by separating the cases when $n$ is odd and even, and then tried a couple of argument, but couldn't arrive at the result with a clear statement, so how can we show this result ?

Note, any help or hint is also appreciated.

Answer 1

Well, let's see, how's this:

For any matrix $B$, we have

$\det(B^T) = \det(B); \tag 1$

if we take

$B = C - \lambda I, \tag 2$

then

$\det((C - \lambda I)^T) = \det(C - \lambda I); \tag 3$

now

$\det((C - \lambda I)^T) = \det (C^T - \lambda I); \tag 4$

since $C^T = - C$, we obtain

$\det(C^T - \lambda I) = \det(-C - \lambda I) = \det(-(C + \lambda I))$ $= (-1)^n \det(C + \lambda I) = (-1)^n \det (C - (-\lambda)I); \tag 5$

combining (3), (4), and (5) we see that

$\det(C - \lambda I) = (-1)^n \det (C - (-\lambda)I), \tag 6$

or

$(-1)^n \det(C - \lambda I) = (-1)^{2n} \det (C - (-\lambda)I) = \det(C - (-\lambda) I). \tag 7$