When proving that the determinant of an odd dimension anti-symmetric matrix is zero, how it it that -det(A) implies that det(A) =0?

Question

So the proof that the determinant of an odd dimension anti-symmetric matrix is zero can be written as:

$$\det(A)=\det(A^T)=\det(−A)=(−1)^{2n+1}\det(A)=−\det(A)$$

$\det(A)=0$

But I don't get the last bit, how exactly is it that $-\det(A)$ implies that $det(A)= 0$? I feel like I'm missing something.

Thanks for you help.

Answer 1

The first line of the proof shows that $\det(A)=-\det(A)$. Adding $\det(A)$ to both sides we get $2\det(A)=0$, so $\det(A)=0$.