Why must n be even? Skew Symmetric matrices Let B be any real, nonsingular $nxn$ matrix, where n is even, and set $A=B-B^T$. Show that A does not admit an LU decomposition without pivoting.

I know that A is a skew symmetric matrix. The 2x2 case is 

0 -a

a 0

It is easy to see we can't eliminate a so we must pivot. What I don't understand is why n must be even? I think even odd cases show this... Thanks.
I don't think any invertible skew-symmetric real matrix admits an LU decomposition without pivoting. 

From $A^T=-A$, if $A=LU$ we get $$LU=-(LU)^T=-U^TL^T.$$ We are assuming that $A$ is invertible, so $L$ and $U$ are also invertible. Thus $$(U^T)^{-1}L=-L^TU^{-1}.$$ Here the left-hand-side is lower triangular, and the right-hand-side is upper triangular; this implies that both are diagonal. So $(U^T)^{-1}L=D$ for an invertible diagonal matrix $D$. We can write $L=U^TD$. We also have $-L^TU^{-1}=D$, and we obtain
$$
L^T=-DU,
$$
which we can write as $L=-U^TD.$ It follows that $U^TD=-U^TD$, which would imply that $U^TD=0$, a contradiction since $U$ and $D$ are invertible.
If $n$ is odd then the characteristic polynomial for $A$ emits at least one real root. And zero happens to be a root which means $A$ is singular. 

As an exercise, try to see if you can prove that $A$ is singular when $n$ is odd. 

Hint: Consider $\det(A) = \det(A^T) = \det(-A)$.