Determinant Equal to $0$ If Product of Two Matrices Equal to $0$ For two $n \times n$ matrices $M$, $N$ ($n$ is odd) s.t. $MN=0$, either $\det(M+M^T)=0$ or $\det(N+N^T)=0$.
I am trying to show this through two cases: 1) WLOG, if $M$=0, then we are done. 2) If the product of two non-zero square matrices is zero, then both factors must be singular. But for 2), it shows that $\det(M)=\det(N)=0$, which does not guarantee that $\det(M+M^T)=0$ or $\det(N+N^T)=0$, and I cannot seem to find a way to use the information that $n$ is odd here.
If $N$ has rank $r$ (that is, $r$ independent columns) then $M$ must send all those columns to $0$, so $M$ has nullity at least $r$, and rank at most $n-r$. In other words, their ranks add up to at most $n$.
Because $n$ is odd, $M$ and $N$ can't both have rank $\frac n2$: one of them has rank strictly less. Say that $M$ has rank less than $\frac n2$. Then so does $M^{\mathsf T}$; as a result, $M + M^{\mathsf T}$ has rank less than $n$, and is singular.
If $MN=0$ taking determinants you conclude either $det(M)=0$ or $det(N)=0$. Assume the first. Then also $det(M^t)=0$ and both $M$ and $M^t$ correspond to transformations with equal rank less than $n$. The maximum possible dimension of the range of $M+M^t$ is thus twice the rank of $M$ so even and consequently less than $n$. Therefore $M+M^t$ is singular. Similar argument if $det(N)=0$.