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.
 A: 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.
A: $\newcommand{\im}{\mathrm{Im}}$
$\newcommand{\ker}{\mathrm{Ker}}$
$\newcommand{\rank}{\mathrm{rank}}$
View $M$ and $N$ as linear operators from $F^n$ to $F^n$. Then $MN = 0$ implies
$\im(N) \subset \ker(M)$, hence $\dim(\im(N)) \leq \dim(\ker(M))$. By the rank-nullity theorem, this is equivalent to say $\rank(N) \leq n - \rank(M)$, i.e.,
\begin{align*}
\rank(M) + \rank(N) \leq n. \tag{1}
\end{align*}
Now by the rank inequality $\rank(A + B) \leq \rank(A) + \rank(B)$ and the rank equality $\rank(A) = \rank(A^T)$, it follows that
\begin{align*}
& \rank(M + M^T) \leq \rank(M) + \rank(M^T) = 2\rank(M), \\
& \rank(N + N^T) \leq \rank(N) + \rank(N^T) = 2\rank(N),
\end{align*}
whence by $(1)$ we can conclude
\begin{align*}
\rank(M + M^T) + \rank(N + N^T) \leq 2(\rank(M) + \rank(N)) \leq 2n. \tag{2}
\end{align*}
If $\rank(M + M^T) + \rank(N + N^T) < 2n$, then $\rank(M + M^T) < n$ or $\rank(N + N^T) < n$, and we are done.
If $\rank(M + M^T) + \rank(N + N^T) = 2n$, then $(2)$ implies
$\rank(M) + \rank(N) = n$. Since $n$ is an odd number, at least one of $\rank(M)$
and $\rank(N)$ is strictly less than $n/2$, say $\rank(M)$. Then $\rank(M + M^T) \leq 2\rank(M) < n$, implying $\det(M + M^T) = 0$.
A: 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 $det(N)=0$.
