Let $Q$ be a $m \times n$ matrix. Prove $\det (Q Q^T) = 0$ if $m > n$. Reading a book on another mathematical subject a proof makes use of linear algebra. Without explanation the author states the following:

Let $Q$ be a $m \times n$ matrix. If $m > n$ then $\det (Q Q^T) = 0$

I've been puzzling with some matrices and in all cases I get $\det(QQ^T) = 0$ for some chosen $Q$.
Can someone tell me if the statement is true ? And if it is give a proof ? Also is it always true that $\det (QQ^T) \neq 0$ if $m \le n$ ?
 A: Rank of $Q$ is less than or equal to $n$ and therefore rank of the matrix $QQ^T$ is also less than or equal to $n$, which is less than $m$. This means null space of $QQ^T$ has non-zero dimension and that zero is an eigenvalue of the matrix and therefore determinant is zero.
However if $m\leq n$, then we cannot say that the determinant of $QQ^T$ is non-zero. Simply assume that $Q$ is a full zero element matrix.
A: As $Q^T$ is $n\times m$, then $Q^T$ has $m$ rows $\boldsymbol{q}_1,\ldots,\boldsymbol{q}_m$ which are $n$-vectors, $m>n$, and thus they are linearly dependent. Thus there exist $c_1,\ldots,c_m$, s.t.
$$
c_1\boldsymbol{q}_1+\cdots+c_m\boldsymbol{q}_m=0,
$$
equivalently
$$
Q^T\boldsymbol{c}=0,
$$
where $\boldsymbol{c}=(c_1,\ldots,c_m)$,
and thus
$$
QQ^T\boldsymbol{c}=0,
$$
which implies that $QQ^T$ is singular, and hence $$\det QQ^T=0.$$
Also, if $m\le n$, then $\det QQ^T\ne 0$ iff $Q$ has rank $n$.
A: The homogeneous system $Q^TX=0$ has a nonzero solution, because $n<m$, i.e. the number of equations is less than the number of variable. Therefore $QQ^TX=0$ has a nonzero solution too, i.e. $QQ^T$ is not invertible, so $det(QQ^T)=0$.
