Can we predict the number of non-zero singular values in this case? If there are two matrices $P$ (dimensioned $m\times 1$) and $Q$ ($n\times1$) and a matrix $M$ is constructed by $M=PQ'$ (where the ' indicates transpose), so $M$ is of size $m\times n$. 
Does $M$ have only one non zero singular value? If so, why? 
All I can think of is that the singular values of $M$ are the square-rooted eigen values of $MM'$ i.e of $(P)(Q^2)(P')$ or $(Q)(P^2)(Q')$. How can one say anything about the number of non zero singular values M has from this?

You have, for any $X$ of size $m\times 1$, 
$$
MM'X=PQ'QP'X=(Q'Q)(P'X)P
$$
(Note that $Q'Q$ and $P'X$ are $1\times1$, i.e. a scalar).
So if $Y$ is any eigenvector of $MM'$ with nonzero eigenvalue, i.e. $MM'Y=\lambda Y$, necessarily $Y$ is colinear with $P$, since we get $\lambda Y=(Q'Q)(P'Y)P$. Thus, $MM'$ can have a single nonzero eigenvalue, with multiplicity one. 
