Exercise 3.F.11

Proof. Suppose that the rank of $A$ is 1. We have that all the columns are multiples of each other. Then $A$ can be written in the following form

Where the first vector is a non-zero scalar multiple of a column in $A$ and the $d$’s are the corresponding scalars of each column such that $d_j$ times the first vector equals the $j$-th column of $A$.

Conversely, suppose there are $(c_1,\dots ,c_m)\in F^m$ and $(d_1,\dots ,d_n)\in F^n$ such that $A_{j,k}=c_j{d}_k$. It is easy to see that $A$ takes the same previous form and thus each column is a scalar multiple of each other which implies that the rank of A is 1. □