Suppose A is an m-by-n matrix with $A\ne0$. Prove that the rank of A is 1 if and only if .... Suppose $A$ is an $m \times n$ matrix with $A\ne 0$. Prove that the rank of A is $1$ if and only if there exist $(c_1,...,c_m) \in {F^m}$ and $(d_1,...,d_n) \in {F^n}$ Such that ${A_{j,k}}$ = ${c_j}{d_k}$ for every $j = 1,...,m$ and every $k = 1,...,n$.
It is problem from linear algebra done right(3.F.11)
I think for firs part if we take all entries $1$ and all i,j ${c_{j}=1}$ , ${d_{k}=1}$  it will work.
but for other direction??
 A: I will give you a proof of one of the implications and leave the other one for you. Assume that $A \neq 0$ and $\text{rank} A = 1$. That means that the column space of $A$ is one-dimensional. Let $\mathbf{c}$ be a vector spanning $\text{Col} (A)$. Then there exists constant $d_1, d_2, \ldots, d_n$ so that
$$A = [d_1 \mathbf{c} \quad d_2 \mathbf{c} \quad \cdots \quad d_n \mathbf{c} ] = \mathbf{c}[d_1 \quad d_2 \quad \cdots \quad d_n]=\mathbf{c} \mathbf{d}^T.$$
$A = \mathbf{c} \mathbf{d}^T$ is equivalent to $A_{j, k} = c_j d_k$ (see why?), which proves the first implication.
A: If $\mathbf{v}_1,\dots,\mathbf{v}_r$ is a basis for the image of $A$ (aka the column space), then for any $\mathbf{x} \in \operatorname{Domain}A$,
$$A\mathbf{x} \in \operatorname{Span}\{\mathbf{v}_1,\dots,\mathbf{v}_r\} \iff A\mathbf{x} = f_1(\mathbf{x})\mathbf{v}_1+\dots+f_r(\mathbf{x})\mathbf{v}_r$$
for some coefficients $f_i(\mathbf{x})$ that depend on $\mathbf{x}$.
I claim, and you should check, these coefficients $f_i(\mathbf{x})$ are linear functionals in $\mathbf{x}$. (i.e. linear maps from $\mathbf{F}^n \to \mathbf{F}$).
If you have a linear functional $f : \mathbf{F}^n \to \mathbf{F}$ then it's associated matrix is a row vector. Because, by definition, the matrix is $[f(\mathbf{e}_1),\dots,f(\mathbf{e}_n)]$. If I call this matrix $\mathbf{w}^\top$ then $f(\mathbf{x}) = \mathbf{w}^\top\mathbf{x}$.
So if $A$ has rank $r$, then I can write $A = \sum_{i = 1}^r \mathbf{v}_i\mathbf{w}_i^\top$. So we can check:
$$A\mathbf{x} = \left(\sum_{i = 1}^r \mathbf{v}_i\mathbf{w}_i^\top \right) \mathbf{x} = \sum_{i = 1}^r \mathbf{v}_i(\mathbf{w}_i^\top\mathbf{x}) = \sum_{i=1}^r f_i(\mathbf{x})\mathbf{v}_i. $$
$r = 1$ is the exercise in question.
A: Suppose $A=M(T)$, where $T \in L(V,W)$ and $rank(A)=1$. Then we have $$rank(A)=dim(range\ T)=1$$
Let $w_1,...,w_m$ be a basis of W, there exists a vector $w$ such that $range\ T=span(w)$ with $w=\sum_{j=1}^m c_jw_j$, where $c_1,...,c_m \in F$. Thus,
$$Tv_k=d_kw=\sum_{j=1}^m c_jd_kw_j $$ where $d_k \in F, k=1...n$, which implies $A_{j,k}=c_jd_k$.
Conversely, assume there exists $(c_1,...,c_m) \in F^m$ and $(d_1,...,d_n) \in F^n$, and there exists a pair of bases on V and W such that
$$T(v_k)=\sum_{j=1}^{m}c_jd_kw_j=d_k\sum_{j=1}^{m}c_jw_j$$,
which implies that $range\ T=span(\sum_{j=1}^{m}c_jw_j)$. Thus, $rank(A)=dim(range\ T)=1$.
