Proof involving matrix rank

Question

Let $A \in M_n$. Prove $r(A)=1$ if and only if there exist matrices $B=\begin{pmatrix} b_1\\ b_2\\ \vdots\\ b_n \end{pmatrix}\in M_{n1}$ and $C=\begin{pmatrix} c_1 &c_2 &\ldots &c_n \end{pmatrix}\in M_{1n}$ such that $A=BC$.

My attempt:

Let $B=\begin{pmatrix} b_1\\ b_2\\ \vdots\\ b_n \end{pmatrix}\in M_{n1}$ and $C=\begin{pmatrix} c_1 &c_2 &\ldots &c_n \end{pmatrix}\in M_{1n}$ such that $A=BC$. Then $r(B)=1$ if $B$ is not a zero matrix, and $r(C)=1$ if $C$ is not a zero matrix. Because of the following theorem: $r(X)=r(Y)=n \Rightarrow r(XY)=n$ we can say $r(B)=r(C)=1\Rightarrow r(BC)=r(A)=1$.

Now I'm not sure about the other direction... What is the assumption here, is it just $r(A)=1$? Or is it $r(A)=1$ and $A=BC$? And then what do we have to prove, that $B\in M_{n1}$ and $C\in M_{1n}?$ If so, then how do we prove that?

Answer 1

The assumption should just be that the rank of $A$ is $1$. Once that is there, this means all rows of $A$ are multiples of one non-zero row (call it the $i^{\text{th}}$ row). $$A=\begin{bmatrix} k_1\mathbf{u}\\ \vdots \\ \mathbf{u}\\ \vdots \\k_{n}\mathbf{u}\end{bmatrix}=\begin{bmatrix} k_1\\ \vdots \\ 1 \\ \vdots \\k_{n}\end{bmatrix}[\mathbf{u}^T].$$

Answer 2

The other direction means that the assumption is simply that $r(A)=1$, and you have to prove that there exists a pair of matrices $B, C$ such that $A=BC$.

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To prove that, you should start by proving that

1. $A$ has a nonzero row.
2. each row of $A$ is a scalar multiple of the nonzero row of $A$.

Once you have that, it's easy to find $B$ and $C$.