Prove that every $m \times n$ matrix of rank $1$ has the form $A=XY^t$, where $X,Y$ are $m$- and $n$-dimensional column vectors. How uniquely determined are these vectors$?$
My attempt: I thought about using the dimension formula somehow. I know that nullity $+$ rank $=$ dim $A$. We know dim $A=mn$ but this got me nowhere. Then I tried doing an example. I took
$$
A=\begin{pmatrix} 1 & 3 \\ 0 & 0 \end{pmatrix}
$$
which has rank $1$. Then I can choose
$$
X=\begin{pmatrix}1 \\ 0 \end{pmatrix}, Y=\begin{pmatrix} 1 \\ 3\end{pmatrix}
$$
or I could choose
$$
X=\begin{pmatrix} 1/2 \\ 0 \end{pmatrix}, Y=\begin{pmatrix}2 \\ 6 \end{pmatrix}
$$
so that they are not uniquely determined. However, I can interchange the entries of $X$ and $Y$ to get ''different'' $X,Y$. But one the entries of $X$ (or $Y$ are chosen) then entries of $Y$ (or $X$) are then forcefully chosen. So I believe the final answer for that will be that they are not unique but choosing one vector determines both.
However, this method works for matrices $A$ with only one nonzero row and certainly many $A$'s have rank $1$ with many nonzero row. I know that $A$ has rank $1$ means I can use a series of elementary matrices to place it in the form I used above, but am unsure of how to use this since I'm not sure how to reconcile this with my ''choices'' for $X,Y$. Is there a way to do this or am I on the wrong track.
Note: I do not need nor want solutions but rather hints or thoughts on how I should proceed.
Best Answer
Hint: as $A$ has rank 1, all the columns of $A$ are proportional to the same vector.