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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.

Rodrigo de Azevedo
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NPH
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2 Answers2

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Hint: as $A$ has rank 1, all the columns of $A$ are proportional to the same vector.

mookid
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  • And at least 1 element of $A$ is not zero. – Alexander Vigodner Oct 22 '14 at 21:39
  • of course. This is not an equivalence. – mookid Oct 22 '14 at 21:39
  • So then I would just take a column $c_i$ and construct say $X$ by taking in each entry $x_i$ the constant needed to scale the column $c_i$ to one of the other columns $c_j$, $i \neq j$. Of course, I could do it the other way around or choose a different column so there is my ``non-uniqueness" I have from above? – NPH Oct 22 '14 at 21:43
  • exactly! can you take it from here? – mookid Oct 22 '14 at 21:48
  • @mookid Indeed! Thank you. We did not discuss of rank in this fashion in class so as soon as you commented that rank could be thought of in that fashion the answer became clear at once. Thank you so much for the help! – NPH Oct 22 '14 at 21:51
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$X$ and $Y$ are unique up to a proportionality constant.

orangeskid
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