Let A be an mxn matrix of rank 1. Show that there exist non zero vectors.

Question

Let A be an $mxn$ matrix of rank 1. Show that there exist non zero vectors $x ε R^m$ and $y ε R^n$ so that $A = xy^T$.(Hint: Try a simple case. Also compute $xy^T$ for some simple choices x,y.)

Attempt:

I know rank = 1 means that there is 1 linearly independent row but I am confused on how to resolve this question.

I would appreciate the help thanks.

Answer 1

Rank one means that $\dim {\cal R} A =1$, hence ${\cal R} A = \operatorname{sp} \{ v \}$ for some $v$.

You have $A e_k = u_k v$ for some $u_k$ (and $v \in {\cal R} A$).

Let $u = (u_1,...,u_n)^T$, then $A = v u^T$.

Answer 2

If two columns are linearly independent then the rank will be 2 or more. This shows that all nonzero columns are scalar multiples of each other. For simplicity call the first column to be nonzero and denote it by $x$.

Take the scalars $a_1,a_2,\ldots,a_n$ such that $a_jx$ is the $j$ th columns of the given matrix. (clearly $a_1=1$;if zero vector is a column of the matrix then corresponding $a_j$ would be zero).

Take $y$ to be the column vector defined by $a_j$'s. Now $xy^t$ would be your rank 1 matrix