Rank-one matrices: a representation theorem

We prove the theorem mentioned here:

Theorem: outer product representation of a rank-one matrix

Every rank-one matrix $A in mathbf{R}^{m times n}$ can be written as an ‘‘outer product’’, or dyad

where $p in mathbf{R}^m$ , $q in mathbf{R}^n$ .

Proof: For any non-zero vectors $p in mathbf{R}^m$ , $q in mathbf{R}^n$ , the matrix pq^T is indeed of rank one: if $x in mathbf{R}^n$ , then

When spans $mathbf{R}^n$ , the scalar q^Tx spans the entire real line (since q ne 0 ), and the vector (q^Tx) p spans the subspace of vectors proportional to . Hence, the range of pq^T is the line

$mathbf{R}(pq^T) = left{ t p ~:~ t in mathbf{R} right},$

which is of dimension .

Conversely, if is of rank one, then its range is of dimension one, hence it must be a line passing through . Hence for any there exist a function t: mathbf{R} rightarrow mathbf{R} such that

Using x = e_i , where e_i is the -th vector of the standard basis, we obtain that there exist numbers t_1,ldots,t_n such that for every :

We can write the above in a single matrix equation:

$A left( begin{array}{ccc} e_1 & ldots & e_n end{array}right) = p left( begin{array}{ccc} t_1 & ldots & t_n end{array}right) .$

(Indeed, the previous equation is the above, written column by column.)

Now letting $q = (t_1,ldots,t_n) in mathbf{R}^n$ , and realizing that the matrix (e_1,ldots,e_n) is simply the identity matrix, we obtain

as desired.