Rank-one matrices: a representation theorem

L. El Ghaoui

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 $A \in \mathbb{R}^{m \times n}$ can be written as an ‘‘outer product’’, or dyad

$A=pq^{T},$

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

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

$p q^T x=\left(q^T x\right) p .$

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

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

which is of dimension 1.
Conversely, if $A$ is of rank one, then its range is of dimension one, hence it must be a line passing through 0 . Hence for any $x$ there exist a function $t: \mathbf{R} \rightarrow \mathbf{R}$ such that

$A x=t(x) p .$

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

$A e_i=t_i p, \quad i=1, \ldots, p .$

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

$A=p q^T,$

as desired.

Rank-one matrices: a representation theorem

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