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We prove the theorem mentioned here:
Theorem: outer product representation of a rank-one matrix
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Every rank-one matrix $A \in \mathbb{R}^{m \times n}$ can be written as an ‘‘outer product’’, or dyad: $$ where $p \in \mathbb{R}^{m}, \; q \in \mathbb{R}^{n}$. |
Proof: For any non-zero vectors $p \in \mathbb{R}^{m}, q \in \mathbb{R}^{n}$, the matrix $pq^T$ is indeed of rank one: if $x \in \mathbb{R}^{n}$, then
$$pq^Tx=(q^Tx)p.$$
When $x$ spans $\mathbb{R}^{n}$, the scalar $q^Tx$ spans the entire real line (since $q \neq 0$), and the vector $(q^Tx)p$ spans the subspace of vectors proportional to $p$. Hence, the range of $pq^T$ is the line:
$$\mathbf{R}(pq^T) = \{tp\; : \;t \in \mathbb{R}\},$$
which is of dimension 1.
Conversely, if $A$ is of rank one, then its range is of dimension one, hence it must be a line passing through $0$. Hence for any $x$ there exist a function $t: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$Ax = t(x)p.$$
Using $x = e_i$, where $e_i$ is the $i$-th vector of the standard basis, we obtain that there exist numbers $t_1, … t_n$ such that for every i:
$$Ae_i = t_ip, \quad i=1,…,n.$$
We can write the above in a single matrix equation:
\[ A \begin{pmatrix} e_1 & \cdots & e_n \end{pmatrix} = p \begin{pmatrix} t_1 & \cdots & t_n \end{pmatrix} \]
Now letting $q = \begin{pmatrix} t_1 \\ \vdots \\ t_n \end{pmatrix} \in \mathbb{R}^{n}$, and realizing that the matrix $(e_1, …, e_n)$ is simply the identity matrix, we obtain $A = pq^T$, as desired.
