Why matrix of rank one can be written as a tensor product of two vectors? I find in Wikipedia that a matrix $A\in\mathbb{M}^{N\times n}$ has rank one if it can be written as the following tensor product
$$A=a\otimes b,$$ with $b\in \mathbb{R}^N$ and $a\in\mathbb{R}^n$.
I know that a matrix has rank one if one is the maximum number of linearly independent colums of the matrix.
Why a rank one matrix can be written in that way?
Thank you!
 A: Let a matrix $A$ have columns $\vec{a}_i$, $i = 1, \dots, N$. A rank one matrix is characterised by the fact that all columns $\vec{a}_i$ are linearly dependent, that is
$$
\vec{a}_i = b_i \vec{a}
$$
for some vector $\vec{a}$ and scalars $b_i$. Thus, the matrix $A$ takes the form
$$
A = (b_1 \vec{a}, \dots, b_N \vec{a}) = \vec{a} \vec{b}^T = \vec{a} \otimes \vec{b},
$$
where $\vec{b}^T = (b_1, \dots, b_N)$.
A: The matrix $A$ is matrix of some linear map $\mathcal{A} \in \{\mathbb R^n \to \mathbb R^N\}$. Having rank $1$ means that in some bases $\{a_1,a_2,...,a_n\}$ and $\{b_1,b_2,...,b_N\}$ of respective spaces its matrix is $$\pmatrix{ 1 & 0 & ... & 0 \\ 0 & 0 &... & 0 \\ ... & ... & ... & ... \\ 0 & 0 & ... & 0 }$$
and thus $\mathcal A = a_1 \otimes b_1$. This equality holds for any numeric representation of $\mathcal A, a_1, b_1$ including initial basis you wrote matrix $A$ in.
A: For example $$\begin{pmatrix}1&1&1\\2&2&2\\3&3&3\end{pmatrix}$$
 has rank one and is $(1,1,1)\otimes (1,2,3)$
A: Look at the first non-zero column of $A$. That is $b$. Then every other column of $A$ is a multiple $b$, since the rank of $A$ is $1$. $a$ consists of these proportionality constants, in order.
