Question: 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!

Answer: 
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 $\vec{b}^T = (b\_1, \dots, b\_N)$.
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.