Why is rank $uv^{T}$ always equal to 1? Let $u \in \mathbb{R}^m$ and $v \in \mathbb{R}^n$ and $u,v \neq 0$. Since $u$ is an $m\times1$ matrix and $v^T$ is a $1\times n$ matrix, the product $uv^{T}$  is a $m\times n$ matrix. Then:

$$\text{rank} (uv^{T}) =1$$

Just to illustrate: pick a random vector $u$ in $\mathbb{R}^4$, and $v$ in $\mathbb{R}^5$, for example: $u= (1, 2, 3, 4),$ and $v= (1, 2, 3, 4, 5)$. The rank of the matrix $m\times n$ is $1$ which can be checked by row-reducing. Chosing arbitrary vectors $u,v \neq 0$, the rank of the product matrix mxn of $uv^{T}$ is always $1$.
Why is this? 
Any help would be greatly appreciated.
 A: Since $v\neq 0$, the range of $uv^T$ is simply $\operatorname{span}\{u\}$, which necessarily has dimension $1$ since $u \neq 0$. 
A: Write $$\vec v=\begin{pmatrix}v_1\\v_2\\\dots\\v_n\end{pmatrix}$$ where $v_i, i=1,\dots,n$ are scalars (numbers) and similarly for $\vec u$. Then, for $u\in \mathbb R^m$, $uv^T$ is indeed an $m\times n$ matrix which may be written as $$\vec{u}v^T=\begin{pmatrix}u_1\\u_2\\\dots\\u_m\end{pmatrix}(v_1,v_2,\dots, v_n)=\begin{pmatrix}v_1\begin{pmatrix}u_1\\u_2\\\dots\\u_m\end{pmatrix},\,v_2\begin{pmatrix}u_1\\u_2\\\dots\\u_m\end{pmatrix},\dots ,\, v_n\begin{pmatrix}u_1\\u_2\\\dots\\u_m\end{pmatrix}\end{pmatrix}$$ Hence any column of $uv^T$ is a multiple of the first non-zero column (the first column for which the corresponfing $v_i\neq 0$) and hence linearly dependent with it. Hence $$\text{rank }(uv^T)=\max {\{\text{rank }(u), \text{rank }(v^T)\}}=1$$ since you assumed that $u,v \neq 0$. 
A: The matrix product $[u]\,[v^\top]$ encodes a composition of linear maps ${\mathbb R}^n\to{\mathbb R}\to{\mathbb R}^m$. Then single ${\mathbb R}$ in the middle enforces ${\rm rank}\bigl([u]\,[v^\top]\bigr)\leq1$. On the other hand the elements of $[u]\,[v^\top]$ are the $m\cdot n$ products $u_iv_j$. By assumption on $u$ and $v$ at least one of these products is $\ne0$, and this excludes $[u]\,[v^\top]=0$.
A: Remember that if $\;AB\;$ is a well defined product of matrices, it is always true that $\;\text{rank}\,AB\le\min(\text{rank}\,A,\,\text{rank}\,B)\;$.
In your case, as matrices, $\;\text{rank}\,u=\text{rank}\,v^t=1\;$ and thus we almost get the claim ("almost" unless, as vectors, these two are orthogonal)
For (counter) example
$$(1\;0)\binom 01=(0)$$
If for you the usual vector is a column one, then $\;uv^t\;$ is an $\;n\times n\;$ matrix instead of a $\;1\times1\;$ as above...but the same applies but this time there is no possibility of "orthogonality" in the way shown above.
