For $v ∈ \mathbb{R}^m$, prove $\operatorname{rank}(vv^T) = 1$, where $v \ne 0$. I have this information from my notes:$\def\rk{\operatorname{rank}}$
Let $A ∈ \mathbb{R}^{m\times n}$. Then 


*

*$\rk(A) = n$

*$\rk(A^TA) = n$

*$A^TA$ is invertible.


In my case, $n = 1$, so I would need to show $\rk(vv^T) = \rk(v^Tv) = \rk(v) = 1$. Suppose $A^TAx = 0$. Because $A^TA$ is invertible, I can multiply both sides by its inverse to get $x = 0$, meaning the nullity of $A^TA$ is $0$. Can I apply the same logic to $AA^T$? i.e. I have some matrix $B = A^T$, so $B^TB$ = $AA^T $ has a nullity of $0$ (and therefore they have the same rank by the rank-nullity theorem)?

Note that $v v^T v = \|v\|^2 v \neq 0$, hence $\operatorname{rk} (v v^T) \ge 1$.
Note that $v v^T x = (v^T x) v \in \operatorname{sp} \{ v \}$ for all $x$. Hence ${R (v v^T)} = \operatorname{sp} \{ v \}$ and hence
$\operatorname{rk} (v v^T) = 1$.
