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)?
 A: You don't need to use any machinery to prove this.
The rank of a matrix is the dimension of the column space of the matrix.
It's easy to see that
\begin{align*}
vv^T =
\begin{pmatrix}
v_1v_1 & v_1v_2 & \cdots & v_1v_n\\
v_2v_1 & v_2v_2 & \cdots & v_2v_n\\
\vdots & \vdots & \ddots & \vdots\\
v_nv_1 & v_nv_2 & \cdots & v_nv_n
\end{pmatrix}, \mathrm{where}\ v = \pmatrix{v_1\\v_2\\\vdots\\v_n}
\end{align*}
Then the column space of $vv^T$ is the span of the column vectors, i.e.
\begin{align*}
\mathrm{span}
\begin{Bmatrix}
\begin{pmatrix}
v_1v_1\\
v_2v_1\\
\vdots\\
v_nv_1\\
\end{pmatrix},
\begin{pmatrix}
v_1v_2\\
v_2v_2\\
\vdots\\
v_nv_2\\
\end{pmatrix}, \cdots,
\begin{pmatrix}
v_1v_n\\
v_2v_n\\
\vdots\\
v_nv_n\\
\end{pmatrix}
\end{Bmatrix}
\end{align*}
But clearly each column vector is a scalar multiple of $v$, for example the first column is equal to $v_1v$ where $v_1$ is a scalar. Thus the span of all of these vectors is just equal to the span of $v$, so the dimension of the column space is $1$.
Hence rank$(vv^T)= 1$.
A: 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
$\operatorname{rk} (v v^T) = 1$.
A: $\def\rk{\operatorname{rank}}$No you cannot apply the same logic to $vv^T$, in other words for $AA^T$ where $A$ is (still) a $1\times m$ matrix, even though the conclusion that it has rank$~1$ is true. The point is that you are trying to apply the implications in your notes with the roles of $m$ and $n$ interchanged with respect to your initial application, in other words with $m=1$ and $n>1$, but the first two points are still talking about $n$ (and necessarily so: if you replace them by $m$ the implication is not valid any more). The first point (which is the hypothesis) could be more clearly stated as "$A$ has full column rank" (meaning all its columns are linearly independent) to emphasise the asymmetry with respect to rows and columns of the statements.
Assuming that you know (or have been told) that the first dotted point implies the two other dotted points (which is true) then substituting $v^T$ for $A$ it says: "if $v^T\in\Bbb R^{1\times n}$ and $\rk(v^T)=n$, then $\rk(vv^T)=n$ and $vv^T$ is invertible". This is valid, but unhelpful since the hypothesis $\rk(v^T)=n$ can never be satisfied if $n>1$, since the rank of an $1\times n$ matrix is at most $1$. A row vector cannot have linearly independent columns (if it has more than one of them). Note by the way that the conclusions that $\rk(vv^T)=n$ and $vv^T$ is invertible would contradict what you are trying so prove (namely $\rk(vv^T)=1$), so it is fortunate that you cannot deduce them in this situation.
The conclusion $\rk(vv^T)=1$ is easy to obtain since $\rk(vv^T)\leq1$ follows from the general fact that the rank of a product of matrices cannot exceed the rank of any factor in the product. All that remains to show is that $vv^T\neq0$, which can be easily done in many ways. I would like to point out that while $\rk(vv^T)=1=\rk(v^Tv)$ in this case, it is not true that $AB$ and $BA$ always have the same rank.
