Let $\displaystyle \bf{x}$ and $\displaystyle \bf{y} \in \mathbb{R}^{n}$ be 2 non zero column vectors. Let $\displaystyle \bf{y}^{T}$ denote the transpose of $\displaystyle \bf{y}$. Let $\displaystyle A=\bf{x} \bf{y}^{T}$. What is the rank of A?
Let $\displaystyle \bf{x}$ and $\displaystyle \bf{y} \in \mathbb{R}^{n}$ be 2 non zero column vectors. Let $\displaystyle \bf{y}^{T}$ denote the transpose of $\displaystyle \bf{y}$. Let $\displaystyle A=\bf{x} \bf{y}^{T}$. What is the rank of A?
What do you mean by $\displaystyle y^T$, because we don't know what "y" is! I think maybe you're referring to the case where x=y. And I think in this case $\displaystyle xx^T$ will have the full column rank if $\displaystyle det(x^Tx) \neq 0$, and it will have the full row rank if $\displaystyle det(xx^T) \neq 0$.
Since $\displaystyle x^Tx$ is square, and it is an nxn matrix, it follows that $\displaystyle x^Tx$ is invertible if and only if $\displaystyle x^Tx$ has rank n. However $\displaystyle x^Tx $has the same rank as x itself, so $\displaystyle x^Tx$ is invertible if rank(x)=n (this means if x has full column rank).
Again I'm assuming x=y.
I see no reason to assume that x= y. The problem makes sense without that. For example, if $\displaystyle x= \begin{bmatrix}a \\ b\end{bmatrix}$ and $\displaystyle y= \begin{bmatrix}c \\ d\end{bmatrix}$, then $\displaystyle y^T= \begin{bmatrix}c & d\end{bmatrix}$ and
$\displaystyle xy^T= \begin{bmatrix}a \\ b\end{bmatrix}\begin{bmatrix}c & d\end{bmatrix}=$ $\displaystyle \begin{bmatrix}ac & ad \\ bc & bd\end{bmatrix}$. The determinant of that is is acbd- adbc= 0 so it does NOT have rank 2. It is not, unless at least one of x and y is the 0 vector, all "0"s so it does not have rank 0. It has rank 1.