If $Ax=0$ for all $x\in \{b\}^{\perp}$ then prove that there exists $a\in \mathbb R^{n}\setminus\{0\}$ such that $A=a\otimes b.$ If $Ax=0$ for all $x\in \{b\}^{\perp}$ then prove that there exists $a\in \mathbb R^{n}\setminus\{0\}$ such that $A=a\otimes b.$ Also it's given that $Ab\neq 0.$
Actually I can find a vector $a\in \mathbb R^n\setminus \{0\}$ such that $A=a\otimes Ab$ because the columns or rows of $A$ span a vector space of rank $1$.
But how can I prove that $A=a \otimes b.$
Any help is appreciated.
Thank you.
Edit: I think if we can get a solution of the equation $Ax=b$ then we are done. So can we get a solution of this equation?

The columns of $A$ span a vector space of dimension $1$, hence the matrix $A$, of rank $1$, can be written $A=uv^T$, for two nonzero vectors $u$ and $v$.
Let $(e_1,e_2,\dots,e_n)$ be an orthogonal base (not necessarily orthonormal), with $e_1=b$. Then, since $Ae_i=0$ for all $i>1$ and $Ae_1\ne0$, we get that the dot product $v^Te_i$ is zero for $i>1$ and nonzero for $i=1$. That is, the only nonzero component of $v$ in the base $(e_i)_i$ is along $e_1=b$. That is, $v=\lambda b$, for some $\lambda\ne0$.
Then, with $a=\lambda u$, we have $A=uv^T=ab^T$, which you can rewrite $A=a\otimes b$.
