Matrices such that $w^\top A (w+v)=0$ for all $v\in w^\perp$

Question

Given a non-zero vector $w$, is there a way of characterizing matrices $A$ such that

$$\forall v\in w^\perp,\ w^\top A (w+v)=0?$$

For example, if $w^\top = [0,\ 1]$, then $v=[\alpha,\ 0]$ and

$$w^\top A (w+v)=\begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \alpha \\ 1 \end{bmatrix} = c+\alpha d$$ which must be zero for all $\alpha$, implying that $c=d=0$. I managed to generalize the result for $w^\top=[w_1,\ w_2]$ with $w_2\neq 0$ using an unelegant approach (expanding the product by hand) and found $c=-a w_1/w_2$ and $d=-bw_1/w_2$. Is there a simple extension in larger dimensions?

Answer 1

Let $z=A^\top w$. Then your condition is simply asking for the hyperplane $\{v : z^\top v = -z^\top w\}$ to be the same as the hyperplane $w^\perp = \{v : w^\top v = 0\}$. This implies both $$z^\top w=0$$ and $$z \in \operatorname{span}(w).$$ Together, we have $z=0$, i.e. $$A^\top w=0.$$ If you check your examples, this exactly characterizes the matrices $A$ you found.