Property of vectors spanned by the columns of a symmetric matrix

Question

Let $z$ be a vector of $\mathbb{R}^d$, and $A$ be a symmetric $d \times d$ real matrix. If $z$ is in the vector space spanned by the columns of $A$, we have that for every vector $x \in \operatorname{ker}(A)$, $x^Tz=0$ (indeed, $z=Av$ for some $v$ implies $x^Tz=x^TAv=v^TAx=0$). I am interested in the converse. If $z$ is not in the vector space spanned by the columns of $A$, are we guaranteed to have the existence of a vector $x \in \operatorname{ker}(A)$ such that $x^Tz \neq 0$ ?

Answer 1

You can use the result in that thread to show that $\operatorname{ker}(A^T)=\operatorname{span}(A)^\perp$. Since $A$ is symmetric we get $\operatorname{ker}(A)=\operatorname{span}(A)^\perp$. Now write $z=z_1+z_2$ with $z_1\in\operatorname{span}(A)$ and $z_2\in\operatorname{span}(A)^\perp$, your assumption makes it such that $z_2\neq 0$. Pick $x=z_2$, then $x^T\cdot z = z_2\cdot (z_1+z_2)=0+\|z_2\| > 0$. Note that $x$ is the orthogonal projection of $z$ onto $\operatorname{ker}(A)$.