Assume $M$ is linear subspace of $\mathbb{R}^{n}$ and define the orthogonal complement of $M$ as $$H=\left\{ x\in\mathbb{R}^{n}:x^{\top}y=0\text{ for all }y\in M\right\}$$ If $z\notin M$, then there exists $x\in H$ such that $x^{\top}z\ne0$.
I have trouble solving this problem. Can anyone help me?
Do an orthogonal decomposition of $z$: $$z = m+m^\bot,$$ for some $m\in M$ and $m^\bot\in H$. Since $z\not\in M$, we must have $m^\bot\neq 0$. Then (here I write $\langle a,b\rangle$ for your $a^\top b$) $$\langle z, m^\bot\rangle = \underbrace{\langle m,m^\bot\rangle}_{=0}+\underbrace{\langle m^\bot,m^\bot\rangle}_{\neq 0 \text{ because }m^\bot\neq 0}\ne0.$$