If $x \in W$ and $x \in W^\perp$, then $x=0$

Question

Let $E=\{e_1,...,e_n\}$ be basis for $W$ and $P=\{p_1,...,p_n\}$ be a basis for $W^\perp$. Since $x$ is in both of these spaces $$x=a_1e_1 + ... +a_ne_n$$ $$x=b_1p_1 + ... + b_np_n$$

multiply $x$ by $x$ $$x \cdot x = a_1b_1(e_1p_1)+...+a_nb_n(e_np_n)$$

Now if $x\neq 0$, that would imply $e_ip_i \neq 0$, which would contradict the fact that they are from orthogonal sets, and their product must by definition be $0$. Thus $x$ must equal zero. Is this correct? Also, does this equivalently show that $W \cap W^\perp = \{0\}$?

Answer 1

The idea of your proof seems correct.

Here's a simpler proof. Suppose $x \in W$ and $x \in W^\perp$, i.e. $x \in W\cap W^\perp$. Since $x \in W^\perp$, for any $y \in W$ we know that $\langle x, y \rangle = 0$. In particular, $x \in W$, so $\langle x, x \rangle = 0$. So $\|x\|^2 = \langle x,x \rangle = 0$, which implies that $x = 0$. Hence $W \cap W^\perp = \{0\}$.

Answer 2

We can do it without a basis too:

$x \in W^\perp$ means that $\forall y \in W: (x|y)=0 $. But if $x \in W$ as well, then we have that $(x|x)=0$, i.e. $x=0$.