T/F :If $W$ is a subspace of $\mathbb{R}^n$ , then $W$ and $W^{\perp}$ have no nonzero-vectors in common

Are the follwing statements true/false?

$1.$If $W$ is a subspace of $\mathbb{R}^n$ , then $W$ and $W^{\perp}$ have no nonzero-vectors in common

$2.$If $W$ is a subspace of $\mathbb{R}^n$ , then $W$ and $W^{\perp}$ have no zero-vectors in common

My attempt : I think both statements $1)$ and $2)$ are true because $W^{\perp}=\mathbb{R}^n -W$

Solution 1:

The definition of $W^\perp$ is not $\mathbb R^n - W$, it is $W^\perp = \{ x \in \mathbb R^n \,|\, \langle x, y \rangle = 0 \text{ for all $y \in W$ }\}$. This is an important distinction, since the set complement of a subspace is never a subspace (for one, it removes $0$, which is required for a subspace). Here by $\langle \cdot, \cdot \rangle$ I mean an inner product/dot product.

With this in mind: 1) is true, since if $x \in W \cap W^\perp$ we must have $\langle x, x \rangle = 0$ by definition of $W^\perp$, but the only element satisfying $\langle x, x \rangle = 0$ is $x = 0$ (why?).

This also tells us for free that 2) is not true, since we just demonstrated that the intersection contains (only) $0$. Note however that even without the above argument 2) has to be false, for reasons of $W$ and $W^\perp$ being subspaces, as in the first paragraph.