If $W$ is a subspace of $\mathbb{R}^n$, then $W^\perp = \overline{W} =
\{v \cdot w = 0, \forall w \in W\}$Prove $W \cap W^\perp = \{\vec{0}\}$.
How do I fully prove this intersection is $\vec{0}$? I showed $\vec{0}$ is in $W$, $\overline{W}$, since they are both subspaces of $\mathbb{R}^n$, thus $\vec{0}$ is a subset of $W \cap \overline{W}$ which is also a subset of $\vec{0}$.
Then I must show $u = \vec{0}$ to complete.
I have let $u$ be an element of $W \cap \overline{W}$ which implies $u$ is an element of the zero vector (because they are subsets of each other). But does this show $u$ must be equal to $\vec{0}$? If not, how do I show that part?
Best Answer
Take ${\bf u} \in W \cap W^\perp$. Then $\langle {\bf u},{\bf u}\rangle = 0$ (why?). So $\bf u = 0$, because $\langle \cdot, \cdot \rangle$ is non-degenerate.