Why is $V^{\perp}={0}$

Question

Let $V$ be an inner product space. I have read a statement saying $V^{\perp}=\{0\}$. Why is this true? It seems trivial to even define an orthogonal complement to $V^{\perp}$ if it is always just $0$.

Thank you!

Answer 1

Let $V$ be an inner product space. For a set $M \subseteq V$, we define $$ M^\perp = \{y \in V: \langle y,x \rangle = 0 \quad \forall x \in M\} $$ So, by that logic, we have $$ V^\perp = \{y \in V: \langle y,x \rangle = 0 \quad \forall x \in V\} $$ however, if $y \in V$, and $\langle y,x \rangle = 0$ for all $x \in V$, then $\langle y,y \rangle = 0$. So, the only element of $V^\perp$ is $0$.