I have the following definition: Let $V$ be an inner product space and let $S \subseteq V$. The orthogonal complement of $S$, denoted $S^{\perp}$, is the set $$S^{\perp} = \left\{\vec{v} \in V \mid \langle\vec{v}, \vec{x}\rangle = 0 \hspace{1mm} \forall\vec{x} \in S\right\}.$$ It seems that it may be possible that if $S^{\perp} = \left\{\vec{0}\right\}$ then $\text{span} \hspace{0.2mm} S = V$, but I'm having trouble proving if this is true or not. I've been trying to use the fact that $S^{\perp} = \left(\text{span} \hspace{0.2mm} S \right)^{\perp}$.


1 Answer 1


In general, we know that $(S^{\perp})^{\perp} \subseteq S$, where we always have equality in the finite-dimensional setting.

If $S^{\perp} = \{0\}$, then what is $(\{0\})^{\perp}$? What vectors satisfy $\langle v, 0\rangle = 0$?

Note that $(S^{\perp})^{\perp} = (\{0\})^{\perp} \subseteq S\subseteq V$, and so $S=\dots$


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