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Let $U$ be a subspace of a linear space $V$ with Euclidean structure. Show that $U \subseteq U^{\perp\perp} $. Moreover, if $V$ is finite dimensional, then $U= U^{\perp\perp}$.

I can prove the last part that $U= U^{\perp\perp}$ I'm just struggling with the first part. Can anyone help me with this or direct me somewhere to help me solve this answer?

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Let $V$ be an inner product space and let $U \subset V$ be a subspace. $U \subset U^{\bot\bot}$ follows immediately from the definition of the annihilator and from the symmetry of the inner product. You have $$ u \in U \,\Rightarrow\, \forall v \in U^\bot \,:\, \langle u,v \rangle = 0 \,\Rightarrow\, u \in U^{\bot\bot} \text{.} $$

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This is an exercise in unfolding definitions.

Your goal, after substituting the definition of $\subseteq$

  • For any $u \in U$, we have $u \in U^{\perp\perp}$

Then, into this, we can substitute in the fact that $u \in U^{\perp \perp}$ means

  • For any $v \in U^{\perp}$, we have $\langle u, v \rangle = 0$

(do this: make this substitution into the statement you are trying to prove)

and so on from there.

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