Let V be an inner product space, W be a subspace of V and
\[ \begin{aligned} W^{\perp} = \{v \in V / &\langle v, w \rangle = 0 \\ &\text{for all } w \in W\} \end{aligned} \]
Then :
MH SET - 2023, SET Exam - 2023,
Answer: C
The statement that holds in every inner product space is \(W \subseteq W^{\perp\perp}\).
Take \(w \in W\). By the definition of \(W^{\perp}\), every \(v \in W^{\perp}\) satisfies \(\langle v, w \rangle = 0\). So \(w\) is orthogonal to all of \(W^{\perp}\), which says exactly that \(w \in (W^{\perp})^{\perp}\). No completeness or dimension assumption is needed.
The other statements need more. \(\dim W^{\perp} = \dim W\) fails already in R3 with W a line (the complement is a plane). \(W^{\perp\perp} = W\) and \(V = W \oplus W^{\perp}\) hold for finite-dimensional V, or for closed subspaces of a Hilbert space, but fail in general: in the space of finite sequences inside \(\ell^{2}\), a dense proper subspace W has \(W^{\perp} = \{0\}\), so \(W^{\perp\perp} = V \neq W\) and \(W \oplus W^{\perp} = W \neq V\).
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