I need to prove the following: If B orthonormal set in a Hilbert space X, then B is maximal if and only if $ B^{\perp} =\{0\} $
I tried the following :
- $ B^{\perp} =\{0\} $ => B maximal
$ B^{\perp} =\{0\} $ + $ B^{\perp} \perp span(B) => 0 \perp span (B) $
i.e B span the space which means that B is maximal
- B maximal => $B^{\perp} =\{0\}$
Assuse B is maximal. Note M=close (span B) and assume M is strict subset of X. We also know that B=${B^{\perp}}^{\perp}$. The strict inclusion forces that $0$ is not at $ B^{\perp} $ . We can take any vector from $ M^{\perp} $ such the union with M is still orthogonal set, which forces that this vector is orthogonal to M , thus contradicting the maximality of B.
What do you think about this proof ?
I assume that $B$ is a maximal orthonormal set iff it is not a proper subset of another orthonormal set.
(1) Let $B^\perp=\{0\}$. Assume $B$ is not maximal. Then there is a larger orthonormal set $\tilde B$ with $B\subset \tilde B$ and $x\in \tilde B\setminus B$ with $x\ne 0$. Then $x$ is orthogonal to all other elements in $\tilde B$, and hence orthogonal to all elements in $B$. Hence $x\in B^\perp$. Contradiction.
(2) Assume $B$ maximal. Assume $B^\perp \ne \{0\}$, i.e. there is $x\ne 0$ with $x\in B^\perp$. Define $\tilde B:= B \cup \{ \frac x{\|x\|} \}$. Then $\tilde B$ is orthonormal, $B$ is a proper subset of $\tilde B$. Contradiction to maximality.