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Let $H$ be a Hilbert space and $H'\le H$ a subspace as Hilbert spaces (I mean, the inner product in $H'$ is the same inner product of $H$ restricted to $H'$).

If we take $f:H\to H$ an automorphism of Hilbert spaces which fixes $H'$ pointwise, notice that if $f(a)=b$ then $\|a\|=\|b\|$ and $dist(a,H')=dist(b,H')$. Is the converse also true? (I mean, if $a,b\in H$ satisfy $\|a\|=\|b\|$ and $dist(a,H')=dist(b,H')$, is there an automorphism $f:H\to H$ which fixes $H'$ pointwise and $f(a)=b$?).

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No, because the projections of $a$ and $b$ to $H'$ might be different.

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Just to expand on this a bit: Let $H'' = (H')^\perp = \{v: (u,v) = 0 \text{ for all } u \in H\}$. The automorphisms (i.e. unitary operators) of $H$ that fix $H'$ pointwise are all of the form $f(u+v) = u + Uv$ for $u \in H', v \in H''$, where $U$ is a unitary operator from $H''$ onto $H''$. In particular, there is an automorphism taking $a$ to $b$ if and only if the orthogonal projections of $a$ and $b$ on $H'$ are equal and the orthogonal projections on $H''$ have the same norm. –  Robert Israel Aug 29 '11 at 17:51
Thanks a lot! Your answers were very useful for me! –  Pedro Aug 30 '11 at 15:20

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