Disjoint isometric copy of countable-dimensional vector subspace of Hilbert space

Question

Let $H$ be a complex separable Hilbert space. Let $A \subset H$ be a vector subspace of countable dimension. Does there exist unitary operator $U$, such that $A \cap U(A) = \{ 0 \} $ ?

Answer 1

Yes. It's a consequence of this fact:

Observation: Let $E$ be an inner product space of countable (possibly infinite) dimension. Then, there is an algebraic basis $\mathcal B=(b^1,b^2,\cdots)$ of $E$ such that $\langle b^i,b^j\rangle=\delta_{ij}$ for all $i,j$.

Such a basis can be obtained by using Gram-Schmidt process on any basis.

With that result, it suffices to find some $A'$ such that $A'\cap A=\{0\}$, $\overline A=\overline{A'}$ and $\dim A'=\aleph_0$. In fact, consider $\mathcal B$ and $\mathcal B'$ orthonormal algebraic basis of $A$ and $A'$ such as the one in the Observation. There is exactly one unitary endomorphism of $\overline A$ such that $Ub^i=b^{\prime i}$ for all $i\in\Bbb N$. Such $U$ can then be extended to an unitary endomorphism $\widehat U$ of $H=\overline A\perp {\overline A}^\perp$ by setting, for instance, $\left.\widehat U\right\rvert_{{\overline A}^\perp}=id$.

Then, $\widehat U(A)=U(A)=U(\operatorname{Span}(b^i\,:\, i\in\Bbb N))=\operatorname{Span}(b^{\prime i}\,:\, i\in\Bbb N)=A'$.

Now, let's provide an $A'$ such as above. Consider $\mathcal B$ an algebraic basis of $A$ (it might be in the form of the Observation, but it isn't necessary) and extend it to an algebraic basis $\mathcal C$ of $\overline A$. Let's also demand that $\langle c,c\rangle=1$ for all $c\in \mathcal C$. Since $\dim\overline A=2^{\aleph_0}$, there is an injective function $c:\Bbb N\times \Bbb N\to \mathcal C\setminus\mathcal B$. Let's now consider the family of vectors $$a^{i,j}=b^i+2^{-j}c^{i,j}$$ and $A'=\operatorname{Span}(a^{i,j}\,:\, i,j\in\Bbb N)$. It is clear that $\overline {A'}$ contains all the $b^i$-s, and therefore $\overline{A'}=\overline A$. On the other hand, let $x\in A\cap A'$. This results in an identity $$\sum_{r=1}^m \alpha_r b^{r}=x=\sum_{h,k=1}^n \gamma_{h,k}a^{h,k}\\\sum_{r=1}^m \alpha_r b^{r}-\sum_{h,k=1}^n \gamma_{h,k}b^{h}=\sum_{h,k=1}^{n}\gamma_{h,k}2^{-k}c^{h,k}$$

Since $\sum_{h,k=1}^n\gamma_{h,k}2^{-k}c^{h,k}\in \operatorname{Span}\mathcal B$ and $c^{h,k}$ are all distinct elements of $\mathcal C\setminus\mathcal B$, we have that $2^{-k}\gamma_{h,k}=0$ for all $h,k$, and thus $\gamma_{h,k}=0$. Therefore, $x=0$.

It's easily proved in similar fashion that all the $a^{i,j}$-s are linearly independent, thus that $\dim A'=\aleph_0$.