Existence of Unitary Map

Question

I've been recently introduced to Unitary operators of a Hilbert space and I've been wondering the following.

Existence of a unitary operator $T$ on a (possibly infinite) Hilbert space $H$ is simple enough. The identity map works and otherwise any "partial isometry" with full initial and final subspace does as well.

However, if I'm given subspaces $U, V \subset H$, then is there any assumption (not directly assuming that $U = V$) I can make that guarantees the existence of unitary operator $T:H\to H$ such that the restriction

$$T\restriction_U:U\to V$$

is a surjective isometry? Would I somehow be able to construct a unitary operator if I assumed $\dim U = \dim V$?

Answer 1

You need exactly $\dim U=\dim V$ and $\dim U^\perp=\dim V^\perp $, in the sense that they have bases with the same cardinality.

Suppose that $\{e_j\}_{j\in J}$ is an orthonormal basis of $U$ and that $\{f_j\}_{j\in J}$ is an orthonormal basis of $V$.

Extend both basis to orthonormal bases of $H$, $\{e_j\}_{j\in K}$, $\{f_j\}_{j\in K}$, where $K\supset J$.

Now define $Te_j=f_j$ and extend by linearity. This map is a unitary that maps $U$ to $V$ (it's an exercise you have to do, to show that $T$ defines a bounded operator and that it is a unitary).

Answer 2

It is not enough to just know that $\dim U=\dim V$. For instance, suppose $U=H$ and $V$ is the orthogonal complement of a single nonzero vector $v\in H$. As long as $H$ is infinite-dimensional, $U$ and $V$ will have the same dimension, but clearly no unitary $T:H\to H$ can map $U$ to $V$, since then $T$ would not be surjective. More generally, one can see that if such a unitary exists, then one must have not only that $\dim U=\dim V$ but also that $\dim U^\perp=\dim V^\perp$.

Conversely, if $\dim U=\dim V$ and $\dim U^\perp=\dim V^\perp$, then there does exist such an isometry. To see this, let $T_0:U\to V$ be any unitary isomorphism and let $T_1:U^\perp\to V^\perp$ be any unitary isomorphism, and let $P:H\to U$ be the orthogonal projection onto $U$. Define $T:H\to H$ by $T(v)=T_0(Pv)+T_1(v-Pv)$. It is straightforward to check that this is unitary, and its restriction to $U$ is $T_0$.