Suppose that $H$ is a Hilbert space such that every isometry $T\colon H\to H$ is also unitary. In other words, $T^{*}T=I\implies TT^{*}=I$ for all $T\in B(H)$. Then why is $H$ necessarily finite dimensional?
I tried to work with the innerproduct and the definition of the adjoint, but I don't see how the "algebraic" assumption translates into a statement about the dimension of $H$.
Best Answer
Hint: if $H$ is infinite-dimensional, you can find a countably infinite orthonormal set $e_1, e_2, \ldots$ Hence we can define an isometry $T$ such that $T(e_i) = T(e_{i+1})$ and $T(x) = x$ for any $x$ that is orthogonal to each $e_i$. $T$ is not surjective, so there is no $S$ such that $TS = I$.
(Note that $H$ will contain elements that are not orthogonal to each $e_i$ but are not in the linear span of the $e_i$. These elements comprise the intersection of all complete subspaces that include each $e_i$. Any such element can be written as $u = \sum_{i=1}^{\infty} t_ie_i$, so that we must define $T(u) = \sum_{i=1}^{\infty} t_ie_{i+1}$.)