Unitary operator between two orthonormal basis’

hilbert-spaceslinear algebraorthogonalityorthonormal

today I started studying the topic of Hilbert spaces. I was solving a problem concerning unitary operators and I got stuck with one of the subquestions which are stated as follows: Imagine we have two Hilbert spaces $H$ and $K$ whose orthonormal basis are given by $(e_n)_{n \in \mathbb{N}}$ and $(f_n)_{n \in \mathbb{N}}$ respectively. I had to show that there exists a unique unitary operator $U: H \rightarrow K$ such that
$$\forall n \in \mathbb{N}: U(e_n) = f_n$$
I know a unitary operator is a bijective linear function for which it holds that
$$\forall x,y \in H: \quad \langle U(x),U(y)\rangle = \langle x,y \rangle$$
However, I am completely stuck and do not know how to start solving this problem. I think I, first of all, have to show that such a unitary operator does exist and after that prove that it is unique. Does anyone have a tip to start solving this problem? Any help would be greatly appreciated :))

Best Answer

The operator you are looking for satisfies $$ U(e_n)=f_n, \quad n\in\mathbb{N}. $$ You have to prove that such correspondence uniquely specifies $U$ and that such a $U$ is unitary. Since any $x\in H$ can expanded as $$ x=\sum_n x_n e_n $$ and $U$ is linear, the action of $U$ is uniquely specified for any $x\in H$ by $$ U(x)=\sum_n x_n U (e_n)=\sum_n x_n f_n\in K. $$ Moreover, $U^{-1}$ exists and it is clearly uniquely specified by the correspondence $$ U^{-1}(f_n)=e_n. $$ In order to prove unitary, it remains to check $U$ preserves inner products. Expanding $x,y\in H$ as $$ x=\sum_n x_n e_n, \quad y=\sum_n y_n e_n $$ we have $$ \langle x,y\rangle=\sum_{n,m} x_n y_m \langle e_n, e_m\rangle= \sum_{n} x_n y_n. $$ where the orthonormality of the basis has been used. Now, it is straightforward to check that $$ \langle U(x),U(y)\rangle=\sum_{n,m} x_n y_n. $$ Therefore $U$ is an isometric and invertible operator, so it is a unitary operator.

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