How can we compute the square root of an operator of the form $Cv=\sum_{n\in\mathbb N}\langle v,e_n\rangle_Ve_n$?

Question

Let

• $\mathbb K\in\left\{\mathbb C,\mathbb R\right\}$
• $U$ and $V$ be $\mathbb K$-Hilbert spaces such that $U\subseteq V$ and that the inclusion $\iota$ is Hilbert-Schmidt
• $C:=\iota^\ast$ denote the adjoint of $\iota$
• $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $U$

We can show that $$Cv=\sum_{n\in\mathbb N}\langle v,e_n\rangle_Ve_n\;\;\;\text{for all }v\in V\;.$$ How can we compute the square root $C^{1/2}$ of $C$? Note that $C^{1/2}$ exists and is uniquely determined, since $C$ is a bounded, linear, nonnegative and symmetric operator on $V$.

Please note that I'm not interested in the case where $\langle\;\cdot\;,\;\cdot\;\rangle_U$ is the restriction of $\langle\;\cdot\;,\;\cdot\;\rangle_V$ to $U\times U$.

Answer 1

This operator is the orthogonal projection onto $U$, it holds $C^2=C$: First we find that $$ \langle Cv,e_n\rangle = \langle v,e_n\rangle, $$ since $(e_n)$ is orthonormal. This implies $$ C^2v = \sum_n \langle Cv,e_n\rangle Ce_n = \sum_n \langle v,e_n\rangle e_n =Cv. $$ Hence, $C^2 = C$, which implies $C^{1/2}=C$.

Edit: I assumed that the scalar product on $U$ is the scalar product of $V$.