If $Q$ is an operator on a Hilbert space with $Qe_n=λ_ne_n$ for all $n$, then $Q^{-\frac 12}e_n=\frac 1{\sqrt{λ_n}}e_n$ for all $n$ with $λ_n>0$ Let


*

*$(U,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a separable Hilbert space

*$\mathfrak L(U)$ be the set of bounded and linear operators on $U$

*$Q\in\mathfrak L(U)$ be nonnegative and symmetric

*$(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $U$ with $$Qe_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N\tag 1$$ for some $(\lambda_n)_{n\ge 0}\subseteq[0,\infty)$



We can prove, that for any nonnegative and symmetric $L\in\mathfrak L(U)$ there is exactly one nonnegative and symmetric $L^{\frac 12}\in\mathfrak L(U)$ with $$L^{\frac 12}L^{\frac 12}=L\;.$$ Now I would like to prove that $$Q^{\frac 12}e_n=\sqrt{\lambda_n}e_n\;\;\;\text{for all }n\in\mathbb N\tag 2$$ and $$Q^{-\frac 12}e_n=\frac 1{\sqrt{\lambda_n}}e_n\;\;\;\text{for all }n\in\mathbb N\text{ with }\lambda_n>0\tag 3\;,$$ where $Q^{-1}$ is the pseudoinverse of $Q$, i.e. $$Q^{-1}v:=\underset{u\in U}{\operatorname{arg min}}\left\{\left\|u\right\|:Qu=v\right\}\;\;\;\text{for }v\in Q(U)\;.$$

Unfortunately, I have no idea how I can prove $(2)$ and $(3)$ and I can't find any reference for such a statement. So, how can we prove it?
 A: Because $Q$ is bounded, then $\{ \lambda_n \}$ must be a bounded sequence. The linear operator $S$ defined by $Se_n = \lambda_n^{1/2}e_n$ is also bounded, and $S^2e_n = Qe_n$ for all $n$, which means that $S^2=Q$. It is not hard to show that $S$ is positive and selfadjoint. So $S=Q^{1/2}$ is the unique positive square root of $Q$.
Because $Q$ is selfadjoint, then $U=\mathcal{N}(Q)\oplus\overline{\mathcal{R}(Q)}$, and each of these subspaces is invariant under $Q$. So $Q : \overline{\mathcal{R}(Q)}\rightarrow\overline{\mathcal{R}(Q)}$ has an inverse $R$ on $\mathcal{R}(Q)$, which is a dense subspace of $\overline{\mathcal{R}(Q)}$. Clearly,
$$
   \{ x \in U : Qx=y\} = \{ Ry+n : n \in \mathcal{N}(Q) \}
$$
And $\|Ry+n\|^2=\|Ry\|^2+\|n\|^2$ because of the orthogonality of $\overline{\mathcal{R}(Q)}$ and $\mathcal{N}(Q)$. So $Ry$ is the unique element of minimal norm for which $QRy=y$. In your vernacular, $R$ is the pseudo inverse of $Q$, and it is not difficult to show that
$$
                Rx = \sum_{\{ n : \lambda_n \ne 0 \}}\lambda_{n}^{-1} (x,e_n)e_n,\;\;\; x\in\mathcal{N}(Q)\oplus\mathcal{R}(Q).
$$
The operator $R$ is densely-defined with domain $\mathcal{D}(R)=\mathcal{N}(Q)\oplus\mathcal{R}(Q)$ (and $\mathcal{D}(R)=U$ iff $\{ \lambda_n^{-1} : \lambda_n \ne 0 \}$ is a bounded subset of $\mathbb{R}$.) Whether densely-defined or everywhere defined, $R$ is selfadjoint with
$$
     R^{1/2}x = \sum_{\{ n : \lambda_n \ne 0\}}\lambda_n^{-1/2}(x,e_n)e_n,
  \\ \mathcal{D}(R^{1/2}) = \left\{ x \in U : \sum_{\{ n : \lambda_n \ne 0 \}}\lambda_n^{-1}|(x,e_n)|^2 < \infty \right\}
$$
