If $Q$ is an operator on a Hilbert space $U$ and $(e_n)_{n\in\mathbb N}$ is an ONB of $U$ with $Qe_n=λ_ne_n$, then $Q^{-1}e_n=\frac 1{λ_n}e_n$ Let


*

*$U$ be a Hilbert space

*$Q$ be a bounded, linear, nonnegative and symmetric operator on $U$

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



Since $Q$ is injective on $\left(\ker Q\right)^\perp$ and $U=\ker Q+\left(\ker Q\right)^\perp$, $$Q^{-1}:=\left(\left.Q\right|_{\left(\ker Q\right)^\perp}\right)^{-1}$$ is a well-defined mapping $Q(U)\to\left(\ker Q\right)^\perp$. Now I would like to show that $$Q^{-1}e_n=\frac 1{\lambda_n}e_n\;\;\;\text{for all }n\in\mathbb N\text{ with }\lambda_n>0\;.$$

How can we do that?
 A: They key observation is that if $\lambda_n=0$ then $Qe_n=0$, and that 
$$
\ker Q=\overline{\text{span}}\,\{e_n:\ \lambda_n=0\},\ \ \ \ (\ker Q)^\perp=\overline{\text{span}}\,\{e_n:\ \lambda_n>0\}.
$$
To check this, let $x\in U$. We can write, since $\{e_n\}$ is an orthonormal basis, 
$$
x=\sum_n\alpha_n\,e_n.
$$
Then 
$$
Qx=\sum_n\alpha_n\lambda_n\,e_n=\sum_{n:\ \lambda_n>0}\alpha_n\lambda_n\,e_n.
$$
If $x\in\ker Q$, then $0=Qx$, so $\alpha_n\lambda_n=0$ for all $n$ such that $\lambda_n>0$. Thus, $\alpha_n=0$ and so 
$$
x=\sum_{n:\ \lambda_n=0}\alpha_ne_n\in\overline{\text{span}}\,\{e_n:\ \lambda_n=0\}.
$$
If $x\in(\ker Q)^\perp$, then $\langle x,e_n\rangle=0$ for all $n$ such that $\lambda_n=0$, so $$
x=\sum_{n:\ \lambda_n>0}\alpha_ne_n\in\overline{\text{span}}\,\{e_n:\ \lambda_n>0\}.
$$
The reverse inclusions are trivial. 
Now, on $(\ker Q)^\perp$, we have 
$$
Qx=\sum_{n:\ \lambda_n>0}\alpha_n\lambda_n\,e_n.
$$
So the map $Q^{-1}:Qx\longmapsto x$ is given by 
$$
Q^\dagger \left(\sum_{n:\ \lambda_n>0}\alpha_n\lambda_n\,e_n\right)=\sum_{n:\ \lambda_n>0}\alpha_n\,e_n.
$$
In particular, $Q^\dagger (\lambda_n e_n)=e_n$, so 
$$
Q^\dagger e_n=\frac1{\lambda_n}\,e_n.
$$
(I have used the notation $Q^\dagger $ for the Moore-Penrose inverse, since $Q^{-1}$ is usually reserved for the actual inverse) 
