Relation between eigenvalues of $0$ in symmetric, non-negative definite operator and its square root

Question

To start off, this question contains similarities to the following questions: If $U_0,V$ are Hilbert spaces, $(e_n)$ is an ONB of $U_0$ and $ι:U_0→V$ is an embedding, can we complete $(ιe_n)$ to an ONB of $V$?, If $Q$ is an operator on a Hilbert space $U$, $(e_n)$ is an ONB of $U$ consisting of eigenvectors of $Q$, then $(Q^{1/2}e_n)$ is an ONB of $Q^{1/2}U$ and Unbounded operator $K\to H$ that is Hilbert--Schmidt from $Q^{1/2}K\to H$, as these questions are about the same topic, but do not cover the exact questions I would like to ask.

If we consider a separable Hilbert space $\mathcal{H}$ and let $Q$ on $\mathcal{H}$ be a symmetric, non-negative definite bounded linear operator with finite trace. (To be complete, symmetric means that $$ \langle Qg, h \rangle_{\mathcal{H}} = \langle g, Qh \rangle_{\mathcal{H}} $$ for every $g,h\in\mathcal{H}$ and non-negative definite means that $\langle Qh, h \rangle_{\mathcal{H}} \geq 0$ for all $h\in\mathcal{H}$.)

Then, $Q$ admits a unique "square root", that is, there is a unique symmetric, non-negative definite bounded linear operator on $\mathcal{H}$, denoted by $Q^{\frac12}$, such that $Q = Q^{\frac12} \circ Q^{\frac12}$.

By separability of the Hilbert space $\mathcal{H}$ and the assumptions on $Q$, there exists an orthonormal basis $\{h_n:n\geq 1\}$ (with respect to $\langle \cdot,\cdot \rangle_{\mathcal{H}}$) and a sequence of non-negative numbers $\{\lambda_n:n\geq1\}$ such that $$ Qh_n = \lambda_nh_n $$ for all $n\in\mathbb{N}$. Hence, we have an orthonormal basis consisting of eigenvectors.

It is known that, when $\lambda_n > 0$ for some $n\geq1$, we have $$ Q^{\frac12}h_n = \sqrt{\lambda_n}h_n. $$ Hence, $h_n$ is an eigenvector of the square root operator, with corresponding eigenvalue $\sqrt{\lambda_n}$.

Regarding this topic I have two questions; the first one is the most important one and the second is good to know, but not essentially necessar

Here are my questions:

1. When $\lambda_n$ is equal to zero, does it also hold that $Q^{\frac12}h_n = \sqrt{\lambda_n} h_n = 0\cdot h_n = \textbf{0}_{\mathcal{H}}$? [Answered, see below.]

2. Moreover, does anyone know how to show that $$ \left(\lambda_n,h_n\right) \text{ eigenpair of $Q$ } \implies \left(\sqrt{\lambda_n},h_n\right) \text{ eigenpair of $Q^{\frac12}$} $$ in the case that $\lambda_n >0$? It might be trivial to prove, but all my attempts to show it are fruitless, as they all assumed some injectivity conditions that I cannot assume to hold.

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The answer to Question 1 can be found in Lemma 1.5 in this document

Answer 1

For question 1, the more general statement is that, for any operator $T$ (not necessarily compact), $$ \ker T=\ker T^*T. $$ The proof is easy: it is clear that $\ker T\subset \ker T^*T$. For the converse, if $T^*Tx=0$, then $$0=\langle T^*Tx,x\rangle=\langle Tx,Tx\rangle=\|Tx\|^2,$$ so $Tx=0$.

For you second question, if $Q^{1/2}h=\lambda^{1/2} h$, then $Qh=Q^{1/2}Q^{1/2}h=(\lambda^{1/2})^2h=\lambda h$. Conversely, if $Qh=\lambda h$, then $Q^2h=\lambda^2 h$, and inductively $Q^nh=\lambda ^nh$. By linearity, $$p(Q)h=p(\lambda)h$$ for all polynomials $p$. Now you use that if $p_n$ is a sequence of polynomials that converge uniformly to $f(t)=t^{1/2}$ on the spectrum of $Q$, then $\lim p_n(Q)=Q^{1/2}$.