Obtain an orthonormal basis of eigenvectors from the Hilbert-Schmidt theorem

Question

Let

• $U$ be a $\mathbb R$-Hilbert space
• $Q$ be a bounded, linear, nonnegative and self-adjoint operator on $U$ with finite trace

By the Hilbert Schmidt theorem as stated on Wikipedia, we obtain the existence of $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $$Qe^n=\lambda_n e^n\;\;\;\text{for all }n\in\mathbb N$$ for some orthonormal basis $(e_n)_{n\in\mathbb N}$ of $QU$.

1. When they talk about an orthonormal basis of $QU$, they consider $QU$ as being a Hilbert space. With which inner product do they equip $QU$? My guess is $$\langle u,v\rangle_{QU}:=\langle Q^{-1}u,Q^{-1}v\rangle_U\;\;\;\text{for all }u,v\in QU\;,$$ where $$Q^{-1}:=\left(\left.Q\right|_{(\ker Q)^\perp}\right)^{-1}$$
2. How can we obtain an eigenpair sequence $(\tilde\lambda_n,\tilde e_n)_{n\in\mathbb N}$ such that $(\tilde e_n)_{n\in\mathbb N}$ is even an orthonormal basis of $U$?

My idea for $(2)$ is the following:

• Since $\lambda_n>0$ for all $n\in\mathbb N$, $$(e^n)_{n\in\mathbb N}\subseteq(\ker Q)^\perp\tag1$$ and hence $$Q^{-1}(\lambda_ne^n)=Q^{-1}(Qe^n)=e^n\;\;\;\text{for all }m,n\in\mathbb N\tag2$$
• Thus, $$\delta_{mn}=\langle e^m,e^n\rangle_{QU}=\langle Q^{-1}e^m,Q^{-1}e^n\rangle_U=\frac1{\lambda_m\lambda_n}\langle e^m,e^n\rangle_U\;\;\;\text{for all }m,n\in\mathbb N\tag3$$

From $(3)$ we obtain $$\left\|e^n\right\|_U=\lambda_n\;\;\;\text{for all }n\in\mathbb N\tag4$$ and that $$\tilde e^n:=\frac1{\left\|e^n\right\|_U}=\frac1{\lambda_n}e^n\;\;\;\text{for }n\in\mathbb N\tag5$$ is an orthonormal system in $U$. Each orthonormal system can be supplemented to an orthonormal basis. However, we've lost the eigenvector property, since $$Q\tilde e^n=e^n\;\;\;\text{for all }n\in\mathbb N\tag6\;.$$

So, what do we need to do?

Answer 1

1. When they talk about an orthonormal basis of $QU$, they consider $QU$ as being a Hilbert space. With which inner product do they equip $QU$?

An operator on a vector space $V$ means a linear map $V\to V$. So $QU$ is a subspace of $U$, and is equipped with the inner product of $U$.

2. How can we obtain an eigenpair sequence $(\tilde\lambda_n,\tilde e_n)_{n\in\mathbb N}$ such that $(\tilde e_n)_{n\in\mathbb N}$ is even an orthonormal basis of $U$?

Because $Q$ is self-adjoint, eigenvectors of distinct eigenvalues will be orthogonal to each other: if $Qv = \lambda v$ and $Qw = \mu w$, then $$\lambda\langle v, w\rangle = \langle \lambda v, w\rangle = \langle Qv, w\rangle = \langle v, Qw\rangle =\langle v, \mu w\rangle= \mu\langle v, w\rangle$$$$(\lambda-\mu) \langle v, w\rangle = 0$$ (If $U$ were a complex Hilbert space, it would be necessary to also show that the eigenvalues of self-adjoint operators are real.)

So for each eigenvalue $\lambda$ of $Q$, choose an arbitrary orthonormal basis for its eigenspace $\ker(Q - \lambda I)$. The union of all these bases for all the eigenvalues of $Q$ will form an orthonormal basis of $U$. (Orthonormality follows from what I've given. I'll leave showing that it spans all $U$ to you.)

Concerning your calculations, let me point out that if $e^n$ is an eigenvector of the eigenvalue $\lambda_n$, then so is $ae^n$ for any non-zero scalar $a$. The eigenvectors of a fixed eigenvalue form an entire subspace of $U$ (if one includes $0$). So if $e^n$ is an eigenvector, then so ie $\bar e^n = \frac {e^n}{\lambda_n}$.