I am working on Stein Real Analysis, Chapter 4, Exercise 35 (a), which is an invariant of the spectral theorem. The exercise is stated as follows:
If $T_{1}$ and $T_{2}$ are two linear self-adjoint and compact operators in $\mathcal{H}$ that commute, i.e. $T_{1}T_{2}=T_{2}T_{1}$, show that there exists an orthonormal basis for $\mathcal{H}$ which consists of eigenvectors for both $T_{1}$ and $T_{2}$.
Some similar post is here: If $S, T \in B(X)$ are self-adjoint, compact, and commuting bounded linear operators on a Hilbert space $X$, they can be simultaneously diagonalized. I have some attempts but I got stuck in the first step (I think I finished all the remaining proof). My idea is to adapt Stein's proof of spectral theorem in the page 193 of the book.
That is, we let $\mathcal{S}$ be the closure of the linear subspace of $\mathcal{H}$ spanned by all common eigenvectors of $T_{1}$ and $T_{2}$. Argue that $\mathcal{S}$ is not empty.
Our goal is then to prove $\mathcal{S}=\mathcal{H}$. It not, since $\mathcal{H}=\mathcal{S}\oplus\mathcal{S}^{\bot},$ $\mathcal{S}^{\bot}$ would be non-empty. We will then have reached a contradiction once we show that $\mathcal{S}^{\bot}$ contains a common eigenvector of $T_{1}$ and $T_{2}$.
However, what should I do to argue that $\mathcal{S}$ is not empty? I understand that for any compact, non-zero, self-adjoint operator, either $\|T\|$ or $-\|T\|$ is an eigenvalue, but I do not know how to show two such operators share at least one eigenvalue to ensure $\mathcal{S}$ not empty.
If I can show this, then here is the remaining proof:
Firstly, we note that both $T_{1}$ and $T_{2}$ respect the decomposition $\mathcal{H}=\mathcal{S}\oplus\mathcal{S}^{\bot}$. In other words, if $f\in\mathcal{S}$, then $T_{i}f\in\mathcal{S}$, which follows from the definitions. Also, if $g\in\mathcal{S}^{\bot}$, then $T_{i}g\in\mathcal{S}^{\bot}$ also holds. This is because $T_{i}$ is self-adjoint and maps $\mathcal{S}$ to itself, and therefore $$\langle T_{i}g,f\rangle=\langle g, T_{i}f\rangle =0\ \ \text{as long as}\ \ g\in\mathcal{S}^{\bot}\ \ \text{and}\ \ f\in\mathcal{S}.$$
Then, by the spectral theorem for one operator, $T_{1}$ must have an eigenvector in $\mathcal{S}^{\bot}$ with the eigenvalue denoted by $\lambda$. Let $E_{\lambda}$ be the eigenspace of $\lambda$ as a subspace of $\mathcal{S}^{\bot}$. Then, for any $v\in E_{\lambda}$, we have $$T_{1}(T_{2}v)=T_{2}(T_{1}v)=T_{2}(\lambda v)=\lambda (T_{2}v),$$ which implies that $T_{2}v\in E_{\lambda}$.
As $T_{2}$ fixes $E_{\lambda}$, it has at least one eigenvector in $E_{\lambda}$. This eigenvector is exactly the common eigenvector of $T_{1}$ and $T_{2}$, but it belongs to $\mathcal{S}^{\bot}$. The contradiction has been reached.
Is my proof for the remaining part correct?
The stein's version of spectral theorem is as follows:
Spectral Theorem. Suppose $T$ is a compact self-adjoint operator on a Hilbert space $\mathcal{H}$. Then there exists an orthonormal basis $\{\varphi_{k}\}_{k=1}^{\infty}$ of $\mathcal{H}$ that consists of eigenvectors of $T$. Moreover, if $T\varphi_{k}=\lambda_{k}\varphi_{k},$ then $\lambda_{k}\in\mathbb{R}$ and $\lambda_{k}\rightarrow 0$ as $k\rightarrow\infty$.
Thank you!
Best Answer
Fix an eigenvalue $\lambda$ of $T_1$ and let $E_1$ be the eigenspace of $T_1$ corresponding to $\lambda_1.$ Observe that $E_1$ is invariant under $T_2$ because if $v\in E_1$ then $T_1(T_2(v))=T_2(\lambda v)=\lambda T_2(v)$ that is $T_2(v)\in E_1.$
Now restrict $T_2$ to $E_1$ and let $f\in E_1$ be an eigenvector of the restriction of $T_2.$ Note that $f$ is then an eigenvector of $T_2$ itself, not just the restriction. And, obviously $f$ is eigenvector of $T_1$ because $f\in E_1.$ Therefore, $f$ is a common eiegnvector of $T_1$ and $T_2.$