# Spectral Theorem for bounded compact, self-adjoint operators as corollary of Hilbert-Schmidt theorem

I'm following Debnath and Mikusinksi's "Introduction to Hilbert Spaces with Applications" and am trying to understand how the spectral theorem for compact self-adjoint operators is a corollary of the Hilbert-Schmidt theorem.

Here is the Hilbert-Schmidt theorem:

Theorem (Hilbert-Schmidt) Let $T:H\to H$ be a bounded, compact, self-adjoint linear operator on a complex Hilbert space $H$. Then there exists an orthonormal set of eigenvectors $\left(w_{n}\right)$ corresponding to non-zero eigenvalues $\left(\lambda_{n}\right)$ s.t. for each $x\in H$ we can write unique $$x=\sum_{n=1}^{\infty}a_{n}w_{n}+v$$ for some $a_{n}\in\mathbb{C}$ and $v\in\mathscr{N}\left(T\right)$.

...and here is the spectral theorem that I wish to prove:

Spectral Theorem Let $T$ be a bounded, compact, self-adjoint linear operator on a complex Hilbert space $H$. Then $H$ has an orthonormal basis $\left\{ v_{n}\right\} _{n\in\mathbb{N}}$ consisting of eigenvectors of $T$. Furthermore, $$Tx=\sum_{k=1}^{\infty}\lambda_{k}\left\langle x,v_{k}\right\rangle v_{k}$$ where $\lambda_{k}$ is the eigenvalue associated with eigenvector $v_{k}$.

Could anyone help me to understand how this comes about from the Hilbert-Schmidt theorem? The explanation in the textbook is not helpful to me.

The explanation is as follows:

"Debnath & Mikusinski's proof of the spectral theorem goes as follows: "To obtain a complete orthonormal system $\left\{v_1 , v_2 , \ldots \right\}$, we need to complement the system $\left\{u_1, u_2, \ldots \right\}$, defined in the proof of the Hilbert-Schmidt theorem, with an arbitrary orthonormal basis of $\mathscr N (T)$. The eigenvalues corresponding to the vectors that form $\mathscr N (T)$ are all equal zero. The desired equality follows from the continuity of $A$."

I can post up the proof of the Hilbert-Schmidt theorem if it is helpful?

• Since most compact operators are not Hilbert-Schmidt (but all Hilbert-Schmidt are compact), I am surprised to hear that the general case can be made to follow from the Hilbert-Schmidt case. For that matter, all variants of proof I know for Hilbert-Schmidt self-adjoint apply equally well to compact self-adjoint. – paul garrett May 8 '13 at 13:13
• I've updated my post with Debnath and Mikusinski's outline proof. I'm somewhat baffled by it. – Harry Williams May 8 '13 at 13:20
• At most, such an idea would prove the spectral theorem only for self-adjoint Hilbert-Schmidt operators... and, even in that case, I'd wager that the o.n.b. "in the proof" was obtained by something like a Rayleigh-Ritz procedure or other discussion that is valid for general compact operators on Hilbert spaces, and is not special to Hilbert-Schmidt. – paul garrett May 8 '13 at 13:35
• But the Hilbert-Schmidt theorem as quoted in the question doesn't require the operator to be Hilbert-Schmidt... It just states "bounded, compact, self-adjoint". That, I think, may be the main point of confusion here... – fgp May 8 '13 at 14:32

First, since the set of $w_n$ is orthonormal, you get (note that the scalar product is continuous!) $$\langle x,w_k \rangle = \underbrace{\langle v,w_k \rangle}_{=0} + \sum_{i=1}^\infty a_n \underbrace{\langle w_i,w_k\rangle}_{=\delta_{i,k}} = a_n \text{.}$$
From Hilbert-Schmidt and the boundedness (i.e. continuity!) of $T$ it follows that $$T(x) = T\left(v + \sum_{k=1}^\infty a_nw_n\right) = \underbrace{T(v)}_{=0} + \sum_{k=1}^\infty \underbrace{a_n}_{=\langle x,w_n\rangle} \underbrace{T(w_n)}_{=\lambda_n w_n} = \sum_{k=1}^\infty \lambda_n\langle x,w_n\rangle w_n \text{.}$$
All that remains is to find an orthonormal basis $B$ of $H$ with $B \supset \{w_n\}$. You get such a thing by picking some orthonormal basis $\{v_n\}$ of $\mathcal{N}(T)$, and setting $B = \{w_n\} \cup \{v_n\}$. This works because $H$ is the direkt sum of $\mathcal{N}(T)$ and $\mathcal{N}(T)^\bot$, since $\mathcal{N}(T)$ is closed.
• Thank you very much. My only confusion with what you wrote (apart from some problems with subscripts - easily done) is that I'm not sure why you've found $B$ like this. Don't we get an orthonormal basis for $H$ 'for free' from the Hilbert-Schmidt theorem? Indeed, the theorem says, "Then $H$ has an orthonormal basis $\left\{ v_{n}\right\} _{n\in\mathbb{N}}$ consisting of eigenvectors of $T$." – Harry Williams May 8 '13 at 17:43
• @HarryWilliams The version of the Hilbert-Schmidth theorem you posted says "Then there exists an orthonormal set of eigenvectors ... corresponding to non-zero eigenvalues". That doesn't directly provide you with an orthonormal basis of $H$, only of $\mathcal{N}(T)^\bot$. But (and that's the point) you can easily extend that to an orthonormal basis of $H$ by just picking enough orthonormal vectors frorm $\mathcal{N}(T)$. You could also make that part of the Hilbert-Schmidt theorem, of course - but in the version you quoted it isn't. – fgp May 9 '13 at 1:14