# 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

## 1 Answer

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
• @HarryWilliams For compact operators, finite-dimensional vector spaces tend to work well as a mental model. You of course get infinite instead of finite sums, but otherwise things mostly work the same. That spectral theorem for compact operators, for example, looks just like the finite-dimensional on, except for the infinite sum. Once you drop the compactness requirement, things change though - the spectral theorem for arbitrary bounded and self-adjoint operators already looks quite different... – fgp May 9 '13 at 1:22