Let $ \mathcal{H} $ be a Hilbert space and $ T: \mathcal{H} \to \mathcal{H} $ a bounded linear operator. The $ n $-th singular number $ {\mu_{n}}(T) $ of $ T $ is defined as the distance from $ T $ to the space of operators of rank at most $ n $. We say that $ T $ is in the trace class if $$ \sum_{n} {\mu_{n}}(T) < \infty. $$ Show that in this case, $ T $ is compact and if $ \{ \lambda_{n} \} $ are its eigenvalues, then $$ \sum_{n} |\lambda_{n}| < \infty. $$ Also, show that, in general, the converse is not true.

I have not seen this definition of ‘trace class’ before. Can anyone give me some hints? Can I approximate $ T $ with finite-rank operators?

  • 4
    $\begingroup$ My "guess" is: because your sum is finite, you must have $\mu_n(T)\rightarrow 0$, hence I think you can use this fact to produce the desired aproximation. $\endgroup$
    – Tomás
    Commented Jan 3, 2013 at 22:33
  • $\begingroup$ To disprove the converse, take a diagonal operator on $\ell_2$ with diagonal entries $\lambda_n$ that converge to zero, but slowly enough so that $\sum \lambda_n=\infty$. $\endgroup$
    – user53153
    Commented Jan 4, 2013 at 1:31
  • 1
    $\begingroup$ @PavelM I dont see this contradiction... isnt it $\sum|\lambda_n| < \infty$ in the converse statement? $\endgroup$
    – Johan
    Commented Jan 4, 2013 at 21:41
  • $\begingroup$ Indeed, I misunderstood what was meant by the converse here. Maybe look at some form of Volterra operator which is compact and has no eigenvalues at all. $\endgroup$
    – user53153
    Commented Jan 4, 2013 at 22:36
  • $\begingroup$ The operator $Te_i=\frac{e_{i+1}}{i}$, where $\{e_i\}$ is a orthonormal sequence seems to work fine as a counter example? $\endgroup$ Commented Jan 5, 2013 at 0:41

1 Answer 1


In order to distinguish the new definition of $ {\mu_{n}}(T) $ from the old one, let us call it $ {\mu^{\text{New}}_{n}}(T) $.

We shall assume throughout this discussion that $ \displaystyle \sum_{n=1}^{\infty} {\mu_{n}}(T) < \infty $.

By the Divergence Test from calculus (it’s hard to believe that something so simple can crop up here!), we have $ \displaystyle \lim_{n \to \infty} {\mu_{n}}(T) = 0 $. Hence, for any $ \epsilon > 0 $, there exists an $ n \in \mathbb{N} $ sufficiently large so that $ {\mu_{n}}(T) < \epsilon $, which means that we can find an $ F \in B(\mathcal{H}) $ of rank $ \leq n $ such that $ \| T - F \|_{B(\mathcal{H})} < \epsilon $. Therefore, $ T $ can be approximated in the operator norm by bounded operators of finite rank, making it a compact operator.

Recall that $ {\mu_{n}}(T) $ is defined as the $ n $-th term of the null sequence that is formed by listing the eigenvalues of the positive compact operator $ |T| $ in decreasing order, taking multiplicity into account. The Minimax Principle then says that $ {\mu^{\text{New}}_{n}}(T) = {\mu_{n}}(T) $ (please click here to access a set of notes on trace-class operators that contains a proof of this result; see Lemma 12). Therefore, ‘trace-class’ in the new sense is the same as ‘trace-class’ in the old sense, and so $$ \text{Tr}(|T|) \stackrel{\text{def}}{=} \sum_{n=1}^{\infty} {\mu_{n}}(T) = \sum_{n=1}^{\infty} {\mu^{\text{New}}_{n}}(T). $$

For any $ T \in K(\mathcal{H}) $, one can find orthonormal sequences $ (\mathbf{v}_{n})_{n \in \mathbb{N}} $ and $ (\mathbf{w}_{n})_{n \in \mathbb{N}} $, not necessarily complete, such that $$ T = \sum_{n=1}^{\infty} {\mu_{n}}(T) \langle \mathbf{v}_{n},\bullet \rangle_{\mathcal{H}} \cdot \mathbf{w}_{n}. $$ We thus obtain a more explicit approximation of $ T $ by bounded operators of finite rank. This is a standard result in the theory of compact operators; please refer to the Wikipedia article on Compact Operator or to Corollary 4 of the notes mentioned above.

  • $\begingroup$ Thanks! can you please expand a little on how this implies the second statement? $\endgroup$
    – Johan
    Commented Jan 4, 2013 at 21:42
  • $\begingroup$ @Johan: I’ve done a revision of my original post. The links will explain some of the details that I haven’t managed to provide. $\endgroup$ Commented Jan 29, 2013 at 4:54
  • $\begingroup$ Great! thank you very much! $\endgroup$
    – Johan
    Commented Jan 29, 2013 at 14:40
  • $\begingroup$ @Johan: One more thing. We already have $ \forall n \in \mathbb{N}: ~ |{\lambda_{n}}(T)| \leq {\mu_{n}}(T) $, so by the Comparison Test, the convergence of $ \displaystyle \sum_{n=1}^{\infty} {\mu_{n}}(T) $ yields the convergence of $ \displaystyle \sum_{n=1}^{\infty} |{\lambda_{n}}(T)| $. $\endgroup$ Commented Jan 29, 2013 at 22:11
  • $\begingroup$ @HaskellCurry: Why does $\mu_n(T)<\varepsilon$ imply that we can find $F\in\mathcal{B}(\mathcal{H})$ of rank $\leq n$ such that $\|T-F\|_{\mathcal{B}(\mathcal{H})}<\varepsilon$? Also why do we assume that $\sum_{n=1}^\infty\mu_n(T)<\infty$. $\endgroup$ Commented Aug 15, 2015 at 12:53

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