# Trace norm of operators obtained by restricting the matrix of a trace class operator

Suppose $$H$$ is a Hilbert space with orthonormal basis $$\{e_i\}_{i\in \mathbb N}$$. To every operator $$T$$, we associate a infinite matrix $$[T_{ij}]$$, where $$T_{ij}=\left$$. We know that for any trace class operator $$T$$, the trace norm is $$||T||_1=\operatorname{Tr}(|T|)$$.

Q). Suppose $$T$$ is a trace class operator and $$S$$ is such that its matrix entries are either equal to the matrix entries of $$T$$ or they vanish (possibly at infinite number of points). Can I say that $$||S||_1\leq ||T||_1$$? If not, is there any finite upper bound to such $$S$$ obtained from $$T$$?

• @SinaBaghal fixed. It was a mistake
– NewB
Jul 13, 2021 at 16:28
• If I remember well there is a counterexample in the paper The main triangle projection in matrix spaces and its applications by A. Pelczynski and S. Kwapien; Studia Math. 34 (1970), 43-67. Jul 13, 2021 at 16:52
• Since the trace class norm and the operator norm are in duality, wouldn't your multiplier have the same norm in either case? Jul 13, 2021 at 21:58
• This is clearly so when $T$ has a finite support and therefore also when $T$ does not have finite support. (The finitely supported matrices / operators are dense in the trace class; and $S_n= Q_n T Q_n$ where $Q_n$ is the orthogonal projection onto $\{e_1,\dots,e_n\}^\perp$.) Jul 18, 2021 at 10:07
• `Also trace class can be identified with projective tensor product of Hilbert space': Once you know that, it follows that $H_0\otimes H_0$ is dense, with $H_0=$ linear span of the basis vectors $e_1, e_2, \dots$. Jul 19, 2021 at 19:34

Here's an algorithm for testing an ad-hoc conjecture $$C$$ about Hilbert space operators. :-)

1. Set up the runtime environment correctly by loading the information "Most conjectures are false" into short term memory.

2. Test $$C$$ against the zero and the identity operator.

3. Test $$C$$ against finite-dimensional diagonal matrices.

4. Test $$C$$ against multiplication operators on $$\ell^2$$ and $$L^2$$.

5. Test $$C$$ against the following $$2 \times 2$$-matrices: $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}.$$

6. Test $$C$$ against simple modifications (appropriate to the setting of $$C$$) of the matrices from Step 4.

7. Write a computer programm to test $$C$$ against randomly generated $$2 \times 2$$-matrices; make sure to restrict the matrices that your random generator creates to the set of matrices that occur in $$C$$.

8. Repeat Step 6 with $$3 \times 3$$-matrices.

9. If you have not found a counterexample yet, there might be a reason to believe that $$C$$ holds.

Of course this should not be taken completely seriously - but often it works. In my experience, for many ad-hoc conjectures the algorithm stops at Step 5 or earlier. The question from the OP adds another data point to this pattern:

The matrix $$T = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$ has trace norm $$2$$, but the matrix $$S = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$ has trace norm $$\frac{1 + \sqrt{5}}{2} + \frac{\lvert 1 - \sqrt{5} \rvert}{2} = \sqrt{5} > 2.$$

• +2 if I could, for the answer, and the humour. Jul 14, 2021 at 8:30
• Also, note that taking suitably large direct sums of this counter-example shows that there is no hope of the conjecture "holding up to a constant" either (which the OP did ask about). Jul 14, 2021 at 8:30
• @MatthewDaws: Thanks for your comments! Yepp, I agree. The direct sum construction is also quite a useful way to built easy counterexamples. :-) I should probably add this to the list, together with a few typical infinite-dimensional candidates for counterexamples. But at the moment I urgently need to go to bed, so I'll postpone it to a moment when I'm really awake... Jul 16, 2021 at 1:34
• @JochenGlueck Thank you! that is a very nice algorithm :) .
– NewB
Jul 16, 2021 at 12:45
• I understand the direct sum argument. Now lets us fix a trace class operator $T$ and define $S_n$ for each $n$ entrywise as $(S_n)_{ij}=T_{ij}$ if $i,j>n$ and 0 else where. Now do you think Is it possible in general to choose $n$ large enough such that $||S_n||_1$ is as small as we wish?
– NewB
Jul 16, 2021 at 18:55