# The trace norm cannot be increased by composing with a unitary operator

$\newcommand{\tr}{\operatorname{tr}}$ I was reading a proof for the statement $|\tr(US)|\leq |\tr(S)|$, for every endomorphism $S$ on a complex vector space $H$ and every unitary operator $U$ on the same space. Though the proof is short it uses the polar decomposition and the cauchy swartz inequality. I came out with the very simple "proof": \begin{eqnarray} \left|\tr(US)\right| &=& \left|\tr\left(\sum_iu_{ii}|i\rangle\langle i|\sum_{jk}s_{jk}|j\rangle\langle k|\right)\right|\\ &=& \left|\tr\left(\sum_i\sum_k u_{ii} s_{ik} |i\rangle\langle k|\right)\right| \\ &=& \left|\sum_{i}u_{ii}s_{ii}\right|\\ &\leq& \left|\sum_i s_{ii}\right| \\ &=& \left|\tr(S)\right| \end{eqnarray} where I just rely in the fact that unitary operators are normal and admit the spectral decomposition theorem (that is $U=\sum_i u_{ii}|i\rangle\langle i|$ for some orthonormal basis $|i\rangle$)

I would like to know if I am missing something very obvious or the proof is indeed correct.

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Perhaps you should explain the notations, they are very common in quantum mechanics but not in mathematics – Mercy King Jul 12 '12 at 15:33
$U$ is a unitary operator doesn't mean that $u_{ii}=1$ for every $i$, therefore your proof is not correct! – Mercy King Jul 12 '12 at 15:35
@Mercy For the inequality it's enough that $|u_{ii}| \le 1$ – Cocopuffs Jul 12 '12 at 15:38
Let's consider the finite-dimensional case. Let $U \in \mathbb{C}^{n\times n}$ be unitary, i.e. $UU^*=(\delta_{ij})=I_n$. Therefore $\sum_{k=1}^nu_{ik}\bar{u}_{jk}=\delta_{ij}$ for every $i,j$, and so $\sum_{k=1}^n|u_{ik}|^2=\sum_{k=1}^nu_{ik}\bar{u}_{ik}=1$ for every $i$. – Mercy King Jul 12 '12 at 16:12
Where you say "where $S$ is an endomorphism on a complex vector space $H$ and $U$ a unitary operator on the same space", did you mean "for every endomorphism $S$ on a complex vector space $H$ and every unitary operator $U$ on the same space"? The latter wouldn't leave any doubts about the two "every"s. – Michael Hardy Jul 12 '12 at 16:32

Also the result is incorrect. Try $$S = U = \pmatrix{1 & 0\cr 0 & -1\cr}$$
That's correct, however there is a typo in my transcription (related to my poor understanding of the issue). The correct statement is: $|tr(US)|\leq tr(|S|)$. Anyway I will accept the answer, since it has clarified my ideas. – Euclean Jul 13 '12 at 9:20
@Euclean: The spectrum of a unitary operator is always on the unit circle. However, in an infinite-dimensional Hilbert space it does not have to consist of eigenvalues. Consider, for example, the operator of multiplication by $e^{ix}$ on $L^2[0,1]$, i.e. $(Uf)(x) = e^{ix} f(x)$. – Robert Israel Jul 13 '12 at 14:56