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$\newcommand{\tr}{\operatorname{tr}}$ 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 mathematics2012-07-12
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    $U$ is a unitary operator doesn't mean that $u_{ii}=1$ for every $i$, therefore your proof is not correct!2012-07-12
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    @Mercy For the inequality it's enough that $|u_{ii}| \le 1$2012-07-12
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    In fact what I assumed is $0\leq u_{ii}\leq 1$. That holds for any non negative operator, but unitary operators are not necessarily non negative.2012-07-12
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    Just ignore my last comment, I missed one line while reading!2012-07-12
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    However, |u_{ii}|=1 for unitary operators (http://en.wikipedia.org/wiki/Unitary_operator), so I guess the derivation stil holds?2012-07-12
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    I guess what you meant is $\sum_j|u_{ij}|^2=1$ for every $i$.2012-07-12
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    If the spectrum lies on the unit circle, and there is a transformation that gets the eigenvalues in the main diagonal, I think it means that $|u_{ii}|=1$2012-07-12
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    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 $i$.2012-07-12
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    That's completely correct, and if we choose the orthonormal basis that diagonalizes the unitary we would get $|u_{ii}|^2=1$ which implies $|u_{ii}|=1$.2012-07-12
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    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.2012-07-12
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    @MichaelHardy Yes, that is exactly what I mean, thanks for the remark.2012-07-13

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Your statement of the "spectral decomposition theorem" is incorrect. Unitary operators don't necessarily have point spectrum.

Also the result is incorrect. Try $$ S = U = \pmatrix{1 & 0\cr 0 & -1\cr}$$

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    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.2012-07-13
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    Do you know of any explicit example for a unitary operator with an eigenvalue outside the unit circle?2012-07-13
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    @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. $(Uf)(x) = e^{ix} f(x)$.2012-07-13
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    I see, thanks for both answers!2012-07-13