3
$\begingroup$

$\newcommand{\tr}{\operatorname{tr}}$ 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.

  • 0
    @MichaelHardy Yes, that is exactly what I mean, thanks for the remark.2012-07-13

1 Answers 1

5

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}$

  • 0
    I see, thanks for both answers!2012-07-13