Where $\Sigma\in M_n$ is a non-negative diagonal matrix, $U$ is any unitary matrix and $\|\bullet\|$ is a unitarily invariant norm.
Is it true that $\|\Sigma-I\|\le\|\Sigma-U\|\le\|\Sigma+I\|$ and why?
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linear-algebra
matrices
1 Answers
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From the definition of norm, we have $\| \Sigma \| - \| U \| \le \| \Sigma - U \| \le \| \Sigma \| + \| U \|$
So if the norm is the spectral norm, which is unitarily invariant, then $\| \Sigma \|=\lambda_{max}, \quad \| U \| =1, \quad \| \Sigma \pm I \| = \lambda_{max} \pm 1,$
where $\lambda_{max}$ is the maximum diagonal element of $\Sigma$. Subsituting these equalities into the above inequality will give the result.
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0$\Sigma =0$ is a non-negative diagonal matrix, but $\|\Sigma -I\| = 1$, not zero. – 2012-10-18