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Let $M$ be a generic nonzero real matrix and $M_{0}$ is constructed by replacing some elements in $M$ with zero. Let $||\cdot||_{*}$ denotes the nuclear norm operator. Is it true that $||M||_{*}\geq ||M_{0}||_{*}$ regardless of the replacement rule?

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No. Consider $$ A=\pmatrix{1&1&1\\ 1&1&1\\ 1&1&1},\ B=\pmatrix{1&0&1\\ 0&1&1\\ 1&1&1}. $$ Since $A$ is positive semidefinite, it nuclear norm is equal to its trace, which is $3$. The matrix $B$ is symmetric but indefinite. Hence its nuclear norm is the sum of absolute values of all its eigenvalues, i.e. $|1|+|1+\sqrt{2}|+|1-\sqrt{2}|=1+2\sqrt{2}>3$.

P.S. I asked a similar question before about the induced $2$-norm (and the answer is also negative).

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  • $\begingroup$ Great example! Many thanks! $\endgroup$ – Ecthelion Feb 7 at 1:47

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