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Let $A$ and $B$ be two square matrices such that $A^\top B = 0$ and $B^\top A = 0$. How can we show that $$ \|A+B\|_{nuc} = \|A\|_{nuc} + \|B\|_{nuc}$$

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Hint: Let $|M| = \sqrt{M^TM}$. Note that $|A|^2$ and $|B|^2$ commute with $|A|^2\;|B|^2 = 0$. Since there exist polynomials $p,q$ with $|A| = p(A^TA)$ and $|B| = q(B^TB)$, we can conclude that $|A|,|B|$ commute with $|A| \, |B| = 0$. So, we have $$ (|A| + |B|)^2 = |A|^2 + |B|^2 = A^TA + B^TB = |A + B|^2. $$ So, $|A+B| = |A| + |B|$.

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  • $\begingroup$ How are you able to conclude the part about |A| and |B| commuting? I'm not familiar with that result. $\endgroup$ – Seth Devlin Apr 13 at 19:32
  • $\begingroup$ @SethDevlin Because $A$ commutes with $B$, $A$ commutes with $q(B) = |B|$. Because $|B|$ commutes with $A$, $|B|$ commutes with $p(A) = |A|$. $\endgroup$ – Ben Grossmann Apr 13 at 19:51

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