# Decompose nuclear norm

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

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, $$|A+B| = |A| + |B|$$.
• @SethDevlin Because $A$ commutes with $B$, $A$ commutes with $q(B) = |B|$. Because $|B|$ commutes with $A$, $|B|$ commutes with $p(A) = |A|$. – Ben Grossmann Apr 13 at 19:51