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Say I have two matrices $A$ and $B$, which necessarily satisfy the triangle inequality for matrix norms \begin{equation} ||A+B||\leq||A||+||B||\;. \end{equation} For a general matrix norm is it possible to give the conditions $A$ and $B$ must satisfy so there is equality in the above equation? If this is not possible for any matrix norm, can anyone provide me the conditions for equality when the norm in question is the nuclear norm?

I don't wish to assume anything about $A$ and $B$. They need not be positive or Hermitian.

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  • This is not a general answer, but a sufficient condition for the nuclear norm and for the operator norm is that $A$ and $B$ share the same singular vectors, so that you can write $A+B = U( \Sigma_A + \Sigma_B) V^T$, where $\Sigma_X$ is the matrix containing the singular values of $X$ on the diagonal. For the nuclear norm this means that the sum of the singular values of A+B is equal to the sum of the singular values of A plus the sum of singular values of B. – justmyfault Nov 08 '21 at 10:33
  • The nuclear norm's triangle inequality is met with equality iff $A$ and $B$ have a common (choice of) unitary matrix for Polar Decomposition. Sufficiency is easy, I gave a proof of necessity after "common choice of orthogonal matrix" here: https://math.stackexchange.com/questions/87982/old-amm-problem/ . That proof was for real matrices but works just as well for complex matrices with the usual minor adjustments. – user8675309 Jun 06 '24 at 05:50
  • and if $A$ and $B$ are not square, you can zero pad $A$ and $B$ with additional rows or columns of zeros until they are square. This doesn't change non-zero singular values but does reduce the problem to the square case. – user8675309 Jun 06 '24 at 06:19

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There may be other conditions, but a somewhat trivial case is when $A=cB$ with $c\geq0.$

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