Skip to main content
1 of 2
Tobias Fritz
  • 6.4k
  • 2
  • 27
  • 52

For hermitian $B$, the inequality is very easy to prove: use that $\|B\| - B$ is positive semidefinite together with the fact that the product of two positive semidefinite matrices has nonnegative trace. This gives $\mathrm{tr}(A(\|B\| - B)) \geq 0$, from which the inequality follows immediately.

For generic non-hermitian $B$, the left-hand side of your inequality is not real. So you need to specify more precisely what you mean for general $B$.

Tobias Fritz
  • 6.4k
  • 2
  • 27
  • 52