- #1
Dragonfall
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URGENT: Can you prove or disprove:
Let A and B be (complex matrices) positive definite with trace 1.
Given A < B, (B-A is pos def )
then
A^2 < AB (AB-A^2 is pos def)
Let A and B be (complex matrices) positive definite with trace 1.
Given A < B, (B-A is pos def )
then
A^2 < AB (AB-A^2 is pos def)