Positive Definite

  • Thread starter Dragonfall
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  • #1
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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)
 

Answers and Replies

  • #2
CompuChip
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Maybe you can use that
A B - A^2 = A(B - A)
and from the result that
One thing that is true is this: if A and B are hermitian (or real
symmetric) with all their eigenvalues in [0, a] and [0, b]
respectively, then A B has all its eigenvalues in [0, a b].
 
  • #3
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A or B might not be real symmetric.
 
  • #4
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This is not homework! But I guess this section will get more viewers.
 
  • #5
morphism
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Usually self-adjointness is included in any notion of positivity for complex operators. How are you defining "positive definite" for a complex matrix A?
 
  • #6
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A matrix M such that for all vectors v, <v, Mv> (inner product, the usual one for complex vector spaces) is a real, positive number.
 
  • #7
morphism
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But if <v, Mv> is real for all v, then M is self-adjoint.
 
  • #8
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M is indeed self-adjoint.
 

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