- #1

- 1,030

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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)

- Thread starter Dragonfall
- Start date

- #1

- 1,030

- 4

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)

- #2

CompuChip

Science Advisor

Homework Helper

- 4,302

- 47

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

- 1,030

- 4

A or B might not be real symmetric.

- #4

- 1,030

- 4

This is not homework! But I guess this section will get more viewers.

- #5

morphism

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- #6

- 1,030

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- #7

morphism

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But if <v, Mv> is real for all v, then M is self-adjoint.

- #8

- 1,030

- 4

M is indeed self-adjoint.

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