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$A,B∈ M_n (R)$ is a positive definite matrix.Then

a)$A^2$ is positive definite.

b)If A and B commutes then $A^k + B^k$ positive definite.

c)$AB + BA$ is positive definite.

d)If A and B commutes then $f(AB)$ is positive definite for any polynomial with positive coefficients.

a is true, for b) I can say that seperately they are positive definite but I am not getting what about their sum! c,d compltely clueless. Thank you for help.

Myshkin
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2 Answers2

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You have to show two things:

  1. If A, B are positiv definite then A+B is positiv definite
  2. If A, B are positive and AB = BA, then AB (and also BA) is positive definite

1: It holds: $x^T(A+B)x = x^TAx + x^TBx > 0$ thus A+B is positiv definite.

Can you cantinue with 2 by yourself?

Hans Meierwurst
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  • well, $(A+B)^2=A^2+B^2+ AB+BA$ so I am not sure whether $A^2+B^2$ will be positive definite or semidefinite.I guess semidefinite. – Myshkin Jun 22 '13 at 11:40
  • @miosaki you have already shown if A is positive definite then A^2 is positive definite. Similarly, B^2 is positive definite. Also, you know that if X and Y are positive definite, then X + Y is positive definite. If you put X as A^2 and Y as B^2, what do you get? – TenaliRaman Jun 22 '13 at 11:50
  • oopss okay okay okay...so $A^k+B^k$ is PD :) – Myshkin Jun 22 '13 at 11:55
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Supplementing Hans, c) is generally false without the additional assumption: Inspired by the answer of this question, you can take $A = \pmatrix{ 1 & 2\cr 2 & 5\cr}$, $B = \pmatrix{1 & -1\cr -1 & 2\cr}$ to obtain a counterexample, as $$\pmatrix{ 1 & 0 } (AB+BA) \pmatrix{1\\0} = -2.$$

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