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Given that $A,B$ are positive definite matrix, are they also PD?

  1. $A+B$

  2. $AB$

  3. $A^2 +I$

  4. $ABA^{*}$

$x^TAx>0, x^TBx>0$ so $1$, is correct, could you tell me about the 2, 3,4?

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1 Answer 1

up vote 2 down vote accepted


in 4. we can argue as follows:

$\langle x, ABA^* x\rangle = \langle A^*x, BA^* x\rangle = \langle B^*A^*x, A^* x\rangle = \langle B Ax, A x\rangle = \langle B y, y\rangle >0 $, where $y=Ax$ and in the two last equalities we use that positive matrix are self-adjoint and since they do not have $0$ as eigenvalue then $Ay\neq 0$.

for 3. the argument is similar.

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@Leonardo, thank you very much –  La Belle Noiseuse Jun 15 '12 at 7:52
@Mex Leandro :) –  Leandro Jun 15 '12 at 7:53
Leandro!!! for (3) $\langle x,(A^2+I)x \langle=\langle x,A^2x \langle +\langle x,x\langle= \langle Ax,Ax \langle+ \langle x,x \langle>0$ right? –  La Belle Noiseuse Jun 15 '12 at 7:56
Yes and worth to mention that in the definition of positivity we only have to verify the inequality for $x\neq 0$. –  Leandro Jun 15 '12 at 8:00
@Tojamaru : Positive definite matrices are symmetric (which is same as self adjoint for real matrix) but for complex matric I am not sure how :O –  Praphulla Koushik Jan 6 '14 at 7:24

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