Given that $A,B$ are positive definite matrix, are they also PD?
$A+B$
$AB$
$A^2 +I$
$ABA^{*}$
$x^TAx>0, x^TBx>0$ so $1$, is correct, could you tell me about the 2, 3,4?
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Mex
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.
Leandro
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@Leonardo, thank you very much – Myshkin Jun 15 '12 at 07:52
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@Mex Leandro :) – Leandro Jun 15 '12 at 07:53
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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? – Myshkin Jun 15 '12 at 07:56
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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 08:00
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positive definite matrices are self adjoint?could you tell me how? – Myshkin May 31 '13 at 07:42
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1@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 – Jan 06 '14 at 07:24