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*Problem:*

Prove that if $A$ and $AB+BA$ are positive definite matrices, then $B$ is positive definite.

I didn't understand some parts of this problem's solution which is given below:

Let $C=AB+BA$. Now, multiply $C$ from right and left by $A^{-\frac{1}{2}}$ to get: 0< A^{-\frac{1}{2}}CA^{-\frac{1}{2}}=A^{\frac{1}{2}}BA^{-\frac{1}{2}}+A^{-\frac{1}{2}}BA^{\frac{1}{2}}=D+D^*

Where $D=A^{\frac{1}{2}}BA^{-\frac{1}{2}}$

Next, the solution says that it is sufficient to show that $D$ is nonsingular.

My first question: Why is 0< A^{-\frac{1}{2}}CA^{-\frac{1}{2}}, i.e positive definite?

My second question: I can't see why is $D$ being non-singular implies that $B$ is positive definite?

1 Answers 1

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Your first question. The reason is $C$ is positive definite, then the congruence is also positive definite (since $A$ is nonsingular).

For your second question. 0< D+D^* implies the real part of the eigenvalues of $D$ are positive, i.e., $B$ is similar to a matrix having real part of the eigenvalues positive. since $B$ is Hermitian, then it is positive definite.

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    @Boyan, the equation relating $D$ and $B$ tells you $B$ is similar to $D$, and Sunni has said why $D$ has eigenvalues with positive real part.2012-04-22