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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?

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up vote 0 down vote accepted

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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Can you elaborate more please? For the first part, what do you mean by the "congruence is also positive definite"? For the second part: How did you figure out that $B$ is similar to a matrix whose eigenvalues have positive real part? – Boyan Klo Apr 22 '12 at 1:27
@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. – Gerry Myerson Apr 22 '12 at 4:19

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