Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

*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?

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.