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I am stuck on the following two questions:

My first question:

If $D$ is a non-singular matrix, such that $(D+D^*)$ is positive definite. From my understanding of the problem's solution, it says that $D$ has to be positive definite. Why is this true?

My second question:

If $A$ and $B$ are positive definite matrices, why is : $A^{\frac{1}{2}}BA^{-\frac{1}{2}}$ positive definite? Now the other way: if $A^{\frac{1}{2}}BA^{-\frac{1}{2}}$ positive definite and $A$ is positive definite, does this imply that $B$ is positive definite?

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For the first question, the spectral theorem gives that $D+D^*$ is at least positive semi-definite. For the second, if all the eigenvalues of $B$ are positive, the same is true for any similar matrix. –  Aaron Apr 21 '12 at 23:19

1 Answer 1

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On the first question: not true. Take $$ D=\begin{bmatrix}1&1\\ 0&1\end{bmatrix} $$

On the second question: again not true . Take $$ A=\begin{bmatrix}1&0\\0&4\end{bmatrix},\ \ \ B=\begin{bmatrix}3&1\\ 1&3\end{bmatrix}; $$ Then $A^{1/2}BA^{-1/2}$ is not even symmetric.

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You don't need symmetric to be positive definite (which is just that all the eigenvalues are positive). –  Aaron Apr 21 '12 at 23:09
    
@Martin Argerami: I must have misunderstood the problem's solution. Anyway, I will post the problem and the solution as it is written in the book in a new topic, and I can tell you where I am stuck. –  Boyan Klo Apr 21 '12 at 23:31
    
@Aaron: I must have misunderstood the problem's solution. Anyway, I will post the problem and the solution as it is written in the book in a new topic, and I can tell you where I am stuck –  Boyan Klo Apr 21 '12 at 23:32
    
@Aaron: I know that symmetric or not is a contentious issue on the definition of positive definite. Now, according to your "definition", $$\begin{bmatrix}1&0\\ 100&1\end{bmatrix}$$ is positive definite. I cannot agree with that. –  Martin Argerami Apr 21 '12 at 23:38
    
@MartinArgerami: The square matrix $A$ represents the bilinear form $(x,y)_A=x^TAy$, and while symmetric bilinear forms are nice, there are other forms worth considering. Positive definite should mean that $(x,x)_A\geq 0$ for all $x$, with equality if and only if $x=0$. However, you are right that my "definition" was an oversimplification of the issue. –  Aaron Apr 23 '12 at 7:11

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