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If $A>0$ and $B>0$ and $A^{-1}

I know these facts (from Matrix Mathematics book) (It seemed to me these might help but I haven't been able to use them in my advantage!),

Let $A,B\in\mathbb{F}^{n\times n}$ (real or complex matrix), and assume that $A$ and $B$ are positive semi-definite. Then, $0\leq A

Let $A,B\in\mathbb{N}^{n\times n}$ (positive semi-definite matrix). Then AB is semi simple, and every eigenvalue of $AB$ is nonnegative. If in addition $A$ and $B$ are positive definite, then every eigenvalue of $AB$ is positive.

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    Could you explain what ordering you are imposing on the set of matrices? For example, it doesn't make sense to say that $w < z$ where $w,z \in \mathbb{C}$, although it does make sense to say that $|w| < |z|.$2012-08-29
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    A and B are real matrices. (A>0 means that A is a symmetric matrix with all eigenvalues greater than zero) and $A>B$ means that $A-B>0$. Is this what you were asking?2012-08-29
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    So $A > 0$ if $\det(A) > 0$ and $A > B$ if $\det(A-B) > 0.$ Is this well defined? If $A$ and $B$ are $2n$-by-$2n$ matrices then $\det(X) = \det(-X),$ meaning that $\det(A-B) > 0$ if and only if $\det(B-A) > 0.$ It follows that $A > B$ is and only if $B > A.$2012-08-29
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    @Fly by Night: No, that's not right. $A>0$ means $A$ is positive definite (this is standard notation in a lot of optimization texts). The negation of the identity matrix is not positive definite, even though it may have positive determinant.2012-08-29
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    Ah, yes of course!2012-08-29
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    Just to put the definition of positive definite , that is, $A>0$, in terms of determinants, straight. It is not enough that $\det(A)>0$, what is needed is that all leading principal minors are positive, that is, $\det(A_{ii})>0$, for all $i$. This is the determinants off the submatrices consisting of first $i$ rows and columns.2012-09-02

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Since $B=A^{-1}+C$ with $C$ symmetric and positive, one asks whether $x^TACx\geqslant0$ for every vector $x$ and every symmetric positive matrices $A$ and $C$.

Consider $A=\begin{pmatrix}5 & 2 \\ 2 & 1\end{pmatrix}$ and $C=\begin{pmatrix}1 & -2 \\ -2 & 10\end{pmatrix}$. Then $A$ and $C$ are symmetric positive, $A^{-1}=\begin{pmatrix}1 & -2 \\ -2 & 5\end{pmatrix}$, $B=\begin{pmatrix}2 & -4 \\ -4 & 15\end{pmatrix}$ and $AC=\begin{pmatrix}1 & 10 \\ 0 & 6\end{pmatrix}$. Hence $x^TACx\lt0$ for $x^T=(1,-1)$.

Thus, the result above does not hold in full generality.

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Do you think this is a counter example for this question?

We know that $\rho(AB)>1$. If we assume $$AB=\begin{bmatrix}2 & -8\\ 0 &2\end{bmatrix}$$ The eigenvalues of this matrix are $(2,2)$

$$\frac{1}{2}(AB+(BA)')=\begin{bmatrix}2 & -4\\ -4 & 2\end{bmatrix}$$ The eigenvalues of this matrix are $(-2,6)$.

Therefore $x^TABx=\frac{1}{2}x^T(AB+(BA)')x$ is not greater or less than $x^Tx$! $(AB+(BA)')$ is an indefinite matrix.

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    What are $A$ and $B$ in your example?2012-09-10