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If $A>0$ and $B>0$ and $A^{-1}$x^T(AB)x\geq x^Tx$? ($A$ and $B$ are real symmetric matrices).

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 if and only if $\rho(AB^{-1})<1$.

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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