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Let $A$ and $B$ be two positive definite Hermitian matrices. Show that the Hermitian matrix $$C\ =\ A^{-1} + B^{-1} - 4(A + B)^{-1}$$ is also positive definite.

Thanks in advance.

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up vote 6 down vote accepted

The statement is not true. Consider $A=B=I$. Then $C=0$ is not positive definite.

Edit: Let $A^{1/2}$ be the positive definite square root of $A$ and $X=A^{-1/2}BA^{-1/2}$. Then $$C=A^{-1/2}[I+X^{-1}-4(I+X)^{-1}]A^{-1/2}.$$ Therefore $C$ is positive definite if and only if $I+X^{-1}-4(I+X)^{-1}$ is positive definite. Since $X$ is positive definite, it is unitarily diagonalizable. Therefore WLOG we may assume that $X$ is a positive diagonal matrix. So it boils down to figuring out when will $1+x^{-1}-4(1+x)^{-1}>0$ for $x>0$. As this inequality is true for all $x\neq1$ and equality holds when $x=1$, we conclude that $C$ is always positive semi-definite and it is positive definite if and only if $1$ is not an eigenvalue of $X=A^{-1/2}BA^{-1/2}$. (By the way, the eigenvalues of $A^{-1/2}BA^{-1/2}$ are identical to the eigenvalues of $A^{-1}B$ or $BA^{-1}$.)

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So, the statement is true if $A\neq B$? And what is WLOG? –  user70195 Mar 31 '13 at 21:06
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@ShanaKugimiya "WLOG" is the standard shorthand of "without loss of generality". And no, the statement can be false if $A\neq B$. As I said in the answer, $C$ is positive definite if and only if all eigenvalues of the matrix product $A^{-1}B$ are not equal to $1$. –  user1551 Mar 31 '13 at 21:11
    
Oh I see. Thanks a lot. By the records, English is not my main languages :P –  user70195 Mar 31 '13 at 21:11
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