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If $A$ is a nonsingular symmetric matrix such that $A-I$ is positive definite, prove that $I-A^{-1}$ is positive definite.

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  • $\begingroup$ hint: this means $A$ is positive definite, so $A^{1/2}$ exists, as does $A^{-1/2}$ $\endgroup$
    – user125932
    Commented Oct 14, 2019 at 4:05
  • $\begingroup$ @user125932 thank you for the hint. I figured it out. $\endgroup$
    – viKinG
    Commented Oct 14, 2019 at 4:18

3 Answers 3

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Without loss of generality we can assume that $A$ is diagonal, then $A_{ii}-1 >0$ for all $i$. Hence ${1 \over A_{ii}} (A_{ii}-1) = 1 - {1 \over A_{ii}} > 0$ for all $i$. Hence $I - A^{-1} >0$.

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Since $A$ is nonsingular, then $A^{-1}$ exists. $AA^{-1}=I$.

Note that a matrix is positive definite if and only if all of its eigenvalues are positive. Since $A-I$ is positive definite, then $\sigma_i(A-I)>0\Longleftrightarrow\sigma_i(A)>1$, where $\sigma_i(A),i=1,\cdots,n$ denotes eigenvalues of A, $n$ is the dimension of $A$.

Since $\sigma_i(A)>1 \Longleftrightarrow \sigma_i(A^{-1})<1$, then we have $\sigma_i(I-A^{-1})>0$, which means that $I-A^{-1}$ is positive definte.

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Let $X$ be a non-zero vector. Then $X^T(A-I)X>0\implies X^TAX>||X||^2\implies X^TA^{-1}X<||X||^2$.

Now $X^T(I-A^{-1})X=||X||^2-X^TA^{-1}X>0$

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