# Prove $I-A^{-1}$ is positive definite

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

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

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