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I am having some little difficulties to show the following:

Let $M$ be a symmetric positive square matrix. I must show that $I-M$ is positive definite matrix if and only if $M^{-1}-I$ is positive definite.

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You can assume that $M$ is diagonal. Can you solve it then? –  user27126 Jan 13 '13 at 5:58

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Note that M is Symmetric Positive Definite, Implying that the eigenvalues are all positive. This argument works backwards too. (This would not be true if M were not Symmetric in general) \begin{align} \rightarrow\\ \because M \text{ is PD, }\\ \lambda_i&>0\quad \forall i\\ eig(I-M)&=(1-\lambda_i)\\\because(I-M)\text{ is PD, }\\ (1-\lambda_i)&>0\\ \implies 1&>\lambda_i>0\quad \forall i\\ eig(M^{-1}-I)&=\dfrac{1}{\lambda_i}-1\\ &>0\\ \implies (M^{-1}-I) \text{ is PD} \end{align} Similarly, you can do the reverse direction.

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thanks a lot. I never noticed that $eig(I-M)=eig(I)-eig(M)$ –  ChuckM Jan 13 '13 at 6:18
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Yeah. Not very difficult to prove : \begin{align} Mx&=\lambda x\\ (M+I-I)x &= \lambda x \\ (M+I)x - Ix &= \lambda x \\ (M+I)x &= (\lambda + 1 ) x \end{align} –  Inquest Jan 13 '13 at 6:34

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