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Suppose A is an $n \times n$ positive definite matrix. Prove that $A+A^{-1}-2I_n$ is positive semidefinite.

I know that the eigenvalues of $A^{-1} = \lambda^{-1}$, and that I have to relate that to the equation $\lambda + \lambda^{-1} -2 =f(\lambda).$ which would mean that $\lambda \geq 0$ A.K.A. semidefinite.

I would also appreciate alternate approaches.

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    $\begingroup$ You can diagonalize both A and it's inverse simultaneously $\endgroup$ – Mark Dec 11 '18 at 20:09
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if $t > 0$

$$ \left( \sqrt t - \frac{1}{\sqrt t} \right)^2 \geq 0 $$ $$ t - 2 + \frac{1}{t} \geq 0 $$

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One alternate approach is as follows: note that $$ A + A^{-1} + 2I = [A^{-1/2}(A - I)][A^{-1/2}(A - I)]^* $$

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  • $\begingroup$ Can you please explain how you got this? $\endgroup$ – mhall14 Dec 11 '18 at 23:53

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