# Proving Semidefinite From Adding Inverse and Subtracting Multiple of Identity Matrix.

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.

• You can diagonalize both A and it's inverse simultaneously – Mark Dec 11 '18 at 20:09

if $$t > 0$$
$$\left( \sqrt t - \frac{1}{\sqrt t} \right)^2 \geq 0$$ $$t - 2 + \frac{1}{t} \geq 0$$
One alternate approach is as follows: note that $$A + A^{-1} + 2I = [A^{-1/2}(A - I)][A^{-1/2}(A - I)]^*$$