Show that if $A$ is positive definite then $A + A^{-1} - 2I$ is positive semidefinite

Question

Let $A$ be a real symmetric positive definite matrix. Show that $$A + A^{-1} -2I$$ is positive semidefinite.

I found that $A^{-1}$ is a positive definite matrix, thus $A + A^{-1}$ is also a positive definite matrix, moreover I know the form of the $z^TIz$ is as follows $(a^2 + b^2 + ... )$, where $a,b, ...$ are the components of vector $z$. I don't know what to do next...

Answer 1

Hint: diagonalize your matrix $A$ in an orthonormal basis and study the real function $f(t)=t+\frac{1}{t}-2$. Or just use functional calculus an spectral mapping if you know about that.

Answer 2

Hint: $A+A^{-1}-2I = A^{-1}(A^2-2A+I)=A^{-1}(A-I)^2$

Answer 3

Hint: Using the fact that $A$ is diagonalizable and the eigenvalues are positive, write $A:=B^2$, where $B$ is symmetric. Then $A+A^{-1}-2I=(B-B^{-1})^2=(B-B^{-1})^t(B-B^{-1})$, which is semi-positive definite (not necessarily positive definite, as $A=I$ shows).

Answer 4

Purely for fun: \begin{align} \min_x \quad &x^T(A+A^{-1}-2I)x\\ f(x)&=x^T(A+A^{-1}-2I)x\\ \nabla f(x)&=2\times (A+A^{-1}-2I)x=0\\ (A+A^{-1}-2I)x&=0\\ \implies x^T(A+A^{-1}-2I)x&\geq0 \qquad \forall x\\ \end{align}