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

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...

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.

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).

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

• Is there a simple argument to see that $A^{-1}(A-I)^2$ is semi-positive definite? – Davide Giraudo Aug 14 '13 at 17:40
• How to make a detailed proof that (A−I)2 is at least semi-positive definite? – mouse Aug 14 '13 at 17:50
• \begin{align} Ax&=\lambda x\\ \lambda_i&>0\\ \text{Let,}\quad \lambda_\min=\alpha>0\\ \min (eig(A-I))&=\alpha -1\\ \min (eig(A-I)^2)&=(\alpha -1)^2\geq0 \end{align} – Inquest Aug 14 '13 at 18:05
• A product of positive semidefinite matrices need not be positive semidefinite. What makes things work in this approach is the supplementary equation $=A^{-1/2}(A-I)^2A^{-1/2}$. And now it is clear, as $P^*BP$ is positive for any $P$, and any positive $B$. Which is clearly what you had in mind. Just answering @Davide Giraudo's comment. – user85486 Aug 14 '13 at 18:09
• Take a basis of eigenvectors $B$ of $A$ and note that each eigenvector of $A$ is an eigenvector of $(A-I)$, then $$v_i^T(A-I)^2v_i = \lambda _i^2 v_i^Tv_i.$$ Thus $(A-I)^2$ is positive semi-definite for a basis. – walcher Aug 14 '13 at 18:10

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}

Since $A$ is symmetric, it has an eigendecomposition $A = Q \Lambda Q^T$. Hence,

$$\begin{array}{rl} A + A^{-1} - 2 I &= Q \Lambda Q^T + Q \Lambda^{-1} Q^T - 2 Q Q^T\\\\ &= Q (\Lambda + \Lambda^{-1} - 2 I) Q^T\\\\ &= Q \Lambda^{-\frac 12} (\Lambda^2 - 2 \Lambda + I) \Lambda^{-\frac 12} Q^T\\\\ &= Q \left(\Lambda^{-\frac 12} (\Lambda - I)^2 \Lambda^{-\frac 12}\right) Q^T\end{array}$$

If $A \succ 0$, then $A + A^{-1} - 2 I$ is at least positive semidefinite. It may be positive definite.