1
$\begingroup$

For invertible symmetric matrices $A$, $B$, I want to show that $$ A \preccurlyeq B \iff A^{-1} \preccurlyeq B^{-1}$$

More concretely, I tried to argue the following. Denote by $\lambda_{S_{min}}$ and $\lambda_{S_{max}}$ the minimum and maximum eigenvalues of $S$. Then \begin{align} A \preccurlyeq B & \iff \forall x, x^\top A x \leq x^\top B x \\ & \iff \lambda_{A_{min}} \leq \lambda_{B_{min}} \\ & \iff \frac{1}{\lambda_{A_{max}}} \leq \frac{1}{\lambda_{B_{max}}}\\ & \iff \lambda_{A^{-1}_{max}} \leq \lambda_{B^{-1}_{max}} \\ & \iff \forall x, x^\top A^{-1} x \leq x^\top B^{-1} x \\ & \iff A^{-1} \preccurlyeq B^{-1} \end{align}

Is this correct?

Update: The result should be $A \preccurlyeq B \iff A^{-1} \succcurlyeq B^{-1}$ for positive definite matrices. The fixed proof: \begin{align} A \preccurlyeq B & \iff \forall x, x^\top A x \leq x^\top B x \\ & \iff \lambda_{A_{max}} \leq \lambda_{B_{min}} \\ & \iff \frac{1}{\lambda_{A_{max}}} \geq \frac{1}{\lambda_{B_{min}}}\\ & \iff \lambda_{A^{-1}_{min}} \geq \lambda_{B^{-1}_{max}} \\ & \iff \forall x, x^\top A^{-1} x \geq x^\top B^{-1} x \\ & \iff A^{-1} \succcurlyeq B^{-1} \end{align}

  • 3
    shouldn't it be $A^{-1}\geq B^{-1}$? Look at the case $A,B\in\Bbb R^{1\times 1}$ – 2017-01-28
  • 0
    yes, that's why I'm confused. Please point out where the proof may have gone wrong. – 2017-01-28
  • 0
    If $A\in\Bbb R^{1\times 1}$, then $\lambda_{A_{min}}=\lambda_{A_{max}}=A$. Same for $B$. Reread your equation for this special case. You will see the mistake (from second to third line). – 2017-01-28
  • 0
    Yes, it looks much better. – 2017-01-28
  • 0
    Related: [Inequality concerning inverses of positive definite matrices](https://math.stackexchange.com/q/214126), [Is $B^{-1}-A^{-1}$ a positive definite matrix?](https://math.stackexchange.com/q/236431) – 2017-01-29

0 Answers 0