# $\kappa(B^{-1}A)=\kappa (AB^{-1}) = \frac{\lambda_{\max}(B^{-1}A)}{\lambda_{\min}(B^{-1}A)}$

Prove or disprove: if $A,B$ are symmetric positive definite (s.p.d.) matrices then operator 2-norm condition number $\kappa(B^{-1}A)=\kappa (AB^{-1}) = \frac{\lambda_{\max}(B^{-1}A)}{\lambda_{\min}(B^{-1}A)}$. Does it hold if $A$ is symmetric and $B$ is s.p.d.?

There is a proof in my lecture note but they used the wrong fact that $B(B^{-1}A)B^{-1} = AB^{-1}$ is an equivalence transformation of $B^{-1}A$ hence $\kappa(B^{-1}A)=\kappa (AB^{-1})$. I have no idea about the correctness of the statement and I am unable to prove it or give a counter example. Can anyone help me? Thank you in advance!

A good first step is to perform some experiments with small matrices! Here tools such as MATLAB can be valuable when disproving identities. You will find that the statement under investigation is false in general, but it is certainly true if the matrices $B^{-1}A$ and $A^{-1}B$ are symmetric! There is condition which occurs frequently in pure mathematics, but less frequently in practical applications, which will ensure that this is the case. Can you find this extra condition?
• Hi Carl, there is nothing to sing if $B^{-1}A$ and $A^{-1}B$ commutes. I don't want to use this. Can you prove or disprove the statement? Jan 27, 2016 at 15:02
• You just need a counter-example. Just make sure that A and B do not commute, so that $B^{-1}A$ and $A^{-1}B$ are not automatically symmetric matrices. Jan 27, 2016 at 15:08
• Then you where extraordinarily unlucky or you did not generate a random matrix. Try $A = \begin{bmatrix} 3 & 2 \\ 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 2 \\ 2 & 4 \end{bmatrix}$. Jan 27, 2016 at 15:30
• No, but your example has $\kappa(B^{-1}A) = \kappa(AB^{-1})$ Jan 27, 2016 at 15:37