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For invertible matrices $A$ and $C$, ($B$ may or may not be invertible) prove or disprove that $(C^{-1}BAB^T + I)$ is invertible.

This problem came up while I was proving an equivalence. I couldn't find a way to prove that it is correct. Can you think of a way to prove it to be correct?

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  • $\begingroup$ Err, don't you get the zero matrix if you take about $C=-I$, $B=A=I$? $\endgroup$ – Nick Alger Mar 24 '16 at 23:51
  • $\begingroup$ @ ndrizza , you are not really an eagle. Yet, to comfort you, be aware that if you randomly choose (with a normal distribution) $A,B,C$ then, with probability $1$, $C$ is invertible and $C^{-1}BAB^T+I$ is invertible. $\endgroup$ – loup blanc Mar 25 '16 at 17:21
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    $\begingroup$ The question is simply dumb (i was really tired at that moment) and nobody will ever find the solution useful. $\endgroup$ – ndrizza Mar 25 '16 at 18:47
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Try, for example, $B=C=I$, $A=-I$.

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  • $\begingroup$ Ok, now you have that your equivalence relation may not be provable,@ndrizza. How about trying to find the smallest equivalence relation it contains? That might help you. $\endgroup$ – астон вілла олоф мэллбэрг Mar 24 '16 at 23:54
  • $\begingroup$ Thanks! I'll see how I can constrain $A$, $B$ and $C$ further s.t. the equivalence holds. $\endgroup$ – ndrizza Mar 24 '16 at 23:59

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