# For invertible matrices $A$ and $C$, prove or disprove that $(C^{-1}BAB^T + I)$ is invertible.

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?

• Err, don't you get the zero matrix if you take about $C=-I$, $B=A=I$? – Nick Alger Mar 24 '16 at 23:51
• @ 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. – loup blanc Mar 25 '16 at 17:21
• The question is simply dumb (i was really tired at that moment) and nobody will ever find the solution useful. – ndrizza Mar 25 '16 at 18:47

Try, for example, $B=C=I$, $A=-I$.
• Thanks! I'll see how I can constrain $A$, $B$ and $C$ further s.t. the equivalence holds. – ndrizza Mar 24 '16 at 23:59