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We are given that a matrix $A$ in $R^{n\times n}$ is invertible. We must show that $A + B$ (also in $R^{n\times n}$) is invertible if and only if $I_n$ + $A^{-1}$$B$ is invertible.

I cannot figure out how to do this proof. It is probably something simple I am missing but I just can't get my head around it. In particular, the $I_n$ (multiplicative identity) is putting me off. Anyone have any suggestions or hints to help me get going?

Thanks!

Helen Byrne
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If $A+B$ is invertible, then

$$(A+B)^{-1}A(I_n + A^{-1}B) = (A+B)^{-1}(A + B) = I$$

By a fundamental theorem in linear algebra, we have $(I_n + A^{-1}B)(A+B)^{-1}A = I$ as well and therefore $(I_n + A^{-1}B)$ is invertible and $(I_n + A^{-1}B)^{-1} = (A+B)^{-1}A$.

If $I_n + A^{-1}B$ is invertible, then

$$(I_n + A^{-1}B)^{-1}A^{-1}(A+B) = (A(I_n + A^{-1}B))^{-1}(A+B) = (A+B)^{-1}(A+B) = I$$

Similarly, $A+B$ is invertible and $(A+B)^{-1} = (I_n + A^{-1}B)^{-1}A^{-1}$.

Empiricist
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  • Thanks for this. What is the 'fundamental theroem' that allows re-arrangement of the terms like that? Is it simply that we the inverse can be applied either side and will still reduce to the identity? (Multiplication by the inverse is commutative) – Helen Byrne Dec 12 '14 at 10:24
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    Please read: http://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i – Empiricist Dec 12 '14 at 10:26