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From following derivation, depending on step, where to substitute $A=I$ one can obtain:

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

From which follows that matrices $B$ and $(I+B)^{-1}$ should commute.

If spectral radius of $B$ is less than $1$, that can be proved through Neumann series expansion ($B$ commute with $I$ and any power of $B$).

If $B^{-1}$ exists: $$ B(I+B)^{-1} = (B^{-1})^{-1}(I+B)^{-1}=((I+B)B^{-1})^{-1}=(B^{-1}+I)^{-1} \\ (I+B)^{-1}B = (I+B)^{-1}(B^{-1})^{-1}=(B^{-1}(I+B))^{-1}=(B^{-1}+I)^{-1} $$

However, aforementioned derivation only require $(I+B)^{-1}$ to exist, thus I wonder how commutation of $B$ and $(I+B)^{-1}$ can be proven in general case?

Kostiantyn Lapchevskyi
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1 Answers1

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Note that $\operatorname{Id}+B$ and $(\operatorname{Id}+B)^{-1}$, since they are the inverse of each other. But $\operatorname{Id}$ commutes with any matrix and, in particular, with $\operatorname{Id}+B$. Therefore, $B$ commutes with it too.

José Carlos Santos
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