Can you help me prove the matrix equality below? $$I+BA = (I+B)(I-(I+B)^{-1}B(I-A)),$$ where $B, A$ are some matrix and $I$ is identity matrix. Should I use matrix inversion lemma?
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The right side looks like $x\left(1-x^{-1}y z\right)$ so it equals $x-y z$ – Lozenges May 08 '17 at 09:27
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Where exactly are you stuck ? – May 08 '17 at 09:58
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Oh, really thanks. I was stupid to try to use matrix inversion at RHS. Your simple answer is good – Hoil Kim May 08 '17 at 12:43
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Simply expanding the RHS: \begin{align*} (I+B)(I-(I+B)^{-1}B(I-A)) & = (I+B) - (I+B)(I+B)^{-1}B(I-A) \\ & = I+B - B(I-A) \\ & = I+B - BI + BA \\ & = I + B - B + BA \\ & = I + BA. \end{align*} Here, we use the fact that $BI=B$ and $(I+B)(I+B)^{-1} = I$.
Chee Han
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If $A=x$, $B=y$, $I=1$, then:
$$1+yx=(1+y)\left(1-(1+y)^{-1}y(1-x)\right)=(1+y)\left(1-\frac{y-yx}{1+y}\right)=$$ $$(1+y)\cdot\frac{1+y-y+yx}{1+y}=1+yx.$$
farruhota
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