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Verify that $(I_n−XY)^{-1}\cdot X=X\cdot (I_m−YX)^{-1}$

The first $I$ is of order $n$ and the second is of order $m$.

$X$ is $n\times m$

$Y$ is $m\times n$

shadow10
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alan213123
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1 Answers1

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$$(I-XY)^{-1}X=X(I-YX)^{-1} \\ \Rightarrow (I-XY)(I-XY)^{-1}X=(I-XY)X(I-YX)^{-1} \\ \Rightarrow X=(I-XY)X(I-YX)^{-1} \\ \Rightarrow X(I-YX)=(I-XY)X(I-YX)^{-1}(I-YX) \\ \Rightarrow X(I-YX)=(I-XY)X \\ \Rightarrow X-XYX=X-XYX $$

which is true.

Therefore, the relation $(I-XY)^{-1}X=X(I-YX)^{-1}$ stands!!

Mary Star
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    Thank you! Is there any tactical approach you take to solving a problem like this, or is it mostly intuition/experience? – alan213123 Sep 20 '14 at 16:03
  • One method to verify this relation is to suppose that the relation is true and then you have to conclude to something that you are sure that it's true. In this case we concluded to a relation in the form $a=a$ which is obviously true. Since at the relation, we wanted to verify, there is the inverse of an element we multiply both sides by the element. You have to pay attention that you multiply both sides at the left or both sides at the right sides of the equation. – Mary Star Sep 20 '14 at 16:37