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I am reading a book on control theory for multiple input multiple output (MIMO) systems. The author claims that for the identity matrix $I$ and the complex transfer function matrices $G_1$ and $G_2$ (let us assume they are square matrices) the following identity holds: $$(I-G_1G_2)^{-1}G_1=G_1(I-G_2G_1)^{-1}.$$

Is this statement true? And how can I derive it?

MrYouMath
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You need, say, $I - G_{2} G_{1}$ to be invertible, and then you can prove that $I - G_{1} G_{2}$ is invertible, and that the identity holds.

In fact, under these hypotheses, you have $$\tag{eq} (I - G_{1} G_{2}) G_{1} (I - G_{2} G_{1})^{-1} = (G_{1} - G_{1} G_{2} G_{1}) (I - G_{2} G_{1})^{-1} = G_{1} (I - G_{2} G_{1}) (I - G_{2} G_{1})^{-1} = G_{1}. $$ In particular $$ G_{1} G_{2} = (I - G_{1} G_{2}) X $$ for some $X$, so that $$ I = G_{1}G_{2} + (I - G_{1} G_{2}) = (I - G_{1} G_{2}) X + (I - G_{1} G_{2}) = (I - G_{1} G_{2}) (X + I), $$ and $I - G_{1} G_{2}$ is invertible. Now multiply by $(I - G_{1} G_{2})^{-1}$ on the left in (eq).

Andreas Caranti
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Yes it is true. It can be derived from $$ G_1(I-G_2G_1) = G_1-G_1G_2G_1 = (I-G_1G_2)G_1 $$ Multiply $(I-G_1G_2)^{-1}$ from the left and $(I-G_2G_1)^{-1}$ from the right and you will get the original identity.

Reinhard Meier
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  • +1 Thank you for the quick and easy answer. Will accept the answer to be correct in 4 minutes :D. – MrYouMath Feb 09 '17 at 08:45
  • There is a caveat: you don't know if $I-G_1G_2$ and $I-G_2G_1$ are simultaneously invertible ! See the answer of Andreas Caranti and (http://math.stackexchange.com/q/1398275). – Jean Marie Feb 09 '17 at 09:12
  • @JeanMarie I would say, this is subject to interpretation. You could argue that the usage of $(I-G_1G_2)^{-1}$ and $(I-G_2G_1)^{-1}$ in the question implies their existence. – Reinhard Meier Feb 09 '17 at 09:19
  • I agree that, in the way MrYouMath has asked his question, there is this implied assumption... – Jean Marie Feb 09 '17 at 10:07