Matrix Identity $(I-G_1G_2)^{-1}G_1=G_1(I-G_2G_1)^{-1}$?

Question

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?

Answer 1

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.

Answer 2

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).