1

Suppose $R_M$ is the resolvent of matrix $M$ defined as $R_M=(sI-M)^{-1}$. How do I show the following?

$$R_{A+B}=(I-R_AB)^{-1}R_A$$

Is this a well known result?

I got this formula by using grouping technique in this post and checked numerically but wondering if theres an easier way.

Yaroslav Bulatov
  • 5,327

1 Answers1

1

By definition of $R_A$, we have $(sI - A)R_A B = B$. Subtract both sides of this from $sI - A$, and we get \begin{align} sI - A - (sI - A)R_A B &= sI - A - B \\ (sI - A)(I - R_A B) &= sI - A - B \\ (I - R_A B)^{-1} (sI - A)^{-1} &= (sI - A - B)^{-1} \\ (I - R_A B)^{-1} R_A &= R_{A+B} \end{align}

Misha Lavrov
  • 142,276