2
$\begingroup$

Consider a matrix that satisfies

$$ \bf M = {\begin{bmatrix} \bf A ^\top & \bf B ^\top \end{bmatrix}} {\begin{bmatrix} \bf A \bf A ^\top + \bf C \bf C ^\top & \bf A \bf B ^\top \\ \bf B \bf A ^\top & \bf B \bf B ^\top \end{bmatrix}}^{-1} {\begin{bmatrix} \bf A \\ \bf B \end{bmatrix}} $$

where $\bf A$, $\bf B$, and $\bf C$ are matrices as well. I am attempting to calculate

$$\lim_{n \to \infty} {\bf{M}}^n$$

Notice that if ${\bf C} = {\bf 0}$ this would be trivial, since then ${\bf{M}}^n={\bf{M}}$ for all $n$. I have tried using the block matrix inversion formula to explicitly invert the middle matrix but I was not able to obtain the result that way. Any suggestion would be much appreciated.

$\endgroup$
0
$\begingroup$

Using the block-matrix inversion formula and simplifying, it follows that

$$ \bf M = \bf E+(E-I)D(E-I), $$

where

\begin{align} \bf D \equiv & \; \bf A^{\top}(A(I-E)A^{\top}+BB^{\top})^{-1}A,\\ \bf E \equiv & \; \bf C^{\top}(CC^{\top})^{-1}C. \end{align}

Notice that $\bf E$ is idempotent. Therefore,

$$ {\bf M}^n = {\bf E+(E-I)(D(E-I))}^n.$$

Hence, for instance, if the eigenvalues of $\bf D(E-I)$ all have modulus less than $1$, then $\lim_{n\to \infty}{\bf M}^n=\bf E$. In general

$$ \lim_{n\to \infty}{\bf M}^n = {\bf E +(E-I)}\lim_{n\to \infty}{\bf (D(E-I))}^n.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.