# A truncated "geometric" matrix series

Let $$A$$ be an $$n\times n$$ matrix, $$B$$ be an $$n\times m$$ matrix, $$C$$ an $$m \times m$$ matrix, and consider the sum $$\sum_{k = 0}^{N-1} A^k B C^k.$$ Is there any smart way to rewrite this sum in a way similar to the partial sums of the geometric series; namely for $$a,b \in \mathbb{R},$$ $$\sum_{k = 0}^{N-1} b a^k = b \frac{1-a^N}{1-a}?$$

• take the simple case $B=I$, $n=3$, what "simplication" of $I+AC+A^2C^2$ would you have in mind? Aug 18, 2022 at 14:17
• No, but you can solve the equation $\sum_{k = 0}^{N-1} A^k X C^k = E$ on a computer, for given $A, C, E$, in $O(m^3+n^3)$ operations. Would you be interested in that? Aug 18, 2022 at 18:20

This is a long comment. Write $$S$$ for this sum. If we just directly imitate the usual geometric series argument we are led to consider

$$ASC = \sum_{k=1}^N A^k B C^k = S - B + A^N B C^N$$

so $$ASC - S = A^N B C^N - B$$. And here we're a bit stuck: $$ASC - S$$ just isn't a multiplication of $$S$$ by any matrix on either the left or the right so we can't hope to perform a simple matrix division here. We just need to talk about the inverse, if it exists, of the linear map $$S \mapsto ASC - S$$. But at least this is a system of linear equations whose solution, which we can hope is unique, gives the coefficients of $$S$$ so that's not too bad.

• Thanks. Indeed I tried a similar idea but I don't need a specific solution, I need something some regularity in the dependency on $N$ because then afterwards I have to plug this into other series. Aug 19, 2022 at 18:45

With meager loss of generality, let's first assume that $$A,C$$ are diagonalizable so that $$A=RDR^{-1},C=SES^{-1}$$ where $$R,S$$ are invertible and $$D,E$$ are diagonal.

Then $$\sum_{k=0}^{N-1}A^kBC^k=\sum_{k=0}^{N-1}RD^kR^{-1}BSE^kS^{-1}$$

$$=R(\sum_{k=0}^{N-1}D^kR^{-1}BSE^k)S^{-1}.$$

Suppose now that $$R^{-1}BS=(\alpha_{i,j})_{i,j}$$ and that the diagonal entries in $$D$$ are $$(\delta_1,\dots,\delta_n)$$ and the diagonal entries in $$E$$ are $$(\epsilon_1,\dots,\epsilon_m)$$.

Then $$\sum_{k=0}^{N-1}A^kBC^k=R(\sum_{k=0}^{N-1}\delta_i^k\alpha_{i,j}\epsilon_j^k)_{i,j}S^{-1}$$

$$=R\cdot\Big(\frac{(\delta_i\epsilon_j)^N-1}{\delta_i\epsilon_j-1}\cdot\alpha_{i,j}\Big)_{i,j}\cdot S^{-1}.$$

It looks like there is a not quite as nice expression for our sum when $$A,C$$ are not diagonalizable, and $$D,E$$ are in Jordan normal form.

Generalization:

Suppose that $$f(x,y)$$ is a polynomial (or an analytic function when we assume convergence), and $$f(x,y)=\sum_{k,l}b_{k,l}x^ky^l$$. Then by using the same reasoning and matrices as before, we have $$\sum_{k,l}b_{k,l}A^kBC^l=R\cdot(f(\delta_i,\epsilon_j)\cdot\alpha_{i,j})_{i,j}\cdot S^{-1}.$$

• Indeed $A$ and $C$ are diagonalizable. I hoped to be able to find a basis-independent formula, but this sounds like a very interesting way to proceed. Aug 19, 2022 at 18:51