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I'm playing a game where I am handed some series in powers of some-handed-to-me matrix $A$. Then I must calculate the sum $\sum_{n=0}^\infty c_n A^n$, where I get to choose the coefficients after seeing the matrix $A$. I must choose positive coefficients, and the series must not diverge. So far, for each $A$ I am handed, assuming $A$ is not the zero matrix, I choose $c_n = \frac{1}{\lvert\lvert 2A \lvert\vert^n}$. Then we have that the sum is just $\sum_{n=0}^\infty \Big(\frac{A}{\lvert\lvert 2A \lvert\vert}\Big)^n = (I-\frac{A}{\lvert\lvert 2A \lvert\vert})^{-1}$. Then the infinite sum becomes an inversion problem for a finite dimensional, invertible matrix, which can be solved in a finite number of steps. If I can solve the series in a finite number of steps, I win, and I have been able to win in this manner for every $A$ my adversary has given me.

In order to make the game more challenging, my adversary will now hand me two matrices $A$ and $B$. I must now sum the following series:$\sum_{n=0}^\infty c_n (A^{\dagger})^n B A^n$. I can still choose the positive coefficients $c_n$ in the series, and I must choose the coefficients so that I can write the value of this sum in a finite number of steps, such as by writing it as a matrix inverse as I did above.

What coefficients can I choose generally so that I can win the modified game?

user196574
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    We can apply the same basic strategy one we note that vectorization lets us identify the map $B \mapsto A^\dagger B A$ with $A^\top \otimes A^\dagger$ ($\otimes$ denotes a Kronecker product). In other words, it suffices to consider series of the form $\sum_n c_n [A^\top \otimes A^\dagger]^n$. – Ben Grossmann May 19 '20 at 05:11
  • @Omnomnomnom Please write those two sentences as an answer and I'll accept it. – user196574 May 19 '20 at 05:54
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    I've rewritten my comment as an answer, as you recommended. For many people, the information that I gave might not be enough of a solution; however, I'm glad to see that this was enough for you to fill in the blanks (or at least to get the sense that you could). – Ben Grossmann May 19 '20 at 06:00

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We can apply the same basic strategy once we note that vectorization lets us identify the map $B \mapsto A^\dagger B A$ with $A^\top \otimes A^\dagger$ ($\otimes$ denotes a Kronecker product). In other words, it suffices to consider series of the form $\sum_n c_n[A^\top \otimes A^\dagger]^n$.

Ben Grossmann
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