Welcome to MathOverflow

Question

1

Is there any known compact expression for the sum $$S_{k} = \sum_{i=1}^{k} A^{i-1} P Q^{k-i}$$ where $A$, $P$ and $Q$ are respectively $m \times m$, $m \times n$ and $n \times n$ matrices?.

You can assume, if needed, that $A$ and $Q$ are invertible.

The trivial relation $$ AS_{k}-S_{k}Q = A^{k}P - P Q^{k}$$ perhaps provides some clues (fo example it is known that if $A$ and $-Q$ have no common eigenvalues then the last equation has unique solution).

flag	asked Jun 15 2010 at 10:27 Jorge 61●2

Would you be satisfied by "no"? Let me take $m=n=2$, $A=[a,0;0,a]$ (a diagonal matrix), $Q=[q,1;0,q]$ (a Jordan cell), and $P$ the identity matrix. I do not see any closed formula for $S_k$ even in this simple case. A standard strategy is to introduce the generating series $f(x)=\sum_{k=1}^\infty S_kx^{k-1}$ but besides a clear but useless functional equation $Af(x)-f(x)Q=(I_n-Ax)^{-1}P-P(I_m-Qx)^{-1}$ nothing special can be said about it. – Wadim Zudilin Jun 15 2010 at 11:24

1

If $T(P)=AP+PQ$, it's $T^{k-1}(P)$ – Homology Jun 15 2010 at 12:15

and the eigenvalues of $T$ are easily deduced from those of $A$ and $Q$ (even the Jordan normal form). – Homology Jun 15 2010 at 12:40

@Homology: you can put your compact expression as answer (although it's not a closed form and doesn't look more helpful than the orginal formula, it's compact). But what do you mean by the eigenvalues of $T$? Do you view it as an operator from the $mn$-dimensional space to itself? – Wadim Zudilin Jun 15 2010 at 12:49

3

@homology isn't $T^2(P)=A^2P+2APQ+PQ^2\neq S_3(P)$ – HenrikRüping Jun 15 2010 at 13:30

show 2 more comments

Ian Morris · Answer 1 · 2010-06-16 14:53:04Z UTC

4

If we define an $(n+m)\times (n+m)$ matrix by $$C:=\left(\begin{array}{cc}A&P\\ 0&Q\end{array}\right),$$ then we have $$C^k=\left(\begin{array}{cc}A^k&\sum_{i=1}^k A^{i-1}PQ^{k-i}\\ 0&Q^k\end{array}\right)=\left(\begin{array}{cc}A^k&S_k\\ 0&Q^k\end{array}\right).$$ I don't think that a simpler expression than this is very likely.

edited Jun 16 2010 at 14:53

answered Jun 16 2010 at 14:32

Ian Morris
1,958●7●16

2

My guess would be that this is exactly the way, the author arrived at his matrix. The other parts of $C^k$ are simple, so he wonders about the complicated one. – Wadim Zudilin Jun 16 2010 at 21:50

Homology · Answer 2 · 2010-06-16 00:32:02Z UTC

As HenrikRüping wrote, my comment is false. Nevertheless, I think the method is interesting (obviously, it isn't mine), although it gives something "explicit", but not "compact". Maybe you could provide us with context? For example, are you interested in the behavior when $k \rightarrow + \infty$ (assuming the field is topological)?

If $XAX^{-1}$ and $YQY^{-1}$ are "nice" (diagonal or Jordan normal form), then make the change of variable (is this English?) $P'=XPY$. Then viewing $S_k$ as a linear function of $P$, $XS_k(P)Y=\sum_{i=1}^k XAX^{-1} P' YQY^{-1}$, so up to a change of base on $M_{m,n}(K)$, the endomorphism $S_k$ of this vector space is given in a nice form (eigenvalues are known). But I'm not sure this is really what you're asking for, and your last comment suggests you already know what I just wrote.

Any known compact expression for

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.