Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|improve this question
    
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 '10 at 11:24
1  
If $T(P)=AP+PQ$, it's $T^{k-1}(P)$ –  Homology Jun 15 '10 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 '10 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 '10 at 12:49
3  
@homology isn't $T^2(P)=A^2P+2APQ+PQ^2\neq S_3(P)$ –  HenrikRüping Jun 15 '10 at 13:30
show 2 more comments

2 Answers

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.

share|improve this answer
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 '10 at 21:50
add comment

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.

share|improve this answer
    
In fact I just need to decide if, given $A$, $P$, $Q$ ($A$ and $Q$ are invertible) and $b$, it exists any $k$ such that the following matrix equation has solution $$x \left(\sum_{i=1}^{k} A^{i-1} P Q^{k-i}\right)=b$$ For example, if I could reduce the sum to a matrix power (in terms of $k$) it would probably be sufficient. –  Jorge Jun 16 '10 at 10:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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