Solving Linear Matrix Recurrences












4












$begingroup$



Question:



Are there standard techniques available for solving the following kind of linear matrix recurrence relations:



$$M_1,cdots,M_k in mathbb{R}^{mtimes n}$$
$$ A_1,cdots,A_k in mathbb{R}^{ntimes n}$$
$$M_{i+k+1} = sum_{j=1}^{k}{M_{i+j}A_j} $$




I need to solve the special case of $k=2$ that appears in the iterative modification of edge weights of complete symmetric graphs.






linear-algebra recurrences







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  • $begingroup$
    Did you mean $n=m=2$?
    $endgroup$
    – FusRoDah
    15 hours ago










  • $begingroup$
    @FusRoDah in the case of symmetric graphs I have $k=2, n=mgt 2$ as the number of vertices can be an arbitrary fixed number with at least 3 edges to rule out the trivial cases.
    $endgroup$
    – Manfred Weis
    15 hours ago


















4












$begingroup$



Question:



Are there standard techniques available for solving the following kind of linear matrix recurrence relations:



$$M_1,cdots,M_k in mathbb{R}^{mtimes n}$$
$$ A_1,cdots,A_k in mathbb{R}^{ntimes n}$$
$$M_{i+k+1} = sum_{j=1}^{k}{M_{i+j}A_j} $$




I need to solve the special case of $k=2$ that appears in the iterative modification of edge weights of complete symmetric graphs.






linear-algebra recurrences







share|cite|improve this question









$endgroup$












  • $begingroup$
    Did you mean $n=m=2$?
    $endgroup$
    – FusRoDah
    15 hours ago










  • $begingroup$
    @FusRoDah in the case of symmetric graphs I have $k=2, n=mgt 2$ as the number of vertices can be an arbitrary fixed number with at least 3 edges to rule out the trivial cases.
    $endgroup$
    – Manfred Weis
    15 hours ago
















4












4








4


1



$begingroup$



Question:



Are there standard techniques available for solving the following kind of linear matrix recurrence relations:



$$M_1,cdots,M_k in mathbb{R}^{mtimes n}$$
$$ A_1,cdots,A_k in mathbb{R}^{ntimes n}$$
$$M_{i+k+1} = sum_{j=1}^{k}{M_{i+j}A_j} $$




I need to solve the special case of $k=2$ that appears in the iterative modification of edge weights of complete symmetric graphs.






linear-algebra recurrences







share|cite|improve this question









$endgroup$





Question:



Are there standard techniques available for solving the following kind of linear matrix recurrence relations:



$$M_1,cdots,M_k in mathbb{R}^{mtimes n}$$
$$ A_1,cdots,A_k in mathbb{R}^{ntimes n}$$
$$M_{i+k+1} = sum_{j=1}^{k}{M_{i+j}A_j} $$




I need to solve the special case of $k=2$ that appears in the iterative modification of edge weights of complete symmetric graphs.






linear-algebra recurrences




linear-algebra recurrences






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked 15 hours ago









Manfred WeisManfred Weis

4,62421542




4,62421542












  • $begingroup$
    Did you mean $n=m=2$?
    $endgroup$
    – FusRoDah
    15 hours ago










  • $begingroup$
    @FusRoDah in the case of symmetric graphs I have $k=2, n=mgt 2$ as the number of vertices can be an arbitrary fixed number with at least 3 edges to rule out the trivial cases.
    $endgroup$
    – Manfred Weis
    15 hours ago




















  • $begingroup$
    Did you mean $n=m=2$?
    $endgroup$
    – FusRoDah
    15 hours ago










  • $begingroup$
    @FusRoDah in the case of symmetric graphs I have $k=2, n=mgt 2$ as the number of vertices can be an arbitrary fixed number with at least 3 edges to rule out the trivial cases.
    $endgroup$
    – Manfred Weis
    15 hours ago


















$begingroup$
Did you mean $n=m=2$?
$endgroup$
– FusRoDah
15 hours ago




$begingroup$
Did you mean $n=m=2$?
$endgroup$
– FusRoDah
15 hours ago












$begingroup$
@FusRoDah in the case of symmetric graphs I have $k=2, n=mgt 2$ as the number of vertices can be an arbitrary fixed number with at least 3 edges to rule out the trivial cases.
$endgroup$
– Manfred Weis
15 hours ago






$begingroup$
@FusRoDah in the case of symmetric graphs I have $k=2, n=mgt 2$ as the number of vertices can be an arbitrary fixed number with at least 3 edges to rule out the trivial cases.
$endgroup$
– Manfred Weis
15 hours ago












1 Answer
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7












$begingroup$

Same method as for scalar equations works. Put matrices $M_{n+k-1},M_{n+k-2},...,M_{n}$
vertically into a big matrix $X_n$ of size $mktimes n$. Then your recurrence becomes
a one-step recurrence $X_{n+1}=AX_{n}$, with some $A$, whose solution is $M^n=A^nX_0$, and this is solved by diagonalizing $A$ (or using its Jordan form).






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    Same method as for scalar equations works. Put matrices $M_{n+k-1},M_{n+k-2},...,M_{n}$
    vertically into a big matrix $X_n$ of size $mktimes n$. Then your recurrence becomes
    a one-step recurrence $X_{n+1}=AX_{n}$, with some $A$, whose solution is $M^n=A^nX_0$, and this is solved by diagonalizing $A$ (or using its Jordan form).






    share|cite|improve this answer









    $endgroup$


















      7












      $begingroup$

      Same method as for scalar equations works. Put matrices $M_{n+k-1},M_{n+k-2},...,M_{n}$
      vertically into a big matrix $X_n$ of size $mktimes n$. Then your recurrence becomes
      a one-step recurrence $X_{n+1}=AX_{n}$, with some $A$, whose solution is $M^n=A^nX_0$, and this is solved by diagonalizing $A$ (or using its Jordan form).






      share|cite|improve this answer









      $endgroup$
















        7












        7








        7





        $begingroup$

        Same method as for scalar equations works. Put matrices $M_{n+k-1},M_{n+k-2},...,M_{n}$
        vertically into a big matrix $X_n$ of size $mktimes n$. Then your recurrence becomes
        a one-step recurrence $X_{n+1}=AX_{n}$, with some $A$, whose solution is $M^n=A^nX_0$, and this is solved by diagonalizing $A$ (or using its Jordan form).






        share|cite|improve this answer









        $endgroup$



        Same method as for scalar equations works. Put matrices $M_{n+k-1},M_{n+k-2},...,M_{n}$
        vertically into a big matrix $X_n$ of size $mktimes n$. Then your recurrence becomes
        a one-step recurrence $X_{n+1}=AX_{n}$, with some $A$, whose solution is $M^n=A^nX_0$, and this is solved by diagonalizing $A$ (or using its Jordan form).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 13 hours ago









        Alexandre EremenkoAlexandre Eremenko

        50.6k6140257




        50.6k6140257






























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