Solving Linear Matrix Recurrences
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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
asked Mar 10 at 10:50
Manfred WeisManfred Weis
4,63121542
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add a comment |
$begingroup$
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
asked Mar 10 at 10:50
Manfred WeisManfred Weis
4,63121542
$endgroup$
$begingroup$
Did you mean $n=m=2$?
$endgroup$
– FusRoDah
Mar 10 at 10:52
$begingroup$
@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
Mar 10 at 10:57
add a comment |
$begingroup$
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
asked Mar 10 at 10:50
Manfred WeisManfred Weis
4,63121542
$endgroup$
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
asked Mar 10 at 10:50
Manfred WeisManfred Weis
4,63121542
asked Mar 10 at 10:50
Manfred WeisManfred Weis
4,63121542
asked Mar 10 at 10:50
Manfred WeisManfred Weis
4,63121542
asked Mar 10 at 10:50
Manfred WeisManfred Weis
4,63121542
asked Mar 10 at 10:50
Manfred WeisManfred Weis
4,63121542
4,63121542
$begingroup$
Did you mean $n=m=2$?
$endgroup$
– FusRoDah
Mar 10 at 10:52
$begingroup$
@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
Mar 10 at 10:57
add a comment |
$begingroup$
Did you mean $n=m=2$?
$endgroup$
– FusRoDah
Mar 10 at 10:52
$begingroup$
@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
Mar 10 at 10:57
$begingroup$
Did you mean $n=m=2$?
$endgroup$
– FusRoDah
Mar 10 at 10:52
$begingroup$
Did you mean $n=m=2$?
$endgroup$
– FusRoDah
Mar 10 at 10:52
$begingroup$
@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
Mar 10 at 10:57
$begingroup$
@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
Mar 10 at 10:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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).
answered Mar 10 at 12:18
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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).
answered Mar 10 at 12:18
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
$endgroup$
add a comment |
$begingroup$
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).
answered Mar 10 at 12:18
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
$endgroup$
add a comment |
$begingroup$
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).
answered Mar 10 at 12:18
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
$endgroup$
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).
answered Mar 10 at 12:18
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
answered Mar 10 at 12:18
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
answered Mar 10 at 12:18
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
answered Mar 10 at 12:18
Alexandre EremenkoAlexandre Eremenko
50.7k6140258
50.7k6140258
add a comment |
add a comment |
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$begingroup$
Did you mean $n=m=2$?
$endgroup$
– FusRoDah
Mar 10 at 10:52
$begingroup$
@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
Mar 10 at 10:57