Recurrence relations in Matrices raised to natural powers
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$begingroup$
Let some matrix be $A$, what methods are there for attaining a recurrence relation which allows us to calculate - for example - the leading term in the matrix $A^n$ when given the leading terms of a sufficient amount of previous matrices $A^{n-1}, A^{n-2},...$
I ask this because I was studying the leading term in the matrix $A^n$ where $$A=
begin{bmatrix}
1 & 2 & 3 \
4 & 5 & 6 \
7 & 8 & 9 \
end{bmatrix}
$$
which can be found in a ridiculous closed form (https://oeis.org/A321045), but has the following recurrence relation between the nth powers of the matrix:
$$a(n+2)=15a(n+1)+18a(n)$$
for $n ge 1$. How does one find a relation like this or equivalently find it's corresponding generating function - $frac{3x^2+14x-1}{18x^2+15x-1}$ ?
matrices recurrence-relations generating-functions
asked Jan 8 at 23:22
Peter ForemanPeter Foreman
2,243114
$endgroup$
add a comment |
$begingroup$
Let some matrix be $A$, what methods are there for attaining a recurrence relation which allows us to calculate - for example - the leading term in the matrix $A^n$ when given the leading terms of a sufficient amount of previous matrices $A^{n-1}, A^{n-2},...$
I ask this because I was studying the leading term in the matrix $A^n$ where $$A=
begin{bmatrix}
1 & 2 & 3 \
4 & 5 & 6 \
7 & 8 & 9 \
end{bmatrix}
$$
which can be found in a ridiculous closed form (https://oeis.org/A321045), but has the following recurrence relation between the nth powers of the matrix:
$$a(n+2)=15a(n+1)+18a(n)$$
for $n ge 1$. How does one find a relation like this or equivalently find it's corresponding generating function - $frac{3x^2+14x-1}{18x^2+15x-1}$ ?
matrices recurrence-relations generating-functions
asked Jan 8 at 23:22
Peter ForemanPeter Foreman
2,243114
$endgroup$
1
$begingroup$
this is just Cayley-Hamilton. In your case, the determinant is zero, so you don't get the full effect, with would be expected to relate $n+3,n+2,n+1,n$
$endgroup$
– Will Jagy
Jan 8 at 23:25
add a comment |
$begingroup$
Let some matrix be $A$, what methods are there for attaining a recurrence relation which allows us to calculate - for example - the leading term in the matrix $A^n$ when given the leading terms of a sufficient amount of previous matrices $A^{n-1}, A^{n-2},...$
I ask this because I was studying the leading term in the matrix $A^n$ where $$A=
begin{bmatrix}
1 & 2 & 3 \
4 & 5 & 6 \
7 & 8 & 9 \
end{bmatrix}
$$
which can be found in a ridiculous closed form (https://oeis.org/A321045), but has the following recurrence relation between the nth powers of the matrix:
$$a(n+2)=15a(n+1)+18a(n)$$
for $n ge 1$. How does one find a relation like this or equivalently find it's corresponding generating function - $frac{3x^2+14x-1}{18x^2+15x-1}$ ?
matrices recurrence-relations generating-functions
asked Jan 8 at 23:22
Peter ForemanPeter Foreman
2,243114
$endgroup$
Let some matrix be $A$, what methods are there for attaining a recurrence relation which allows us to calculate - for example - the leading term in the matrix $A^n$ when given the leading terms of a sufficient amount of previous matrices $A^{n-1}, A^{n-2},...$
I ask this because I was studying the leading term in the matrix $A^n$ where $$A=
begin{bmatrix}
1 & 2 & 3 \
4 & 5 & 6 \
7 & 8 & 9 \
end{bmatrix}
$$
which can be found in a ridiculous closed form (https://oeis.org/A321045), but has the following recurrence relation between the nth powers of the matrix:
$$a(n+2)=15a(n+1)+18a(n)$$
for $n ge 1$. How does one find a relation like this or equivalently find it's corresponding generating function - $frac{3x^2+14x-1}{18x^2+15x-1}$ ?
matrices recurrence-relations generating-functions
matrices recurrence-relations generating-functions
asked Jan 8 at 23:22
Peter ForemanPeter Foreman
2,243114
asked Jan 8 at 23:22
Peter ForemanPeter Foreman
2,243114
asked Jan 8 at 23:22
Peter ForemanPeter Foreman
2,243114
asked Jan 8 at 23:22
Peter ForemanPeter Foreman
2,243114
asked Jan 8 at 23:22
Peter ForemanPeter Foreman
2,243114
2,243114
1
$begingroup$
this is just Cayley-Hamilton. In your case, the determinant is zero, so you don't get the full effect, with would be expected to relate $n+3,n+2,n+1,n$
$endgroup$
– Will Jagy
Jan 8 at 23:25
add a comment |
1
$begingroup$
this is just Cayley-Hamilton. In your case, the determinant is zero, so you don't get the full effect, with would be expected to relate $n+3,n+2,n+1,n$
$endgroup$
– Will Jagy
Jan 8 at 23:25
1
1
$begingroup$
this is just Cayley-Hamilton. In your case, the determinant is zero, so you don't get the full effect, with would be expected to relate $n+3,n+2,n+1,n$
$endgroup$
– Will Jagy
Jan 8 at 23:25
$begingroup$
this is just Cayley-Hamilton. In your case, the determinant is zero, so you don't get the full effect, with would be expected to relate $n+3,n+2,n+1,n$
$endgroup$
– Will Jagy
Jan 8 at 23:25
add a comment |
1 Answer
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First, find some relation between “small” powers of $A$, eg here I guess $A^3=15A^2+18A$ (For this, you can try and test or use Cayley-Hamilton). Then multiply everything by $A^{n-1}$.
answered Jan 8 at 23:27
MindlackMindlack
4,740210
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add a comment |
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$begingroup$
First, find some relation between “small” powers of $A$, eg here I guess $A^3=15A^2+18A$ (For this, you can try and test or use Cayley-Hamilton). Then multiply everything by $A^{n-1}$.
answered Jan 8 at 23:27
MindlackMindlack
4,740210
$endgroup$
add a comment |
$begingroup$
First, find some relation between “small” powers of $A$, eg here I guess $A^3=15A^2+18A$ (For this, you can try and test or use Cayley-Hamilton). Then multiply everything by $A^{n-1}$.
answered Jan 8 at 23:27
MindlackMindlack
4,740210
$endgroup$
add a comment |
$begingroup$
First, find some relation between “small” powers of $A$, eg here I guess $A^3=15A^2+18A$ (For this, you can try and test or use Cayley-Hamilton). Then multiply everything by $A^{n-1}$.
answered Jan 8 at 23:27
MindlackMindlack
4,740210
$endgroup$
First, find some relation between “small” powers of $A$, eg here I guess $A^3=15A^2+18A$ (For this, you can try and test or use Cayley-Hamilton). Then multiply everything by $A^{n-1}$.
answered Jan 8 at 23:27
MindlackMindlack
4,740210
answered Jan 8 at 23:27
MindlackMindlack
4,740210
answered Jan 8 at 23:27
MindlackMindlack
4,740210
answered Jan 8 at 23:27
MindlackMindlack
4,740210
4,740210
add a comment |
add a comment |
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1
$begingroup$
this is just Cayley-Hamilton. In your case, the determinant is zero, so you don't get the full effect, with would be expected to relate $n+3,n+2,n+1,n$
$endgroup$
– Will Jagy
Jan 8 at 23:25