Powers of Bidiagonal Toeplitz Matrix
Consider the following bidiagonal $n times n$ Toeplitz matrix $A$, where $0 < p < 1$:
$$A=begin{bmatrix}
1-p & 0 & 0 & cdots & 0\
p & 1-p & 0 && vdots \
0 & ddots & ddots & ddots & 0 \
vdots && p & 1-p & 0\
0 & cdots & 0 & p & 1-p
end{bmatrix}$$
What is $A^m$ for any $m ge 2$? It's easy to show what the matrix is when $n = 2$ for all $m$, but not for general $n$. I've seen several papers on powers of tridiagonal Toeplitz matrices but they assume that the off-by-1 diagonals are all nonzero, but the "upper" diagonal here is all 0s.
matrices
asked Jan 3 at 14:37
RyanRyan
3811219
add a comment |
Consider the following bidiagonal $n times n$ Toeplitz matrix $A$, where $0 < p < 1$:
$$A=begin{bmatrix}
1-p & 0 & 0 & cdots & 0\
p & 1-p & 0 && vdots \
0 & ddots & ddots & ddots & 0 \
vdots && p & 1-p & 0\
0 & cdots & 0 & p & 1-p
end{bmatrix}$$
What is $A^m$ for any $m ge 2$? It's easy to show what the matrix is when $n = 2$ for all $m$, but not for general $n$. I've seen several papers on powers of tridiagonal Toeplitz matrices but they assume that the off-by-1 diagonals are all nonzero, but the "upper" diagonal here is all 0s.
matrices
asked Jan 3 at 14:37
RyanRyan
3811219
add a comment |
Consider the following bidiagonal $n times n$ Toeplitz matrix $A$, where $0 < p < 1$:
$$A=begin{bmatrix}
1-p & 0 & 0 & cdots & 0\
p & 1-p & 0 && vdots \
0 & ddots & ddots & ddots & 0 \
vdots && p & 1-p & 0\
0 & cdots & 0 & p & 1-p
end{bmatrix}$$
What is $A^m$ for any $m ge 2$? It's easy to show what the matrix is when $n = 2$ for all $m$, but not for general $n$. I've seen several papers on powers of tridiagonal Toeplitz matrices but they assume that the off-by-1 diagonals are all nonzero, but the "upper" diagonal here is all 0s.
matrices
asked Jan 3 at 14:37
RyanRyan
3811219
Consider the following bidiagonal $n times n$ Toeplitz matrix $A$, where $0 < p < 1$:
$$A=begin{bmatrix}
1-p & 0 & 0 & cdots & 0\
p & 1-p & 0 && vdots \
0 & ddots & ddots & ddots & 0 \
vdots && p & 1-p & 0\
0 & cdots & 0 & p & 1-p
end{bmatrix}$$
What is $A^m$ for any $m ge 2$? It's easy to show what the matrix is when $n = 2$ for all $m$, but not for general $n$. I've seen several papers on powers of tridiagonal Toeplitz matrices but they assume that the off-by-1 diagonals are all nonzero, but the "upper" diagonal here is all 0s.
matrices
matrices
asked Jan 3 at 14:37
RyanRyan
3811219
asked Jan 3 at 14:37
RyanRyan
3811219
asked Jan 3 at 14:37
RyanRyan
3811219
asked Jan 3 at 14:37
RyanRyan
3811219
asked Jan 3 at 14:37
RyanRyan
3811219
3811219
add a comment |
add a comment |
2 Answers
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You can rewrite $A$ as
$$A = (1-p)I + p D $$
and note that $D^k$ corresponds to non-null elements on the $k^{th}$ sub-diagonal.
Then, noting that multiplication is commutative for $I$ and the $D^k$ matrices:
$$A^m = sum_k {m choose k} (1-p)^k p^{m-k} D^{m-k} = sum_k {m choose k} (1-p)^{m-k} p^{k} D^{k}$$
Note: $D^k = 0$ for $k ge n$
answered Jan 3 at 15:03
DamienDamien
52714
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You can use the spectral symbol that generates $A_n$, to get the symbol that generates $A_n^m$
Symbol that generates $A_n$: $f(theta)=1-p+pe^{mathbf{i}theta}$
Symbol that generates $A_n^m$: $(f(theta))^m$
For example $m=3$ we have
$(1-p+pe^{mathbf{i}theta})^3=-(p - 1)^3+3p(p - 1)^2e^{mathbf{i}theta}-3p^2(p - 1)e^{2mathbf{i}theta}+p^3e^{3mathbf{i}theta}$.
Thus you have $-(p - 1)^3$ on the main diagonal, $3p(p - 1)^2$ on first sub diagonal, $-3p^2(p - 1)$ on second sub diagonal, and finally $p^3$ on the third sub diagonal.
answered Jan 5 at 22:20
santasanta
112
add a comment |
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2 Answers
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2 Answers
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active
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You can rewrite $A$ as
$$A = (1-p)I + p D $$
and note that $D^k$ corresponds to non-null elements on the $k^{th}$ sub-diagonal.
Then, noting that multiplication is commutative for $I$ and the $D^k$ matrices:
$$A^m = sum_k {m choose k} (1-p)^k p^{m-k} D^{m-k} = sum_k {m choose k} (1-p)^{m-k} p^{k} D^{k}$$
Note: $D^k = 0$ for $k ge n$
answered Jan 3 at 15:03
DamienDamien
52714
add a comment |
You can rewrite $A$ as
$$A = (1-p)I + p D $$
and note that $D^k$ corresponds to non-null elements on the $k^{th}$ sub-diagonal.
Then, noting that multiplication is commutative for $I$ and the $D^k$ matrices:
$$A^m = sum_k {m choose k} (1-p)^k p^{m-k} D^{m-k} = sum_k {m choose k} (1-p)^{m-k} p^{k} D^{k}$$
Note: $D^k = 0$ for $k ge n$
answered Jan 3 at 15:03
DamienDamien
52714
add a comment |
You can rewrite $A$ as
$$A = (1-p)I + p D $$
and note that $D^k$ corresponds to non-null elements on the $k^{th}$ sub-diagonal.
Then, noting that multiplication is commutative for $I$ and the $D^k$ matrices:
$$A^m = sum_k {m choose k} (1-p)^k p^{m-k} D^{m-k} = sum_k {m choose k} (1-p)^{m-k} p^{k} D^{k}$$
Note: $D^k = 0$ for $k ge n$
answered Jan 3 at 15:03
DamienDamien
52714
You can rewrite $A$ as
$$A = (1-p)I + p D $$
and note that $D^k$ corresponds to non-null elements on the $k^{th}$ sub-diagonal.
Then, noting that multiplication is commutative for $I$ and the $D^k$ matrices:
$$A^m = sum_k {m choose k} (1-p)^k p^{m-k} D^{m-k} = sum_k {m choose k} (1-p)^{m-k} p^{k} D^{k}$$
Note: $D^k = 0$ for $k ge n$
answered Jan 3 at 15:03
DamienDamien
52714
edited Jan 3 at 15:09
answered Jan 3 at 15:03
DamienDamien
52714
answered Jan 3 at 15:03
DamienDamien
52714
answered Jan 3 at 15:03
DamienDamien
52714
52714
add a comment |
add a comment |
You can use the spectral symbol that generates $A_n$, to get the symbol that generates $A_n^m$
Symbol that generates $A_n$: $f(theta)=1-p+pe^{mathbf{i}theta}$
Symbol that generates $A_n^m$: $(f(theta))^m$
For example $m=3$ we have
$(1-p+pe^{mathbf{i}theta})^3=-(p - 1)^3+3p(p - 1)^2e^{mathbf{i}theta}-3p^2(p - 1)e^{2mathbf{i}theta}+p^3e^{3mathbf{i}theta}$.
Thus you have $-(p - 1)^3$ on the main diagonal, $3p(p - 1)^2$ on first sub diagonal, $-3p^2(p - 1)$ on second sub diagonal, and finally $p^3$ on the third sub diagonal.
answered Jan 5 at 22:20
santasanta
112
add a comment |
You can use the spectral symbol that generates $A_n$, to get the symbol that generates $A_n^m$
Symbol that generates $A_n$: $f(theta)=1-p+pe^{mathbf{i}theta}$
Symbol that generates $A_n^m$: $(f(theta))^m$
For example $m=3$ we have
$(1-p+pe^{mathbf{i}theta})^3=-(p - 1)^3+3p(p - 1)^2e^{mathbf{i}theta}-3p^2(p - 1)e^{2mathbf{i}theta}+p^3e^{3mathbf{i}theta}$.
Thus you have $-(p - 1)^3$ on the main diagonal, $3p(p - 1)^2$ on first sub diagonal, $-3p^2(p - 1)$ on second sub diagonal, and finally $p^3$ on the third sub diagonal.
answered Jan 5 at 22:20
santasanta
112
add a comment |
You can use the spectral symbol that generates $A_n$, to get the symbol that generates $A_n^m$
Symbol that generates $A_n$: $f(theta)=1-p+pe^{mathbf{i}theta}$
Symbol that generates $A_n^m$: $(f(theta))^m$
For example $m=3$ we have
$(1-p+pe^{mathbf{i}theta})^3=-(p - 1)^3+3p(p - 1)^2e^{mathbf{i}theta}-3p^2(p - 1)e^{2mathbf{i}theta}+p^3e^{3mathbf{i}theta}$.
Thus you have $-(p - 1)^3$ on the main diagonal, $3p(p - 1)^2$ on first sub diagonal, $-3p^2(p - 1)$ on second sub diagonal, and finally $p^3$ on the third sub diagonal.
answered Jan 5 at 22:20
santasanta
112
You can use the spectral symbol that generates $A_n$, to get the symbol that generates $A_n^m$
Symbol that generates $A_n$: $f(theta)=1-p+pe^{mathbf{i}theta}$
Symbol that generates $A_n^m$: $(f(theta))^m$
For example $m=3$ we have
$(1-p+pe^{mathbf{i}theta})^3=-(p - 1)^3+3p(p - 1)^2e^{mathbf{i}theta}-3p^2(p - 1)e^{2mathbf{i}theta}+p^3e^{3mathbf{i}theta}$.
Thus you have $-(p - 1)^3$ on the main diagonal, $3p(p - 1)^2$ on first sub diagonal, $-3p^2(p - 1)$ on second sub diagonal, and finally $p^3$ on the third sub diagonal.
answered Jan 5 at 22:20
santasanta
112
answered Jan 5 at 22:20
santasanta
112
answered Jan 5 at 22:20
santasanta
112
answered Jan 5 at 22:20
santasanta
112
112
add a comment |
add a comment |
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