Inverse of a circulant matrix with a specific pattern

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I'm trying to invert the following circulant matrix:

$$beginbmatrix1 & -1/4 & 0&0 &0&cdots&0&-1/4\ -1/4 & 1 & -1/4 & 0&0&cdots&0&0\0 & -1/4 & 1 & -1/4&0&cdots&0&0\vdots&vdots&vdots&vdots&vdots&ddots&vdots&vdots
\-1/4 & 0 & 0 & 0 & 0&cdots&-1/4 & 1 endbmatrix$$

As it turns out, Fuyong (2011) "The inverse of circulant matrix" proposes the following method:

1-) Find the roots of the polynomials $g(z)=g^prime(z)=1 - frac14z- frac14z^-1$ that are inside the unit circle:

$z_1 = z_1^prime= 2-sqrt3$ and $z_2 = z_2^prime= 2+sqrt3$

Only $z_1$ and $z_1^prime$ are inside the unit circle.

2-) Compute $g_1(z_1)=-frac14z_1^-1(z_1-z_2)$ and $g_1^prime(z^prime_1)=-frac14z_1^prime-1(z_1^prime-z_2^prime)$:

$g_1(z_1)=g_1^prime(z^prime_1)= fracsqrt32(2-sqrt3)$

3-) The elements of the inverse are given by:

$b_k= dfracz_1^k_1g_1(z_1)(1-z_1^n) +
dfracz_1^prime k_2z_1^prime sg_1^prime(z^prime_1)(1-z_1^prime n) $ , $k=0,ldots,n-1$

where:

$k_1 = mathrmmod,e (k-1,n) $

$k_2 = mathrmmod,e (n-k-1+s,n) $

I don't know how $s$ is determined and what the operator $mathrmmod,e$ does. Can anyone help?

edited Jul 19 at 5:28

Rodrigo de Azevedo

12.6k41751

asked Jul 18 at 10:10

capadocia

204

Do you just want to invert a circulant matrix or you want to solve matrix equation of the form $Cx=b$, where $C$ is your circulant matrix? For the latter, you can use the FFT to do it very fast.
â€“Â pedroszattoni
Jul 22 at 15:52

add a commentÂ |Â

up vote
1
down vote

favorite

I'm trying to invert the following circulant matrix:

As it turns out, Fuyong (2011) "The inverse of circulant matrix" proposes the following method:

1-) Find the roots of the polynomials $g(z)=g^prime(z)=1 - frac14z- frac14z^-1$ that are inside the unit circle:

$z_1 = z_1^prime= 2-sqrt3$ and $z_2 = z_2^prime= 2+sqrt3$

Only $z_1$ and $z_1^prime$ are inside the unit circle.

2-) Compute $g_1(z_1)=-frac14z_1^-1(z_1-z_2)$ and $g_1^prime(z^prime_1)=-frac14z_1^prime-1(z_1^prime-z_2^prime)$:

$g_1(z_1)=g_1^prime(z^prime_1)= fracsqrt32(2-sqrt3)$

3-) The elements of the inverse are given by:

$b_k= dfracz_1^k_1g_1(z_1)(1-z_1^n) +
dfracz_1^prime k_2z_1^prime sg_1^prime(z^prime_1)(1-z_1^prime n) $ , $k=0,ldots,n-1$

where:

$k_1 = mathrmmod,e (k-1,n) $

$k_2 = mathrmmod,e (n-k-1+s,n) $

I don't know how $s$ is determined and what the operator $mathrmmod,e$ does. Can anyone help?

edited Jul 19 at 5:28

Rodrigo de Azevedo

12.6k41751

asked Jul 18 at 10:10

capadocia

204

Do you just want to invert a circulant matrix or you want to solve matrix equation of the form $Cx=b$, where $C$ is your circulant matrix? For the latter, you can use the FFT to do it very fast.
â€“Â pedroszattoni
Jul 22 at 15:52

add a commentÂ |Â

up vote
1
down vote

favorite

I'm trying to invert the following circulant matrix:

As it turns out, Fuyong (2011) "The inverse of circulant matrix" proposes the following method:

1-) Find the roots of the polynomials $g(z)=g^prime(z)=1 - frac14z- frac14z^-1$ that are inside the unit circle:

$z_1 = z_1^prime= 2-sqrt3$ and $z_2 = z_2^prime= 2+sqrt3$

Only $z_1$ and $z_1^prime$ are inside the unit circle.

2-) Compute $g_1(z_1)=-frac14z_1^-1(z_1-z_2)$ and $g_1^prime(z^prime_1)=-frac14z_1^prime-1(z_1^prime-z_2^prime)$:

$g_1(z_1)=g_1^prime(z^prime_1)= fracsqrt32(2-sqrt3)$

3-) The elements of the inverse are given by:

$b_k= dfracz_1^k_1g_1(z_1)(1-z_1^n) +
dfracz_1^prime k_2z_1^prime sg_1^prime(z^prime_1)(1-z_1^prime n) $ , $k=0,ldots,n-1$

where:

$k_1 = mathrmmod,e (k-1,n) $

$k_2 = mathrmmod,e (n-k-1+s,n) $

I don't know how $s$ is determined and what the operator $mathrmmod,e$ does. Can anyone help?

edited Jul 19 at 5:28

Rodrigo de Azevedo

12.6k41751

asked Jul 18 at 10:10

capadocia

204

I'm trying to invert the following circulant matrix:

As it turns out, Fuyong (2011) "The inverse of circulant matrix" proposes the following method:

1-) Find the roots of the polynomials $g(z)=g^prime(z)=1 - frac14z- frac14z^-1$ that are inside the unit circle:

$z_1 = z_1^prime= 2-sqrt3$ and $z_2 = z_2^prime= 2+sqrt3$

Only $z_1$ and $z_1^prime$ are inside the unit circle.

2-) Compute $g_1(z_1)=-frac14z_1^-1(z_1-z_2)$ and $g_1^prime(z^prime_1)=-frac14z_1^prime-1(z_1^prime-z_2^prime)$:

$g_1(z_1)=g_1^prime(z^prime_1)= fracsqrt32(2-sqrt3)$

3-) The elements of the inverse are given by:

$b_k= dfracz_1^k_1g_1(z_1)(1-z_1^n) +
dfracz_1^prime k_2z_1^prime sg_1^prime(z^prime_1)(1-z_1^prime n) $ , $k=0,ldots,n-1$

where:

$k_1 = mathrmmod,e (k-1,n) $

$k_2 = mathrmmod,e (n-k-1+s,n) $

I don't know how $s$ is determined and what the operator $mathrmmod,e$ does. Can anyone help?

edited Jul 19 at 5:28

Rodrigo de Azevedo

12.6k41751

asked Jul 18 at 10:10

capadocia

204

edited Jul 19 at 5:28

Rodrigo de Azevedo

12.6k41751

edited Jul 19 at 5:28

Rodrigo de Azevedo

12.6k41751

edited Jul 19 at 5:28

Rodrigo de Azevedo

12.6k41751

asked Jul 18 at 10:10

capadocia

204

asked Jul 18 at 10:10

capadocia

204

asked Jul 18 at 10:10

capadocia

204

Do you just want to invert a circulant matrix or you want to solve matrix equation of the form $Cx=b$, where $C$ is your circulant matrix? For the latter, you can use the FFT to do it very fast.
â€“Â pedroszattoni
Jul 22 at 15:52

add a commentÂ |Â

Do you just want to invert a circulant matrix or you want to solve matrix equation of the form $Cx=b$, where $C$ is your circulant matrix? For the latter, you can use the FFT to do it very fast.
â€“Â pedroszattoni
Jul 22 at 15:52

Do you just want to invert a circulant matrix or you want to solve matrix equation of the form $Cx=b$, where $C$ is your circulant matrix? For the latter, you can use the FFT to do it very fast.
â€“Â pedroszattoni
Jul 22 at 15:52

add a commentÂ |Â

1 Answer
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The polynomial $g$ is defined as $g(z) = a_0 + a_1 z + cdots + a_s x^s$. Thus, $s=-1$.

Let $B=A^-1$:

beginalign*
&B= beginbmatrixb_0 & b_1 & b_2&b_3 &b_4&cdots &b_n-2&b_n-1\ b_n-1 & b_0 & b_1 & b_2&b_3&cdots & b_n-3&b_n-2\b_n-2 & b_n-1 & b_0 & b_1&b_2&cdots&b_n-4&b_n-3\vdots&vdots&vdots&vdots&vdots&ddots&vdots&vdots
\b_1 & b_2 & b_3 & b_4 & b_5 &cdots&b_n-1 & b_0 endbmatrix&
endalign*

beginalign*
b_k = dfrac2bigl(2-sqrt3bigr)^ksqrt3bigl[1-bigl(2-sqrt3bigr)^nbigr] + dfrac2bigl(2-sqrt3bigr)^n-ksqrt3bigl[1-bigl(2-sqrt3bigr)^nbigr]qquad k = 0,ldots, n-1
endalign*

answered Jul 23 at 18:10

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1 Answer
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active

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1 Answer
1

active

oldest

votes

up vote
0
down vote

accepted

The polynomial $g$ is defined as $g(z) = a_0 + a_1 z + cdots + a_s x^s$. Thus, $s=-1$.

Let $B=A^-1$:

beginalign*
b_k = dfrac2bigl(2-sqrt3bigr)^ksqrt3bigl[1-bigl(2-sqrt3bigr)^nbigr] + dfrac2bigl(2-sqrt3bigr)^n-ksqrt3bigl[1-bigl(2-sqrt3bigr)^nbigr]qquad k = 0,ldots, n-1
endalign*

answered Jul 23 at 18:10

capadocia

204

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up vote
0
down vote

accepted

The polynomial $g$ is defined as $g(z) = a_0 + a_1 z + cdots + a_s x^s$. Thus, $s=-1$.

Let $B=A^-1$:

beginalign*
b_k = dfrac2bigl(2-sqrt3bigr)^ksqrt3bigl[1-bigl(2-sqrt3bigr)^nbigr] + dfrac2bigl(2-sqrt3bigr)^n-ksqrt3bigl[1-bigl(2-sqrt3bigr)^nbigr]qquad k = 0,ldots, n-1
endalign*

answered Jul 23 at 18:10

capadocia

204

add a commentÂ |Â

up vote
0
down vote

accepted

The polynomial $g$ is defined as $g(z) = a_0 + a_1 z + cdots + a_s x^s$. Thus, $s=-1$.

Let $B=A^-1$:

beginalign*
b_k = dfrac2bigl(2-sqrt3bigr)^ksqrt3bigl[1-bigl(2-sqrt3bigr)^nbigr] + dfrac2bigl(2-sqrt3bigr)^n-ksqrt3bigl[1-bigl(2-sqrt3bigr)^nbigr]qquad k = 0,ldots, n-1
endalign*

answered Jul 23 at 18:10

capadocia

204

The polynomial $g$ is defined as $g(z) = a_0 + a_1 z + cdots + a_s x^s$. Thus, $s=-1$.

Let $B=A^-1$:

beginalign*
b_k = dfrac2bigl(2-sqrt3bigr)^ksqrt3bigl[1-bigl(2-sqrt3bigr)^nbigr] + dfrac2bigl(2-sqrt3bigr)^n-ksqrt3bigl[1-bigl(2-sqrt3bigr)^nbigr]qquad k = 0,ldots, n-1
endalign*

answered Jul 23 at 18:10

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answered Jul 23 at 18:10

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answered Jul 23 at 18:10

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answered Jul 23 at 18:10

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