Boundedness of the spectral radius of matrices $A^h$ as $hto 0.$
$begingroup$
I need to know if the following matrix has a bounded spectral radius $rho(A)$, as $hto 0:$
$$A^h=frac{1}{h^2}begin{pmatrix}
h^2-2h-2&2&0&0dots &0\
1&h^2-2&1&0dots &0\
0&1&h^2-2&1dots &0\
dots&dots&dots&dots\
dots&dots& 1&h^2-2&1\
dots&dots&dots& 2&h^2-2h-2
end{pmatrix}$$
where this is $(n+2)times (n+2)$ matrix with $h = dfrac{1}{n+1}.$ If anyone is curious, this came from an attempted finite difference method on the equation:
$$u''+u = f(x),quad 0<x<1$$
$$u'(0) = u(0), u'(1) = u(1).$$
As this is neither circulant, nor Toeplitz matrix, I have not found a nice way to compute its eigenvalues; hence the spectral radius.
This matrix is almost tridiagonal, so when I wrote it as $A^h = B+C$ with $B = frac{1}{h^2}text{tridiag}(1,-2,1)$ and $C$ some really sparse matrix, $rho(B)$ is bounded so I have some reason to hope that $rho(A)$ also stays bounded.
linear-algebra ordinary-differential-equations numerical-methods numerical-linear-algebra
asked Dec 7 '18 at 22:41
dezdichadodezdichado
6,3711929
$endgroup$
add a comment |
$begingroup$
I need to know if the following matrix has a bounded spectral radius $rho(A)$, as $hto 0:$
$$A^h=frac{1}{h^2}begin{pmatrix}
h^2-2h-2&2&0&0dots &0\
1&h^2-2&1&0dots &0\
0&1&h^2-2&1dots &0\
dots&dots&dots&dots\
dots&dots& 1&h^2-2&1\
dots&dots&dots& 2&h^2-2h-2
end{pmatrix}$$
where this is $(n+2)times (n+2)$ matrix with $h = dfrac{1}{n+1}.$ If anyone is curious, this came from an attempted finite difference method on the equation:
$$u''+u = f(x),quad 0<x<1$$
$$u'(0) = u(0), u'(1) = u(1).$$
As this is neither circulant, nor Toeplitz matrix, I have not found a nice way to compute its eigenvalues; hence the spectral radius.
This matrix is almost tridiagonal, so when I wrote it as $A^h = B+C$ with $B = frac{1}{h^2}text{tridiag}(1,-2,1)$ and $C$ some really sparse matrix, $rho(B)$ is bounded so I have some reason to hope that $rho(A)$ also stays bounded.
linear-algebra ordinary-differential-equations numerical-methods numerical-linear-algebra
asked Dec 7 '18 at 22:41
dezdichadodezdichado
6,3711929
$endgroup$
$begingroup$
@user1551I know that, but this situation is a bit tricky. Norms are equivalent up to a constant that's dependent on the size of the matrix, but in my case the matrix is getting bigger as $hto 0.$
$endgroup$
– dezdichado
Dec 8 '18 at 16:21
add a comment |
$begingroup$
I need to know if the following matrix has a bounded spectral radius $rho(A)$, as $hto 0:$
$$A^h=frac{1}{h^2}begin{pmatrix}
h^2-2h-2&2&0&0dots &0\
1&h^2-2&1&0dots &0\
0&1&h^2-2&1dots &0\
dots&dots&dots&dots\
dots&dots& 1&h^2-2&1\
dots&dots&dots& 2&h^2-2h-2
end{pmatrix}$$
where this is $(n+2)times (n+2)$ matrix with $h = dfrac{1}{n+1}.$ If anyone is curious, this came from an attempted finite difference method on the equation:
$$u''+u = f(x),quad 0<x<1$$
$$u'(0) = u(0), u'(1) = u(1).$$
As this is neither circulant, nor Toeplitz matrix, I have not found a nice way to compute its eigenvalues; hence the spectral radius.
This matrix is almost tridiagonal, so when I wrote it as $A^h = B+C$ with $B = frac{1}{h^2}text{tridiag}(1,-2,1)$ and $C$ some really sparse matrix, $rho(B)$ is bounded so I have some reason to hope that $rho(A)$ also stays bounded.
linear-algebra ordinary-differential-equations numerical-methods numerical-linear-algebra
asked Dec 7 '18 at 22:41
dezdichadodezdichado
6,3711929
$endgroup$
I need to know if the following matrix has a bounded spectral radius $rho(A)$, as $hto 0:$
$$A^h=frac{1}{h^2}begin{pmatrix}
h^2-2h-2&2&0&0dots &0\
1&h^2-2&1&0dots &0\
0&1&h^2-2&1dots &0\
dots&dots&dots&dots\
dots&dots& 1&h^2-2&1\
dots&dots&dots& 2&h^2-2h-2
end{pmatrix}$$
where this is $(n+2)times (n+2)$ matrix with $h = dfrac{1}{n+1}.$ If anyone is curious, this came from an attempted finite difference method on the equation:
$$u''+u = f(x),quad 0<x<1$$
$$u'(0) = u(0), u'(1) = u(1).$$
As this is neither circulant, nor Toeplitz matrix, I have not found a nice way to compute its eigenvalues; hence the spectral radius.
This matrix is almost tridiagonal, so when I wrote it as $A^h = B+C$ with $B = frac{1}{h^2}text{tridiag}(1,-2,1)$ and $C$ some really sparse matrix, $rho(B)$ is bounded so I have some reason to hope that $rho(A)$ also stays bounded.
linear-algebra ordinary-differential-equations numerical-methods numerical-linear-algebra
linear-algebra ordinary-differential-equations numerical-methods numerical-linear-algebra
asked Dec 7 '18 at 22:41
dezdichadodezdichado
6,3711929
asked Dec 7 '18 at 22:41
dezdichadodezdichado
6,3711929
edited Dec 8 '18 at 17:38
dezdichado
asked Dec 7 '18 at 22:41
dezdichadodezdichado
6,3711929
asked Dec 7 '18 at 22:41
dezdichadodezdichado
6,3711929
asked Dec 7 '18 at 22:41
dezdichadodezdichado
6,3711929
6,3711929
$begingroup$
@user1551I know that, but this situation is a bit tricky. Norms are equivalent up to a constant that's dependent on the size of the matrix, but in my case the matrix is getting bigger as $hto 0.$
$endgroup$
– dezdichado
Dec 8 '18 at 16:21
add a comment |
$begingroup$
@user1551I know that, but this situation is a bit tricky. Norms are equivalent up to a constant that's dependent on the size of the matrix, but in my case the matrix is getting bigger as $hto 0.$
$endgroup$
– dezdichado
Dec 8 '18 at 16:21
$begingroup$
@user1551I know that, but this situation is a bit tricky. Norms are equivalent up to a constant that's dependent on the size of the matrix, but in my case the matrix is getting bigger as $hto 0.$
$endgroup$
– dezdichado
Dec 8 '18 at 16:21
$begingroup$
@user1551I know that, but this situation is a bit tricky. Norms are equivalent up to a constant that's dependent on the size of the matrix, but in my case the matrix is getting bigger as $hto 0.$
$endgroup$
– dezdichado
Dec 8 '18 at 16:21
add a comment |
1 Answer
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oldest
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Your matrix belongs to the class of Generalized Locally Toeplitz (GLT) sequences (see for example https://www.springer.com/us/book/9783319536781).
The sequence of matrices ${h^2A_n}_nsim_{GLT}f$ where $f$ is the spectral symbol.
We have $h^2A^h=h^2A_n=T_n(f)+R_n+N_n$ where
$T_n(f)$ is the Toeplitz matrix of order $n$ generated by $f(theta)=-2+2cos(theta)$.
$R_n$ is a low rank matrix $o(n)$ having 1 in the two elements $(1,2)$ and $(n,n-1)$.
$N_n$ is a low norm matrix $|N_n|to 0$ with the $h$ terms.
${T_n(f)}_nsim_{GLT}f$, ${R_n}_nsim_{GLT}0$, ${N_n}_nsim_{GLT}0$.
thus ${T_n(f)+R_n+N_n}_nsim_{GLT}f+0+0=f$
The spectral radius is 4 (maximum of $|f(theta)|, theta in [-pi,pi]$ ) for $h^2A^h$.
Edit: You do have one outlier eigenvalue, as theory predicts $o(n)$ of them, but as $ntoinfty$ this eigenvalue will tend to $-4$.
Edit 2: Added the scaling $h^2$.
answered Dec 23 '18 at 10:11
santasanta
112
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1 Answer
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1 Answer
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active
oldest
votes
$begingroup$
Your matrix belongs to the class of Generalized Locally Toeplitz (GLT) sequences (see for example https://www.springer.com/us/book/9783319536781).
The sequence of matrices ${h^2A_n}_nsim_{GLT}f$ where $f$ is the spectral symbol.
We have $h^2A^h=h^2A_n=T_n(f)+R_n+N_n$ where
$T_n(f)$ is the Toeplitz matrix of order $n$ generated by $f(theta)=-2+2cos(theta)$.
$R_n$ is a low rank matrix $o(n)$ having 1 in the two elements $(1,2)$ and $(n,n-1)$.
$N_n$ is a low norm matrix $|N_n|to 0$ with the $h$ terms.
${T_n(f)}_nsim_{GLT}f$, ${R_n}_nsim_{GLT}0$, ${N_n}_nsim_{GLT}0$.
thus ${T_n(f)+R_n+N_n}_nsim_{GLT}f+0+0=f$
The spectral radius is 4 (maximum of $|f(theta)|, theta in [-pi,pi]$ ) for $h^2A^h$.
Edit: You do have one outlier eigenvalue, as theory predicts $o(n)$ of them, but as $ntoinfty$ this eigenvalue will tend to $-4$.
Edit 2: Added the scaling $h^2$.
answered Dec 23 '18 at 10:11
santasanta
112
$endgroup$
add a comment |
$begingroup$
Your matrix belongs to the class of Generalized Locally Toeplitz (GLT) sequences (see for example https://www.springer.com/us/book/9783319536781).
The sequence of matrices ${h^2A_n}_nsim_{GLT}f$ where $f$ is the spectral symbol.
We have $h^2A^h=h^2A_n=T_n(f)+R_n+N_n$ where
$T_n(f)$ is the Toeplitz matrix of order $n$ generated by $f(theta)=-2+2cos(theta)$.
$R_n$ is a low rank matrix $o(n)$ having 1 in the two elements $(1,2)$ and $(n,n-1)$.
$N_n$ is a low norm matrix $|N_n|to 0$ with the $h$ terms.
${T_n(f)}_nsim_{GLT}f$, ${R_n}_nsim_{GLT}0$, ${N_n}_nsim_{GLT}0$.
thus ${T_n(f)+R_n+N_n}_nsim_{GLT}f+0+0=f$
The spectral radius is 4 (maximum of $|f(theta)|, theta in [-pi,pi]$ ) for $h^2A^h$.
Edit: You do have one outlier eigenvalue, as theory predicts $o(n)$ of them, but as $ntoinfty$ this eigenvalue will tend to $-4$.
Edit 2: Added the scaling $h^2$.
answered Dec 23 '18 at 10:11
santasanta
112
$endgroup$
add a comment |
$begingroup$
Your matrix belongs to the class of Generalized Locally Toeplitz (GLT) sequences (see for example https://www.springer.com/us/book/9783319536781).
The sequence of matrices ${h^2A_n}_nsim_{GLT}f$ where $f$ is the spectral symbol.
We have $h^2A^h=h^2A_n=T_n(f)+R_n+N_n$ where
$T_n(f)$ is the Toeplitz matrix of order $n$ generated by $f(theta)=-2+2cos(theta)$.
$R_n$ is a low rank matrix $o(n)$ having 1 in the two elements $(1,2)$ and $(n,n-1)$.
$N_n$ is a low norm matrix $|N_n|to 0$ with the $h$ terms.
${T_n(f)}_nsim_{GLT}f$, ${R_n}_nsim_{GLT}0$, ${N_n}_nsim_{GLT}0$.
thus ${T_n(f)+R_n+N_n}_nsim_{GLT}f+0+0=f$
The spectral radius is 4 (maximum of $|f(theta)|, theta in [-pi,pi]$ ) for $h^2A^h$.
Edit: You do have one outlier eigenvalue, as theory predicts $o(n)$ of them, but as $ntoinfty$ this eigenvalue will tend to $-4$.
Edit 2: Added the scaling $h^2$.
answered Dec 23 '18 at 10:11
santasanta
112
$endgroup$
Your matrix belongs to the class of Generalized Locally Toeplitz (GLT) sequences (see for example https://www.springer.com/us/book/9783319536781).
The sequence of matrices ${h^2A_n}_nsim_{GLT}f$ where $f$ is the spectral symbol.
We have $h^2A^h=h^2A_n=T_n(f)+R_n+N_n$ where
$T_n(f)$ is the Toeplitz matrix of order $n$ generated by $f(theta)=-2+2cos(theta)$.
$R_n$ is a low rank matrix $o(n)$ having 1 in the two elements $(1,2)$ and $(n,n-1)$.
$N_n$ is a low norm matrix $|N_n|to 0$ with the $h$ terms.
${T_n(f)}_nsim_{GLT}f$, ${R_n}_nsim_{GLT}0$, ${N_n}_nsim_{GLT}0$.
thus ${T_n(f)+R_n+N_n}_nsim_{GLT}f+0+0=f$
The spectral radius is 4 (maximum of $|f(theta)|, theta in [-pi,pi]$ ) for $h^2A^h$.
Edit: You do have one outlier eigenvalue, as theory predicts $o(n)$ of them, but as $ntoinfty$ this eigenvalue will tend to $-4$.
Edit 2: Added the scaling $h^2$.
answered Dec 23 '18 at 10:11
santasanta
112
edited Dec 23 '18 at 16:44
answered Dec 23 '18 at 10:11
santasanta
112
answered Dec 23 '18 at 10:11
santasanta
112
answered Dec 23 '18 at 10:11
santasanta
112
112
add a comment |
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$begingroup$
@user1551I know that, but this situation is a bit tricky. Norms are equivalent up to a constant that's dependent on the size of the matrix, but in my case the matrix is getting bigger as $hto 0.$
$endgroup$
– dezdichado
Dec 8 '18 at 16:21