Boundedness of the spectral radius of matrices $A^h$ as $hto 0.$












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







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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
















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







share|cite|improve this question











$endgroup$












  • $begingroup$
    @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














0












0








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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 17:38







dezdichado

















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


















  • $begingroup$
    @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










1 Answer
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$begingroup$

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 $h^2$.






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    1 Answer
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    $begingroup$

    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 $h^2$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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 $h^2$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $h^2$.






        share|cite|improve this answer











        $endgroup$



        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 $h^2$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 23 '18 at 16:44

























        answered Dec 23 '18 at 10:11









        santasanta

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