Show the spectral radius of a matrix is smaller than 1












3












$begingroup$


Let $hat{bf H}$ be a $phat{N}times p hat{N}$ sparse matrix consisting of $ptimes p$ blocks, where each block is of size $hat{N}timeshat{N}$. The values in $hat{bf H}$ is illustrated below (empty places are zero):




enter image description here




The infinite norm of $hat{bf H}$ is obviously 1, and I know spectral radius is no larger than any natural norm. My question is how do I prove the spectral radius of this matrix is smaller than 1?



I did a simple numerical experiment, and found the claim should hold. If $p to infty$ and $hat{N} to infty$, then the spectral radius should approach to 1.



If we fix $p = 5$ and let $hat{N}$ go from 5 to 100, we have




enter image description here




If we fix $hat{N} = 10$ and let $p$ go from 5 to 100, we have




enter image description here







linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't understand the pattern of the matrix $hat H$, could you give and example with $N=1$ and $p=6$ ? and then with $N=2$ and $p=2$ ?
    $endgroup$
    – P. Quinton
    Dec 10 '18 at 7:15










  • $begingroup$
    Thanks. If $hat{N}=1$ and $p=6$, then $hat{mathbf{H}} = left( {begin{array}{*{20}{c}} {}&{rm{1}}&{}&{}&{}&{}\ {rm{1}}&{}&{rm{1}}&{}&{}&{}\ {}&{rm{1}}&{}&{rm{1}}&{}&{}\ {}&{}&{rm{1}}&{}&{rm{1}}&{}\ {}&{}&{}&{rm{1}}&{}&{rm{1}}\ {}&{}&{}&{}&{rm{1}}&{} end{array}} right)$. If $hat{N}=2$ and $p=2$, I think the matrix is $left( {begin{array}{*{20}{c}} {}&{}&{rm{1}}&{}\ {}&{}&{rm{2}}&{}\ {}&{rm{2}}&{}&{}\ {}&{rm{1}}&{}&{} end{array}} right)$
    $endgroup$
    – Tony
    Dec 10 '18 at 7:21












  • $begingroup$
    Ok, perfect for the p=6 example, for the other one, I'm still confused, could you place all the zeroes ? Are the blocs diagonal ?
    $endgroup$
    – P. Quinton
    Dec 10 '18 at 7:25










  • $begingroup$
    Sure. I think it is $left( {begin{array}{*{20}{c}} {rm{0}}&{rm{0}}&{rm{1}}&{rm{0}}\ {rm{0}}&{rm{0}}&{rm{2}}&{rm{0}}\ {rm{0}}&{rm{2}}&{rm{0}}&{rm{0}}\ {rm{0}}&{rm{1}}&{rm{0}}&{rm{0}} end{array}} right)$ with zeros. Looks not exactly block diagonal.
    $endgroup$
    – Tony
    Dec 10 '18 at 7:26


















3












$begingroup$


Let $hat{bf H}$ be a $phat{N}times p hat{N}$ sparse matrix consisting of $ptimes p$ blocks, where each block is of size $hat{N}timeshat{N}$. The values in $hat{bf H}$ is illustrated below (empty places are zero):




enter image description here




The infinite norm of $hat{bf H}$ is obviously 1, and I know spectral radius is no larger than any natural norm. My question is how do I prove the spectral radius of this matrix is smaller than 1?



I did a simple numerical experiment, and found the claim should hold. If $p to infty$ and $hat{N} to infty$, then the spectral radius should approach to 1.



If we fix $p = 5$ and let $hat{N}$ go from 5 to 100, we have




enter image description here




If we fix $hat{N} = 10$ and let $p$ go from 5 to 100, we have




enter image description here







linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't understand the pattern of the matrix $hat H$, could you give and example with $N=1$ and $p=6$ ? and then with $N=2$ and $p=2$ ?
    $endgroup$
    – P. Quinton
    Dec 10 '18 at 7:15










  • $begingroup$
    Thanks. If $hat{N}=1$ and $p=6$, then $hat{mathbf{H}} = left( {begin{array}{*{20}{c}} {}&{rm{1}}&{}&{}&{}&{}\ {rm{1}}&{}&{rm{1}}&{}&{}&{}\ {}&{rm{1}}&{}&{rm{1}}&{}&{}\ {}&{}&{rm{1}}&{}&{rm{1}}&{}\ {}&{}&{}&{rm{1}}&{}&{rm{1}}\ {}&{}&{}&{}&{rm{1}}&{} end{array}} right)$. If $hat{N}=2$ and $p=2$, I think the matrix is $left( {begin{array}{*{20}{c}} {}&{}&{rm{1}}&{}\ {}&{}&{rm{2}}&{}\ {}&{rm{2}}&{}&{}\ {}&{rm{1}}&{}&{} end{array}} right)$
    $endgroup$
    – Tony
    Dec 10 '18 at 7:21












  • $begingroup$
    Ok, perfect for the p=6 example, for the other one, I'm still confused, could you place all the zeroes ? Are the blocs diagonal ?
    $endgroup$
    – P. Quinton
    Dec 10 '18 at 7:25










  • $begingroup$
    Sure. I think it is $left( {begin{array}{*{20}{c}} {rm{0}}&{rm{0}}&{rm{1}}&{rm{0}}\ {rm{0}}&{rm{0}}&{rm{2}}&{rm{0}}\ {rm{0}}&{rm{2}}&{rm{0}}&{rm{0}}\ {rm{0}}&{rm{1}}&{rm{0}}&{rm{0}} end{array}} right)$ with zeros. Looks not exactly block diagonal.
    $endgroup$
    – Tony
    Dec 10 '18 at 7:26
















3












3








3


1



$begingroup$


Let $hat{bf H}$ be a $phat{N}times p hat{N}$ sparse matrix consisting of $ptimes p$ blocks, where each block is of size $hat{N}timeshat{N}$. The values in $hat{bf H}$ is illustrated below (empty places are zero):




enter image description here




The infinite norm of $hat{bf H}$ is obviously 1, and I know spectral radius is no larger than any natural norm. My question is how do I prove the spectral radius of this matrix is smaller than 1?



I did a simple numerical experiment, and found the claim should hold. If $p to infty$ and $hat{N} to infty$, then the spectral radius should approach to 1.



If we fix $p = 5$ and let $hat{N}$ go from 5 to 100, we have




enter image description here




If we fix $hat{N} = 10$ and let $p$ go from 5 to 100, we have




enter image description here







linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




Let $hat{bf H}$ be a $phat{N}times p hat{N}$ sparse matrix consisting of $ptimes p$ blocks, where each block is of size $hat{N}timeshat{N}$. The values in $hat{bf H}$ is illustrated below (empty places are zero):




enter image description here




The infinite norm of $hat{bf H}$ is obviously 1, and I know spectral radius is no larger than any natural norm. My question is how do I prove the spectral radius of this matrix is smaller than 1?



I did a simple numerical experiment, and found the claim should hold. If $p to infty$ and $hat{N} to infty$, then the spectral radius should approach to 1.



If we fix $p = 5$ and let $hat{N}$ go from 5 to 100, we have




enter image description here




If we fix $hat{N} = 10$ and let $p$ go from 5 to 100, we have




enter image description here







linear-algebra eigenvalues-eigenvectors




linear-algebra eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 6:15







Tony

















asked Dec 10 '18 at 6:06









TonyTony

1,7641828




1,7641828












  • $begingroup$
    I don't understand the pattern of the matrix $hat H$, could you give and example with $N=1$ and $p=6$ ? and then with $N=2$ and $p=2$ ?
    $endgroup$
    – P. Quinton
    Dec 10 '18 at 7:15










  • $begingroup$
    Thanks. If $hat{N}=1$ and $p=6$, then $hat{mathbf{H}} = left( {begin{array}{*{20}{c}} {}&{rm{1}}&{}&{}&{}&{}\ {rm{1}}&{}&{rm{1}}&{}&{}&{}\ {}&{rm{1}}&{}&{rm{1}}&{}&{}\ {}&{}&{rm{1}}&{}&{rm{1}}&{}\ {}&{}&{}&{rm{1}}&{}&{rm{1}}\ {}&{}&{}&{}&{rm{1}}&{} end{array}} right)$. If $hat{N}=2$ and $p=2$, I think the matrix is $left( {begin{array}{*{20}{c}} {}&{}&{rm{1}}&{}\ {}&{}&{rm{2}}&{}\ {}&{rm{2}}&{}&{}\ {}&{rm{1}}&{}&{} end{array}} right)$
    $endgroup$
    – Tony
    Dec 10 '18 at 7:21












  • $begingroup$
    Ok, perfect for the p=6 example, for the other one, I'm still confused, could you place all the zeroes ? Are the blocs diagonal ?
    $endgroup$
    – P. Quinton
    Dec 10 '18 at 7:25










  • $begingroup$
    Sure. I think it is $left( {begin{array}{*{20}{c}} {rm{0}}&{rm{0}}&{rm{1}}&{rm{0}}\ {rm{0}}&{rm{0}}&{rm{2}}&{rm{0}}\ {rm{0}}&{rm{2}}&{rm{0}}&{rm{0}}\ {rm{0}}&{rm{1}}&{rm{0}}&{rm{0}} end{array}} right)$ with zeros. Looks not exactly block diagonal.
    $endgroup$
    – Tony
    Dec 10 '18 at 7:26




















  • $begingroup$
    I don't understand the pattern of the matrix $hat H$, could you give and example with $N=1$ and $p=6$ ? and then with $N=2$ and $p=2$ ?
    $endgroup$
    – P. Quinton
    Dec 10 '18 at 7:15










  • $begingroup$
    Thanks. If $hat{N}=1$ and $p=6$, then $hat{mathbf{H}} = left( {begin{array}{*{20}{c}} {}&{rm{1}}&{}&{}&{}&{}\ {rm{1}}&{}&{rm{1}}&{}&{}&{}\ {}&{rm{1}}&{}&{rm{1}}&{}&{}\ {}&{}&{rm{1}}&{}&{rm{1}}&{}\ {}&{}&{}&{rm{1}}&{}&{rm{1}}\ {}&{}&{}&{}&{rm{1}}&{} end{array}} right)$. If $hat{N}=2$ and $p=2$, I think the matrix is $left( {begin{array}{*{20}{c}} {}&{}&{rm{1}}&{}\ {}&{}&{rm{2}}&{}\ {}&{rm{2}}&{}&{}\ {}&{rm{1}}&{}&{} end{array}} right)$
    $endgroup$
    – Tony
    Dec 10 '18 at 7:21












  • $begingroup$
    Ok, perfect for the p=6 example, for the other one, I'm still confused, could you place all the zeroes ? Are the blocs diagonal ?
    $endgroup$
    – P. Quinton
    Dec 10 '18 at 7:25










  • $begingroup$
    Sure. I think it is $left( {begin{array}{*{20}{c}} {rm{0}}&{rm{0}}&{rm{1}}&{rm{0}}\ {rm{0}}&{rm{0}}&{rm{2}}&{rm{0}}\ {rm{0}}&{rm{2}}&{rm{0}}&{rm{0}}\ {rm{0}}&{rm{1}}&{rm{0}}&{rm{0}} end{array}} right)$ with zeros. Looks not exactly block diagonal.
    $endgroup$
    – Tony
    Dec 10 '18 at 7:26


















$begingroup$
I don't understand the pattern of the matrix $hat H$, could you give and example with $N=1$ and $p=6$ ? and then with $N=2$ and $p=2$ ?
$endgroup$
– P. Quinton
Dec 10 '18 at 7:15




$begingroup$
I don't understand the pattern of the matrix $hat H$, could you give and example with $N=1$ and $p=6$ ? and then with $N=2$ and $p=2$ ?
$endgroup$
– P. Quinton
Dec 10 '18 at 7:15












$begingroup$
Thanks. If $hat{N}=1$ and $p=6$, then $hat{mathbf{H}} = left( {begin{array}{*{20}{c}} {}&{rm{1}}&{}&{}&{}&{}\ {rm{1}}&{}&{rm{1}}&{}&{}&{}\ {}&{rm{1}}&{}&{rm{1}}&{}&{}\ {}&{}&{rm{1}}&{}&{rm{1}}&{}\ {}&{}&{}&{rm{1}}&{}&{rm{1}}\ {}&{}&{}&{}&{rm{1}}&{} end{array}} right)$. If $hat{N}=2$ and $p=2$, I think the matrix is $left( {begin{array}{*{20}{c}} {}&{}&{rm{1}}&{}\ {}&{}&{rm{2}}&{}\ {}&{rm{2}}&{}&{}\ {}&{rm{1}}&{}&{} end{array}} right)$
$endgroup$
– Tony
Dec 10 '18 at 7:21






$begingroup$
Thanks. If $hat{N}=1$ and $p=6$, then $hat{mathbf{H}} = left( {begin{array}{*{20}{c}} {}&{rm{1}}&{}&{}&{}&{}\ {rm{1}}&{}&{rm{1}}&{}&{}&{}\ {}&{rm{1}}&{}&{rm{1}}&{}&{}\ {}&{}&{rm{1}}&{}&{rm{1}}&{}\ {}&{}&{}&{rm{1}}&{}&{rm{1}}\ {}&{}&{}&{}&{rm{1}}&{} end{array}} right)$. If $hat{N}=2$ and $p=2$, I think the matrix is $left( {begin{array}{*{20}{c}} {}&{}&{rm{1}}&{}\ {}&{}&{rm{2}}&{}\ {}&{rm{2}}&{}&{}\ {}&{rm{1}}&{}&{} end{array}} right)$
$endgroup$
– Tony
Dec 10 '18 at 7:21














$begingroup$
Ok, perfect for the p=6 example, for the other one, I'm still confused, could you place all the zeroes ? Are the blocs diagonal ?
$endgroup$
– P. Quinton
Dec 10 '18 at 7:25




$begingroup$
Ok, perfect for the p=6 example, for the other one, I'm still confused, could you place all the zeroes ? Are the blocs diagonal ?
$endgroup$
– P. Quinton
Dec 10 '18 at 7:25












$begingroup$
Sure. I think it is $left( {begin{array}{*{20}{c}} {rm{0}}&{rm{0}}&{rm{1}}&{rm{0}}\ {rm{0}}&{rm{0}}&{rm{2}}&{rm{0}}\ {rm{0}}&{rm{2}}&{rm{0}}&{rm{0}}\ {rm{0}}&{rm{1}}&{rm{0}}&{rm{0}} end{array}} right)$ with zeros. Looks not exactly block diagonal.
$endgroup$
– Tony
Dec 10 '18 at 7:26






$begingroup$
Sure. I think it is $left( {begin{array}{*{20}{c}} {rm{0}}&{rm{0}}&{rm{1}}&{rm{0}}\ {rm{0}}&{rm{0}}&{rm{2}}&{rm{0}}\ {rm{0}}&{rm{2}}&{rm{0}}&{rm{0}}\ {rm{0}}&{rm{1}}&{rm{0}}&{rm{0}} end{array}} right)$ with zeros. Looks not exactly block diagonal.
$endgroup$
– Tony
Dec 10 '18 at 7:26












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