Matrix norms - question about the inequality $||A||<rho(A)+epsilon$
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Since for any $epsilon>0$ there exists a matrix norm $||cdot||$ such that
$$||A||<rho(A)+epsilon$$ holds, why does that imply that there exists a matrix norm $||cdot||$ such that $||A||<1$ if the spectral radius $rho(A)<1$ ?
Is this because if $rho(A)<1$ and if $||A||<rho(A)+epsilon$ has to hold for any $epsilon>0$, however small, then however close $rho(A)$ is to $1$, if we add an infinitesimally small $epsilon$ the sum $rho(A)+epsilon$ will still be less than $1$ since there are infinitely many numbers between $rho(A)$ and $1$?
linear-algebra matrices inequality norm
asked Dec 11 '18 at 22:36
user619360user619360
12
$endgroup$
add a comment |
$begingroup$
Since for any $epsilon>0$ there exists a matrix norm $||cdot||$ such that
$$||A||<rho(A)+epsilon$$ holds, why does that imply that there exists a matrix norm $||cdot||$ such that $||A||<1$ if the spectral radius $rho(A)<1$ ?
Is this because if $rho(A)<1$ and if $||A||<rho(A)+epsilon$ has to hold for any $epsilon>0$, however small, then however close $rho(A)$ is to $1$, if we add an infinitesimally small $epsilon$ the sum $rho(A)+epsilon$ will still be less than $1$ since there are infinitely many numbers between $rho(A)$ and $1$?
linear-algebra matrices inequality norm
asked Dec 11 '18 at 22:36
user619360user619360
12
$endgroup$
$begingroup$
Yes. In particular, let $epsilon = dfrac{1}{2}(1-rho(A))$
$endgroup$
– BaronVT
Dec 11 '18 at 22:54
add a comment |
$begingroup$
Since for any $epsilon>0$ there exists a matrix norm $||cdot||$ such that
$$||A||<rho(A)+epsilon$$ holds, why does that imply that there exists a matrix norm $||cdot||$ such that $||A||<1$ if the spectral radius $rho(A)<1$ ?
Is this because if $rho(A)<1$ and if $||A||<rho(A)+epsilon$ has to hold for any $epsilon>0$, however small, then however close $rho(A)$ is to $1$, if we add an infinitesimally small $epsilon$ the sum $rho(A)+epsilon$ will still be less than $1$ since there are infinitely many numbers between $rho(A)$ and $1$?
linear-algebra matrices inequality norm
asked Dec 11 '18 at 22:36
user619360user619360
12
$endgroup$
Since for any $epsilon>0$ there exists a matrix norm $||cdot||$ such that
$$||A||<rho(A)+epsilon$$ holds, why does that imply that there exists a matrix norm $||cdot||$ such that $||A||<1$ if the spectral radius $rho(A)<1$ ?
Is this because if $rho(A)<1$ and if $||A||<rho(A)+epsilon$ has to hold for any $epsilon>0$, however small, then however close $rho(A)$ is to $1$, if we add an infinitesimally small $epsilon$ the sum $rho(A)+epsilon$ will still be less than $1$ since there are infinitely many numbers between $rho(A)$ and $1$?
linear-algebra matrices inequality norm
linear-algebra matrices inequality norm
asked Dec 11 '18 at 22:36
user619360user619360
12
asked Dec 11 '18 at 22:36
user619360user619360
12
asked Dec 11 '18 at 22:36
user619360user619360
12
asked Dec 11 '18 at 22:36
user619360user619360
12
asked Dec 11 '18 at 22:36
user619360user619360
12
12
$begingroup$
Yes. In particular, let $epsilon = dfrac{1}{2}(1-rho(A))$
$endgroup$
– BaronVT
Dec 11 '18 at 22:54
add a comment |
$begingroup$
Yes. In particular, let $epsilon = dfrac{1}{2}(1-rho(A))$
$endgroup$
– BaronVT
Dec 11 '18 at 22:54
$begingroup$
Yes. In particular, let $epsilon = dfrac{1}{2}(1-rho(A))$
$endgroup$
– BaronVT
Dec 11 '18 at 22:54
$begingroup$
Yes. In particular, let $epsilon = dfrac{1}{2}(1-rho(A))$
$endgroup$
– BaronVT
Dec 11 '18 at 22:54
add a comment |
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$begingroup$
Yes. In particular, let $epsilon = dfrac{1}{2}(1-rho(A))$
$endgroup$
– BaronVT
Dec 11 '18 at 22:54