Prove that the spectral radius $rho(A)$ is a continuous function, where $A$ is a square matrix.












4












$begingroup$


Let $||. ||$ be some norm on $mathbb{R^n}.$ Interpret the real-valued matrices as a Euclidean space, $mathbb{R^{n^2}}=Mat_{ntimes n},$ and prove that the following are continuous functions of the $n^2$ matrix entries $A_{ij}$



(a) The spectral radius of $A$, $rho(A)=max{|lambda| ~~| text{such that}~lambda
~text{is an eigenvalue of}~A }.$



(b) The spectral radius of the inverse, $f(A)=rho (A^{-1}),$ on the domain of invertible matrices.




For part (b), I think the goal is $displaystylelim_{Delta A to 0} rho(A+Delta A)=rho(A).$
I tried to use Gelfand formula. Then I have $rho(A)=displaystylelim_{Delta A to 0} rho(A+Delta A)=displaystylelim_{Delta A to 0} lim_{kto infty}||(A+Delta A)^k||^{1/k},$ which becomes more complicated. In addition, I tried to use the operator norm to bound the spectral radius, but I felt that I was still far away from the proof.



Why does the part (b) need to prove if the part (a) is true? $f(A)=rho(A) $ restrict on the domain of invertable matrices. I was confuled about part(b).




I know how to prove this statement using a theorem in complex analysis: The roots of a complex-valued polynomial are continuous wrt the coefficients of the polynomial.



By using this theorem, (b) is very easy to prove.



I am looking for another proof of (b) without the theorem in complex analysis.






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    (b) follows from (a) because $Amapsto A^{-1}$ is continuous on $GL_n(mathbb{R})$ regardless of norm (all of them are equivalent anyway).
    $endgroup$
    – freakish
    Dec 5 '18 at 14:15












  • $begingroup$
    @freakish Thanks. But how to prove part(b)?
    $endgroup$
    – learner
    Dec 5 '18 at 14:17










  • $begingroup$
    @kimchilover Sorry, part (c)=part (b). It is a typo.
    $endgroup$
    – learner
    Dec 5 '18 at 16:16










  • $begingroup$
    @kimchilover I already edited my question.
    $endgroup$
    – learner
    Dec 5 '18 at 16:20










  • $begingroup$
    Does anyone know whether the functions $f_k:M_{ntimes n}rightarrow mathbb{R}$, $f_k(A) = |A^k|^{frac{1}{k}}$, have derivatives $frac{partial f_k}{partial a_{ij}}$ uniformly bounded (in $k$) when $f_k$'s are restricted to some compact set of matrices? Then one could use limit theorems to show spectral radius is continuous.
    $endgroup$
    – user25959
    Dec 5 '18 at 19:14


















4












$begingroup$


Let $||. ||$ be some norm on $mathbb{R^n}.$ Interpret the real-valued matrices as a Euclidean space, $mathbb{R^{n^2}}=Mat_{ntimes n},$ and prove that the following are continuous functions of the $n^2$ matrix entries $A_{ij}$



(a) The spectral radius of $A$, $rho(A)=max{|lambda| ~~| text{such that}~lambda
~text{is an eigenvalue of}~A }.$



(b) The spectral radius of the inverse, $f(A)=rho (A^{-1}),$ on the domain of invertible matrices.




For part (b), I think the goal is $displaystylelim_{Delta A to 0} rho(A+Delta A)=rho(A).$
I tried to use Gelfand formula. Then I have $rho(A)=displaystylelim_{Delta A to 0} rho(A+Delta A)=displaystylelim_{Delta A to 0} lim_{kto infty}||(A+Delta A)^k||^{1/k},$ which becomes more complicated. In addition, I tried to use the operator norm to bound the spectral radius, but I felt that I was still far away from the proof.



Why does the part (b) need to prove if the part (a) is true? $f(A)=rho(A) $ restrict on the domain of invertable matrices. I was confuled about part(b).




I know how to prove this statement using a theorem in complex analysis: The roots of a complex-valued polynomial are continuous wrt the coefficients of the polynomial.



By using this theorem, (b) is very easy to prove.



I am looking for another proof of (b) without the theorem in complex analysis.






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    (b) follows from (a) because $Amapsto A^{-1}$ is continuous on $GL_n(mathbb{R})$ regardless of norm (all of them are equivalent anyway).
    $endgroup$
    – freakish
    Dec 5 '18 at 14:15












  • $begingroup$
    @freakish Thanks. But how to prove part(b)?
    $endgroup$
    – learner
    Dec 5 '18 at 14:17










  • $begingroup$
    @kimchilover Sorry, part (c)=part (b). It is a typo.
    $endgroup$
    – learner
    Dec 5 '18 at 16:16










  • $begingroup$
    @kimchilover I already edited my question.
    $endgroup$
    – learner
    Dec 5 '18 at 16:20










  • $begingroup$
    Does anyone know whether the functions $f_k:M_{ntimes n}rightarrow mathbb{R}$, $f_k(A) = |A^k|^{frac{1}{k}}$, have derivatives $frac{partial f_k}{partial a_{ij}}$ uniformly bounded (in $k$) when $f_k$'s are restricted to some compact set of matrices? Then one could use limit theorems to show spectral radius is continuous.
    $endgroup$
    – user25959
    Dec 5 '18 at 19:14
















4












4








4


1



$begingroup$


Let $||. ||$ be some norm on $mathbb{R^n}.$ Interpret the real-valued matrices as a Euclidean space, $mathbb{R^{n^2}}=Mat_{ntimes n},$ and prove that the following are continuous functions of the $n^2$ matrix entries $A_{ij}$



(a) The spectral radius of $A$, $rho(A)=max{|lambda| ~~| text{such that}~lambda
~text{is an eigenvalue of}~A }.$



(b) The spectral radius of the inverse, $f(A)=rho (A^{-1}),$ on the domain of invertible matrices.




For part (b), I think the goal is $displaystylelim_{Delta A to 0} rho(A+Delta A)=rho(A).$
I tried to use Gelfand formula. Then I have $rho(A)=displaystylelim_{Delta A to 0} rho(A+Delta A)=displaystylelim_{Delta A to 0} lim_{kto infty}||(A+Delta A)^k||^{1/k},$ which becomes more complicated. In addition, I tried to use the operator norm to bound the spectral radius, but I felt that I was still far away from the proof.



Why does the part (b) need to prove if the part (a) is true? $f(A)=rho(A) $ restrict on the domain of invertable matrices. I was confuled about part(b).




I know how to prove this statement using a theorem in complex analysis: The roots of a complex-valued polynomial are continuous wrt the coefficients of the polynomial.



By using this theorem, (b) is very easy to prove.



I am looking for another proof of (b) without the theorem in complex analysis.






linear-algebra matrices







share|cite|improve this question











$endgroup$




Let $||. ||$ be some norm on $mathbb{R^n}.$ Interpret the real-valued matrices as a Euclidean space, $mathbb{R^{n^2}}=Mat_{ntimes n},$ and prove that the following are continuous functions of the $n^2$ matrix entries $A_{ij}$



(a) The spectral radius of $A$, $rho(A)=max{|lambda| ~~| text{such that}~lambda
~text{is an eigenvalue of}~A }.$



(b) The spectral radius of the inverse, $f(A)=rho (A^{-1}),$ on the domain of invertible matrices.




For part (b), I think the goal is $displaystylelim_{Delta A to 0} rho(A+Delta A)=rho(A).$
I tried to use Gelfand formula. Then I have $rho(A)=displaystylelim_{Delta A to 0} rho(A+Delta A)=displaystylelim_{Delta A to 0} lim_{kto infty}||(A+Delta A)^k||^{1/k},$ which becomes more complicated. In addition, I tried to use the operator norm to bound the spectral radius, but I felt that I was still far away from the proof.



Why does the part (b) need to prove if the part (a) is true? $f(A)=rho(A) $ restrict on the domain of invertable matrices. I was confuled about part(b).




I know how to prove this statement using a theorem in complex analysis: The roots of a complex-valued polynomial are continuous wrt the coefficients of the polynomial.



By using this theorem, (b) is very easy to prove.



I am looking for another proof of (b) without the theorem in complex analysis.






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 16:19







learner

















asked Dec 5 '18 at 14:11









learnerlearner

14117




14117












  • $begingroup$
    (b) follows from (a) because $Amapsto A^{-1}$ is continuous on $GL_n(mathbb{R})$ regardless of norm (all of them are equivalent anyway).
    $endgroup$
    – freakish
    Dec 5 '18 at 14:15












  • $begingroup$
    @freakish Thanks. But how to prove part(b)?
    $endgroup$
    – learner
    Dec 5 '18 at 14:17










  • $begingroup$
    @kimchilover Sorry, part (c)=part (b). It is a typo.
    $endgroup$
    – learner
    Dec 5 '18 at 16:16










  • $begingroup$
    @kimchilover I already edited my question.
    $endgroup$
    – learner
    Dec 5 '18 at 16:20










  • $begingroup$
    Does anyone know whether the functions $f_k:M_{ntimes n}rightarrow mathbb{R}$, $f_k(A) = |A^k|^{frac{1}{k}}$, have derivatives $frac{partial f_k}{partial a_{ij}}$ uniformly bounded (in $k$) when $f_k$'s are restricted to some compact set of matrices? Then one could use limit theorems to show spectral radius is continuous.
    $endgroup$
    – user25959
    Dec 5 '18 at 19:14




















  • $begingroup$
    (b) follows from (a) because $Amapsto A^{-1}$ is continuous on $GL_n(mathbb{R})$ regardless of norm (all of them are equivalent anyway).
    $endgroup$
    – freakish
    Dec 5 '18 at 14:15












  • $begingroup$
    @freakish Thanks. But how to prove part(b)?
    $endgroup$
    – learner
    Dec 5 '18 at 14:17










  • $begingroup$
    @kimchilover Sorry, part (c)=part (b). It is a typo.
    $endgroup$
    – learner
    Dec 5 '18 at 16:16










  • $begingroup$
    @kimchilover I already edited my question.
    $endgroup$
    – learner
    Dec 5 '18 at 16:20










  • $begingroup$
    Does anyone know whether the functions $f_k:M_{ntimes n}rightarrow mathbb{R}$, $f_k(A) = |A^k|^{frac{1}{k}}$, have derivatives $frac{partial f_k}{partial a_{ij}}$ uniformly bounded (in $k$) when $f_k$'s are restricted to some compact set of matrices? Then one could use limit theorems to show spectral radius is continuous.
    $endgroup$
    – user25959
    Dec 5 '18 at 19:14


















$begingroup$
(b) follows from (a) because $Amapsto A^{-1}$ is continuous on $GL_n(mathbb{R})$ regardless of norm (all of them are equivalent anyway).
$endgroup$
– freakish
Dec 5 '18 at 14:15






$begingroup$
(b) follows from (a) because $Amapsto A^{-1}$ is continuous on $GL_n(mathbb{R})$ regardless of norm (all of them are equivalent anyway).
$endgroup$
– freakish
Dec 5 '18 at 14:15














$begingroup$
@freakish Thanks. But how to prove part(b)?
$endgroup$
– learner
Dec 5 '18 at 14:17




$begingroup$
@freakish Thanks. But how to prove part(b)?
$endgroup$
– learner
Dec 5 '18 at 14:17












$begingroup$
@kimchilover Sorry, part (c)=part (b). It is a typo.
$endgroup$
– learner
Dec 5 '18 at 16:16




$begingroup$
@kimchilover Sorry, part (c)=part (b). It is a typo.
$endgroup$
– learner
Dec 5 '18 at 16:16












$begingroup$
@kimchilover I already edited my question.
$endgroup$
– learner
Dec 5 '18 at 16:20




$begingroup$
@kimchilover I already edited my question.
$endgroup$
– learner
Dec 5 '18 at 16:20












$begingroup$
Does anyone know whether the functions $f_k:M_{ntimes n}rightarrow mathbb{R}$, $f_k(A) = |A^k|^{frac{1}{k}}$, have derivatives $frac{partial f_k}{partial a_{ij}}$ uniformly bounded (in $k$) when $f_k$'s are restricted to some compact set of matrices? Then one could use limit theorems to show spectral radius is continuous.
$endgroup$
– user25959
Dec 5 '18 at 19:14






$begingroup$
Does anyone know whether the functions $f_k:M_{ntimes n}rightarrow mathbb{R}$, $f_k(A) = |A^k|^{frac{1}{k}}$, have derivatives $frac{partial f_k}{partial a_{ij}}$ uniformly bounded (in $k$) when $f_k$'s are restricted to some compact set of matrices? Then one could use limit theorems to show spectral radius is continuous.
$endgroup$
– user25959
Dec 5 '18 at 19:14












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