Are the sets ${X: max_i text{Re}lambda_i (B+AX) < 0}$ and ${X: rho(B+AX) < 1}$ homeomorphic?












2












$begingroup$


Suppose $A in M_{n times m}(mathbb R)$ and $B in M_n(mathbb R)$ are fixed with $m < n$. Denote two sets $mathcal E, mathcal F$ by
begin{align*}
mathcal E &= {X in M_{m times n}(mathbb R): max_i text{Re} lambda_i(B+AX) < 0 }, \
mathcal F &= {X in M_{m times n}(mathbb R): rho(B+AX) < 1 },
end{align*}
where $lambda_i$ denotes the eigenvalues and $rho(cdot)$ denotes spectral radius of a square matrix. In other words, $mathcal E$ contains matrix $X$ such that $B+AX$ has eigenvalues on the left open half plane of $mathbb C$ and $mathcal F$ contains matrix $X$ such that $B+AX$ has spectral radius bounded by $1$. Let us further assume $mathcal E$ and $mathcal F$ are both nonempty. My question is: could we determine whether they are homeomorphic or not?



I know the transformation $A mapsto (A-I)^{-1}(A+I)$ is a diffeomorphism between the square matrices with spectral radius bounded by $1$ and those whose eigenvalues on the left half plane of $mathbb C$. But it seems to me this map does not necessarily work for $mathcal E$ and $mathcal F$. If $X in mathcal F$, then $(B+AX-I)^{-1} (B+AX+I) = - sum_{j=0}^{infty} (B+AX)^j(I + B+AX) := C$. I am not sure $C-B$ can be represented by $AY$ for some $Y$. Or there are some other maps.
An example to show the non-existence would also be nice.






linear-algebra general-topology matrix-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    The first set is a void set.
    $endgroup$
    – loup blanc
    Feb 1 at 9:26












  • $begingroup$
    Thanks. I translated both sets to make them nonvoid.
    $endgroup$
    – user1101010
    Feb 1 at 9:35










  • $begingroup$
    @user1101010 I now edited my answer and it should now provide an example where these sets are not homeomorphic.
    $endgroup$
    – James
    Feb 7 at 12:48
















2












$begingroup$


Suppose $A in M_{n times m}(mathbb R)$ and $B in M_n(mathbb R)$ are fixed with $m < n$. Denote two sets $mathcal E, mathcal F$ by
begin{align*}
mathcal E &= {X in M_{m times n}(mathbb R): max_i text{Re} lambda_i(B+AX) < 0 }, \
mathcal F &= {X in M_{m times n}(mathbb R): rho(B+AX) < 1 },
end{align*}
where $lambda_i$ denotes the eigenvalues and $rho(cdot)$ denotes spectral radius of a square matrix. In other words, $mathcal E$ contains matrix $X$ such that $B+AX$ has eigenvalues on the left open half plane of $mathbb C$ and $mathcal F$ contains matrix $X$ such that $B+AX$ has spectral radius bounded by $1$. Let us further assume $mathcal E$ and $mathcal F$ are both nonempty. My question is: could we determine whether they are homeomorphic or not?



I know the transformation $A mapsto (A-I)^{-1}(A+I)$ is a diffeomorphism between the square matrices with spectral radius bounded by $1$ and those whose eigenvalues on the left half plane of $mathbb C$. But it seems to me this map does not necessarily work for $mathcal E$ and $mathcal F$. If $X in mathcal F$, then $(B+AX-I)^{-1} (B+AX+I) = - sum_{j=0}^{infty} (B+AX)^j(I + B+AX) := C$. I am not sure $C-B$ can be represented by $AY$ for some $Y$. Or there are some other maps.
An example to show the non-existence would also be nice.






linear-algebra general-topology matrix-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    The first set is a void set.
    $endgroup$
    – loup blanc
    Feb 1 at 9:26












  • $begingroup$
    Thanks. I translated both sets to make them nonvoid.
    $endgroup$
    – user1101010
    Feb 1 at 9:35










  • $begingroup$
    @user1101010 I now edited my answer and it should now provide an example where these sets are not homeomorphic.
    $endgroup$
    – James
    Feb 7 at 12:48














2












2








2


1



$begingroup$


Suppose $A in M_{n times m}(mathbb R)$ and $B in M_n(mathbb R)$ are fixed with $m < n$. Denote two sets $mathcal E, mathcal F$ by
begin{align*}
mathcal E &= {X in M_{m times n}(mathbb R): max_i text{Re} lambda_i(B+AX) < 0 }, \
mathcal F &= {X in M_{m times n}(mathbb R): rho(B+AX) < 1 },
end{align*}
where $lambda_i$ denotes the eigenvalues and $rho(cdot)$ denotes spectral radius of a square matrix. In other words, $mathcal E$ contains matrix $X$ such that $B+AX$ has eigenvalues on the left open half plane of $mathbb C$ and $mathcal F$ contains matrix $X$ such that $B+AX$ has spectral radius bounded by $1$. Let us further assume $mathcal E$ and $mathcal F$ are both nonempty. My question is: could we determine whether they are homeomorphic or not?



I know the transformation $A mapsto (A-I)^{-1}(A+I)$ is a diffeomorphism between the square matrices with spectral radius bounded by $1$ and those whose eigenvalues on the left half plane of $mathbb C$. But it seems to me this map does not necessarily work for $mathcal E$ and $mathcal F$. If $X in mathcal F$, then $(B+AX-I)^{-1} (B+AX+I) = - sum_{j=0}^{infty} (B+AX)^j(I + B+AX) := C$. I am not sure $C-B$ can be represented by $AY$ for some $Y$. Or there are some other maps.
An example to show the non-existence would also be nice.






linear-algebra general-topology matrix-analysis







share|cite|improve this question











$endgroup$




Suppose $A in M_{n times m}(mathbb R)$ and $B in M_n(mathbb R)$ are fixed with $m < n$. Denote two sets $mathcal E, mathcal F$ by
begin{align*}
mathcal E &= {X in M_{m times n}(mathbb R): max_i text{Re} lambda_i(B+AX) < 0 }, \
mathcal F &= {X in M_{m times n}(mathbb R): rho(B+AX) < 1 },
end{align*}
where $lambda_i$ denotes the eigenvalues and $rho(cdot)$ denotes spectral radius of a square matrix. In other words, $mathcal E$ contains matrix $X$ such that $B+AX$ has eigenvalues on the left open half plane of $mathbb C$ and $mathcal F$ contains matrix $X$ such that $B+AX$ has spectral radius bounded by $1$. Let us further assume $mathcal E$ and $mathcal F$ are both nonempty. My question is: could we determine whether they are homeomorphic or not?



I know the transformation $A mapsto (A-I)^{-1}(A+I)$ is a diffeomorphism between the square matrices with spectral radius bounded by $1$ and those whose eigenvalues on the left half plane of $mathbb C$. But it seems to me this map does not necessarily work for $mathcal E$ and $mathcal F$. If $X in mathcal F$, then $(B+AX-I)^{-1} (B+AX+I) = - sum_{j=0}^{infty} (B+AX)^j(I + B+AX) := C$. I am not sure $C-B$ can be represented by $AY$ for some $Y$. Or there are some other maps.
An example to show the non-existence would also be nice.






linear-algebra general-topology matrix-analysis




linear-algebra general-topology matrix-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 1 at 23:51









Paul Frost

12.6k31035




12.6k31035










asked Feb 1 at 6:39









user1101010user1101010

9011830




9011830












  • $begingroup$
    The first set is a void set.
    $endgroup$
    – loup blanc
    Feb 1 at 9:26












  • $begingroup$
    Thanks. I translated both sets to make them nonvoid.
    $endgroup$
    – user1101010
    Feb 1 at 9:35










  • $begingroup$
    @user1101010 I now edited my answer and it should now provide an example where these sets are not homeomorphic.
    $endgroup$
    – James
    Feb 7 at 12:48


















  • $begingroup$
    The first set is a void set.
    $endgroup$
    – loup blanc
    Feb 1 at 9:26












  • $begingroup$
    Thanks. I translated both sets to make them nonvoid.
    $endgroup$
    – user1101010
    Feb 1 at 9:35










  • $begingroup$
    @user1101010 I now edited my answer and it should now provide an example where these sets are not homeomorphic.
    $endgroup$
    – James
    Feb 7 at 12:48
















$begingroup$
The first set is a void set.
$endgroup$
– loup blanc
Feb 1 at 9:26






$begingroup$
The first set is a void set.
$endgroup$
– loup blanc
Feb 1 at 9:26














$begingroup$
Thanks. I translated both sets to make them nonvoid.
$endgroup$
– user1101010
Feb 1 at 9:35




$begingroup$
Thanks. I translated both sets to make them nonvoid.
$endgroup$
– user1101010
Feb 1 at 9:35












$begingroup$
@user1101010 I now edited my answer and it should now provide an example where these sets are not homeomorphic.
$endgroup$
– James
Feb 7 at 12:48




$begingroup$
@user1101010 I now edited my answer and it should now provide an example where these sets are not homeomorphic.
$endgroup$
– James
Feb 7 at 12:48










1 Answer
1






active

oldest

votes


















2





+100







$begingroup$

Before voting this answer down, give me a chance to improve the answer or to delete it if it is complete nonsense.



EDIT: Example is now working as far as I see.



This being said, let's consider the simplest example. Take $m=1$, $n=2$, $B=begin{pmatrix}frac{1}{2} &0\ 0& -frac{1}{2} end{pmatrix}$, $A=begin{pmatrix}1 \ 0end{pmatrix}$. Take an arbitrary $X=begin{pmatrix}x & yend{pmatrix}$. We have $AX=begin{pmatrix}x & y\ 0 & 0end{pmatrix}$. Now the Eigenvalues of $B+AX$ are the roots of the characteristic polynomial, i.e. $$p(lambda)=detbegin{pmatrix}frac{1}{2}+x-lambda & y \ 0 & -frac{1}{2}-lambdaend{pmatrix}=(frac{1}{2}+x-lambda)(-frac{1}{2}-lambda)$$ with roots $frac{1}{2}+x$ and $-frac{1}{2}$.



The maximal real part of all eigenvalues is $max{Re(frac{1}{2}+x),-frac{1}{2}}=max{frac{1}{2}+x,-frac{1}{2}}$, since we assumed the matrices to be real. As you can see the answer does not depend on the choice of $y$. We may write the first set $$mathcal E={(x,y)in mathbb R^2mid max{frac{1}{2}+x,-frac{1}{2}}<0}.$$ If $xleq -1$, the maximum always is $-frac{1}{2}$ and strictly smaller than $0$. If $-1<x<-frac{1}{2}$, the maximum is $x+frac{1}{2}<0$.
Thus, $$mathcal E={(x,y)in mathbb R^2mid x<-1}=(-infty,-frac{1}{2})times mathbb R.$$



Now for the spectral radius. $rho(B+AX)=max{|frac{1}{2}+x|,|-frac{1}{2}|}=max{sqrt{frac{1}{4}+x^2},frac{1}{2}}$.
The maximum is $frac{1}{2}$ iff $sqrt{frac{1}{4}+x^2}leq frac{1}{2}$ which is the case iff $frac{1}{4}+x^2leq frac{1}{4}$ i.e. for $x=0$.



$sqrt{frac{1}{4}+x^2}<1$ iff $frac{1}{4}+x^2<1$ iff $x^2<frac{3}{4}$. Since for $xgeq 0$ the maximum is $frac{1}{2}$, $sqrt{frac{1}{4}+x^2}$ is the maximum and $<1$ precisely for $x<0$.



Together this yields $$mathcal{F}=(-infty, 0]times mathbb R$$



Now we would have to find a continuous map between an open interval and an interval that is closed on one side. This is not possible because the inverse image of the open interval would have to be open again by continuity.



Thus these sets are not homeomorphic in general.






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    active

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    +100







    $begingroup$

    Before voting this answer down, give me a chance to improve the answer or to delete it if it is complete nonsense.



    EDIT: Example is now working as far as I see.



    This being said, let's consider the simplest example. Take $m=1$, $n=2$, $B=begin{pmatrix}frac{1}{2} &0\ 0& -frac{1}{2} end{pmatrix}$, $A=begin{pmatrix}1 \ 0end{pmatrix}$. Take an arbitrary $X=begin{pmatrix}x & yend{pmatrix}$. We have $AX=begin{pmatrix}x & y\ 0 & 0end{pmatrix}$. Now the Eigenvalues of $B+AX$ are the roots of the characteristic polynomial, i.e. $$p(lambda)=detbegin{pmatrix}frac{1}{2}+x-lambda & y \ 0 & -frac{1}{2}-lambdaend{pmatrix}=(frac{1}{2}+x-lambda)(-frac{1}{2}-lambda)$$ with roots $frac{1}{2}+x$ and $-frac{1}{2}$.



    The maximal real part of all eigenvalues is $max{Re(frac{1}{2}+x),-frac{1}{2}}=max{frac{1}{2}+x,-frac{1}{2}}$, since we assumed the matrices to be real. As you can see the answer does not depend on the choice of $y$. We may write the first set $$mathcal E={(x,y)in mathbb R^2mid max{frac{1}{2}+x,-frac{1}{2}}<0}.$$ If $xleq -1$, the maximum always is $-frac{1}{2}$ and strictly smaller than $0$. If $-1<x<-frac{1}{2}$, the maximum is $x+frac{1}{2}<0$.
    Thus, $$mathcal E={(x,y)in mathbb R^2mid x<-1}=(-infty,-frac{1}{2})times mathbb R.$$



    Now for the spectral radius. $rho(B+AX)=max{|frac{1}{2}+x|,|-frac{1}{2}|}=max{sqrt{frac{1}{4}+x^2},frac{1}{2}}$.
    The maximum is $frac{1}{2}$ iff $sqrt{frac{1}{4}+x^2}leq frac{1}{2}$ which is the case iff $frac{1}{4}+x^2leq frac{1}{4}$ i.e. for $x=0$.



    $sqrt{frac{1}{4}+x^2}<1$ iff $frac{1}{4}+x^2<1$ iff $x^2<frac{3}{4}$. Since for $xgeq 0$ the maximum is $frac{1}{2}$, $sqrt{frac{1}{4}+x^2}$ is the maximum and $<1$ precisely for $x<0$.



    Together this yields $$mathcal{F}=(-infty, 0]times mathbb R$$



    Now we would have to find a continuous map between an open interval and an interval that is closed on one side. This is not possible because the inverse image of the open interval would have to be open again by continuity.



    Thus these sets are not homeomorphic in general.






    share|cite|improve this answer











    $endgroup$


















      2





      +100







      $begingroup$

      Before voting this answer down, give me a chance to improve the answer or to delete it if it is complete nonsense.



      EDIT: Example is now working as far as I see.



      This being said, let's consider the simplest example. Take $m=1$, $n=2$, $B=begin{pmatrix}frac{1}{2} &0\ 0& -frac{1}{2} end{pmatrix}$, $A=begin{pmatrix}1 \ 0end{pmatrix}$. Take an arbitrary $X=begin{pmatrix}x & yend{pmatrix}$. We have $AX=begin{pmatrix}x & y\ 0 & 0end{pmatrix}$. Now the Eigenvalues of $B+AX$ are the roots of the characteristic polynomial, i.e. $$p(lambda)=detbegin{pmatrix}frac{1}{2}+x-lambda & y \ 0 & -frac{1}{2}-lambdaend{pmatrix}=(frac{1}{2}+x-lambda)(-frac{1}{2}-lambda)$$ with roots $frac{1}{2}+x$ and $-frac{1}{2}$.



      The maximal real part of all eigenvalues is $max{Re(frac{1}{2}+x),-frac{1}{2}}=max{frac{1}{2}+x,-frac{1}{2}}$, since we assumed the matrices to be real. As you can see the answer does not depend on the choice of $y$. We may write the first set $$mathcal E={(x,y)in mathbb R^2mid max{frac{1}{2}+x,-frac{1}{2}}<0}.$$ If $xleq -1$, the maximum always is $-frac{1}{2}$ and strictly smaller than $0$. If $-1<x<-frac{1}{2}$, the maximum is $x+frac{1}{2}<0$.
      Thus, $$mathcal E={(x,y)in mathbb R^2mid x<-1}=(-infty,-frac{1}{2})times mathbb R.$$



      Now for the spectral radius. $rho(B+AX)=max{|frac{1}{2}+x|,|-frac{1}{2}|}=max{sqrt{frac{1}{4}+x^2},frac{1}{2}}$.
      The maximum is $frac{1}{2}$ iff $sqrt{frac{1}{4}+x^2}leq frac{1}{2}$ which is the case iff $frac{1}{4}+x^2leq frac{1}{4}$ i.e. for $x=0$.



      $sqrt{frac{1}{4}+x^2}<1$ iff $frac{1}{4}+x^2<1$ iff $x^2<frac{3}{4}$. Since for $xgeq 0$ the maximum is $frac{1}{2}$, $sqrt{frac{1}{4}+x^2}$ is the maximum and $<1$ precisely for $x<0$.



      Together this yields $$mathcal{F}=(-infty, 0]times mathbb R$$



      Now we would have to find a continuous map between an open interval and an interval that is closed on one side. This is not possible because the inverse image of the open interval would have to be open again by continuity.



      Thus these sets are not homeomorphic in general.






      share|cite|improve this answer











      $endgroup$
















        2





        +100







        2





        +100



        2




        +100



        $begingroup$

        Before voting this answer down, give me a chance to improve the answer or to delete it if it is complete nonsense.



        EDIT: Example is now working as far as I see.



        This being said, let's consider the simplest example. Take $m=1$, $n=2$, $B=begin{pmatrix}frac{1}{2} &0\ 0& -frac{1}{2} end{pmatrix}$, $A=begin{pmatrix}1 \ 0end{pmatrix}$. Take an arbitrary $X=begin{pmatrix}x & yend{pmatrix}$. We have $AX=begin{pmatrix}x & y\ 0 & 0end{pmatrix}$. Now the Eigenvalues of $B+AX$ are the roots of the characteristic polynomial, i.e. $$p(lambda)=detbegin{pmatrix}frac{1}{2}+x-lambda & y \ 0 & -frac{1}{2}-lambdaend{pmatrix}=(frac{1}{2}+x-lambda)(-frac{1}{2}-lambda)$$ with roots $frac{1}{2}+x$ and $-frac{1}{2}$.



        The maximal real part of all eigenvalues is $max{Re(frac{1}{2}+x),-frac{1}{2}}=max{frac{1}{2}+x,-frac{1}{2}}$, since we assumed the matrices to be real. As you can see the answer does not depend on the choice of $y$. We may write the first set $$mathcal E={(x,y)in mathbb R^2mid max{frac{1}{2}+x,-frac{1}{2}}<0}.$$ If $xleq -1$, the maximum always is $-frac{1}{2}$ and strictly smaller than $0$. If $-1<x<-frac{1}{2}$, the maximum is $x+frac{1}{2}<0$.
        Thus, $$mathcal E={(x,y)in mathbb R^2mid x<-1}=(-infty,-frac{1}{2})times mathbb R.$$



        Now for the spectral radius. $rho(B+AX)=max{|frac{1}{2}+x|,|-frac{1}{2}|}=max{sqrt{frac{1}{4}+x^2},frac{1}{2}}$.
        The maximum is $frac{1}{2}$ iff $sqrt{frac{1}{4}+x^2}leq frac{1}{2}$ which is the case iff $frac{1}{4}+x^2leq frac{1}{4}$ i.e. for $x=0$.



        $sqrt{frac{1}{4}+x^2}<1$ iff $frac{1}{4}+x^2<1$ iff $x^2<frac{3}{4}$. Since for $xgeq 0$ the maximum is $frac{1}{2}$, $sqrt{frac{1}{4}+x^2}$ is the maximum and $<1$ precisely for $x<0$.



        Together this yields $$mathcal{F}=(-infty, 0]times mathbb R$$



        Now we would have to find a continuous map between an open interval and an interval that is closed on one side. This is not possible because the inverse image of the open interval would have to be open again by continuity.



        Thus these sets are not homeomorphic in general.






        share|cite|improve this answer











        $endgroup$



        Before voting this answer down, give me a chance to improve the answer or to delete it if it is complete nonsense.



        EDIT: Example is now working as far as I see.



        This being said, let's consider the simplest example. Take $m=1$, $n=2$, $B=begin{pmatrix}frac{1}{2} &0\ 0& -frac{1}{2} end{pmatrix}$, $A=begin{pmatrix}1 \ 0end{pmatrix}$. Take an arbitrary $X=begin{pmatrix}x & yend{pmatrix}$. We have $AX=begin{pmatrix}x & y\ 0 & 0end{pmatrix}$. Now the Eigenvalues of $B+AX$ are the roots of the characteristic polynomial, i.e. $$p(lambda)=detbegin{pmatrix}frac{1}{2}+x-lambda & y \ 0 & -frac{1}{2}-lambdaend{pmatrix}=(frac{1}{2}+x-lambda)(-frac{1}{2}-lambda)$$ with roots $frac{1}{2}+x$ and $-frac{1}{2}$.



        The maximal real part of all eigenvalues is $max{Re(frac{1}{2}+x),-frac{1}{2}}=max{frac{1}{2}+x,-frac{1}{2}}$, since we assumed the matrices to be real. As you can see the answer does not depend on the choice of $y$. We may write the first set $$mathcal E={(x,y)in mathbb R^2mid max{frac{1}{2}+x,-frac{1}{2}}<0}.$$ If $xleq -1$, the maximum always is $-frac{1}{2}$ and strictly smaller than $0$. If $-1<x<-frac{1}{2}$, the maximum is $x+frac{1}{2}<0$.
        Thus, $$mathcal E={(x,y)in mathbb R^2mid x<-1}=(-infty,-frac{1}{2})times mathbb R.$$



        Now for the spectral radius. $rho(B+AX)=max{|frac{1}{2}+x|,|-frac{1}{2}|}=max{sqrt{frac{1}{4}+x^2},frac{1}{2}}$.
        The maximum is $frac{1}{2}$ iff $sqrt{frac{1}{4}+x^2}leq frac{1}{2}$ which is the case iff $frac{1}{4}+x^2leq frac{1}{4}$ i.e. for $x=0$.



        $sqrt{frac{1}{4}+x^2}<1$ iff $frac{1}{4}+x^2<1$ iff $x^2<frac{3}{4}$. Since for $xgeq 0$ the maximum is $frac{1}{2}$, $sqrt{frac{1}{4}+x^2}$ is the maximum and $<1$ precisely for $x<0$.



        Together this yields $$mathcal{F}=(-infty, 0]times mathbb R$$



        Now we would have to find a continuous map between an open interval and an interval that is closed on one side. This is not possible because the inverse image of the open interval would have to be open again by continuity.



        Thus these sets are not homeomorphic in general.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 8 at 8:41

























        answered Feb 7 at 9:08









        JamesJames

        2,112422




        2,112422






























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            ts Property 'filter' does not exist on type '{}'

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            1 I'm trying to follow Angular Material guide to create a custom filter for autocomplete input text https://material.angular.io/components/autocomplete/overview This is what I have TS import { Component, OnInit } from '@angular/core'; import { TemplateRef } from '@angular/core'; import { FormControl } from "@angular/forms"; import { Observable } from "rxjs"; import { map, startWith, filter } from 'rxjs/operators'; (...) export class AppComponent implements OnInit { title = 'Your Profile'; displayedColumns: string = ['selected', 'name', 'description', 'pnr']; dataSource = ELEMENT_DATA; myControl = new FormControl(); options: any = ['One', 'Two', 'Three']; filteredOptions:
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            mat-slide-toggle shouldn't change it's state when I click cancel in confirmation window

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            1 When I checked the slide toggle it should work as expected. But when I try to uncheck it there is a confirmation window asking if you are sure. When I click cancel in that confirmation window, still the toggle changes to unchecked, which shouldn't happen. It should stay in the same state. Here is my Html and TS Code // html <mat-slide-toggle (change)="change($event)" [checked]="isChecked()" > To-pay </mat-slide-toggle> TS code: // ts change(e) { if(this.checked) { if(confirm("Are you sure")) { console.log("toggle") } else { e.stopPropagation(); console.log("toggle should not change if I click the cancel button") }
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