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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            0 I'm trying to update an entry in my database (postgresql). I'm having a problem with updating the query: INSERT INTO prices (price, product_id) SELECT product_price, commn_id FROM products_temp as prod WHERE NOT EXISTS (SELECT * FROM prices od WHERE prod.commn_id = od.product_id) OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) Query works fine when i delete: OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) So it seems to me that it's looping through this operation. My question is, how can I fix it? sql postgresql
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            WPF add header to Image with URL pettitions [duplicate]

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            up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:
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