Is the closure $overline{ {X in mathbb{R}^{m times n} : rho(M-NX) < 1} }$ equal to $ {X in mathbb{R}^{m...
$begingroup$
Suppose $M in mathcal M(n times n; mathbb R)$ and $N in mathcal M(n times m; mathbb R)$ are fixed with $N neq 0$. Let
begin{align*}
E = {X in mathcal{M}(m times n; mathbb R) : rho(M-NX) < 1},
end{align*}
where $rho(cdot)$ denotes the spectral radius of a matrix.
I want to know whether the closure $bar{E}$ of $E$ is equal to
$$F = {X in mathcal{M}(m times n; mathbb R) : rho(M-NX) le 1}.$$
We will assume $E$ is not empty.
If we define following composition of continuous maps
begin{align*}
f: X mapsto M-NX mapsto (lambda_1(M-NX), dots, lambda_n(M-NX)) mapsto (|lambda_1(M-NX)|, dots, |lambda_n(M-NX)|) \
mapsto max( |lambda_1(M-NX)|, dots, |lambda_n(M-NX)|).
end{align*}
Then $f$ is continuous into $[0, infty)$. We note $E = {X: f^{-1}([0,1)}$ and $F = {X: f^{-1}([0,1])}$. So $E$ is open and $F$ is closed. Clearly $bar{E} subset F$. But I could not show the other direction (or possibly $bar{E}$ is a proper subset of $F$). I tried to construct a sequence ${X_n}$ converging to $X in F setminus E$ by multiplying a factor $1-varepsilon$ to $X$ but apparently to conclude $bar{E} = F$ we need some kind of sublinearatily of spectral radius which is not true in general.
linear-algebra general-topology spectral-radius
asked May 25 '18 at 23:48
user1101010user1101010
7391730
$endgroup$
|
show 5 more comments
$begingroup$
Suppose $M in mathcal M(n times n; mathbb R)$ and $N in mathcal M(n times m; mathbb R)$ are fixed with $N neq 0$. Let
begin{align*}
E = {X in mathcal{M}(m times n; mathbb R) : rho(M-NX) < 1},
end{align*}
where $rho(cdot)$ denotes the spectral radius of a matrix.
I want to know whether the closure $bar{E}$ of $E$ is equal to
$$F = {X in mathcal{M}(m times n; mathbb R) : rho(M-NX) le 1}.$$
We will assume $E$ is not empty.
If we define following composition of continuous maps
begin{align*}
f: X mapsto M-NX mapsto (lambda_1(M-NX), dots, lambda_n(M-NX)) mapsto (|lambda_1(M-NX)|, dots, |lambda_n(M-NX)|) \
mapsto max( |lambda_1(M-NX)|, dots, |lambda_n(M-NX)|).
end{align*}
Then $f$ is continuous into $[0, infty)$. We note $E = {X: f^{-1}([0,1)}$ and $F = {X: f^{-1}([0,1])}$. So $E$ is open and $F$ is closed. Clearly $bar{E} subset F$. But I could not show the other direction (or possibly $bar{E}$ is a proper subset of $F$). I tried to construct a sequence ${X_n}$ converging to $X in F setminus E$ by multiplying a factor $1-varepsilon$ to $X$ but apparently to conclude $bar{E} = F$ we need some kind of sublinearatily of spectral radius which is not true in general.
linear-algebra general-topology spectral-radius
asked May 25 '18 at 23:48
user1101010user1101010
7391730
$endgroup$
1
$begingroup$
If $N=0$ and $rho(M)=1$, then the LHS is empty but the RHS is the whole matrix space.
$endgroup$
– user1551
May 26 '18 at 10:17
$begingroup$
Actually I made the assumption $E$ is not empty. But sorry, the assumption was in the middle which made it not clear enough.
$endgroup$
– user1101010
May 26 '18 at 14:08
$begingroup$
The problem can be restated as follows: "if $1$ is a local minimum of $f$, must it also be a global minimum?" That is certainly true if $N$ has rank $m$, as then the mapping $X mapsto M - NX$ is open and any local minimum of $f$ must be a local minimum of $rho$, which can only be $0$. For lower ranks the answer is less obvious.
$endgroup$
– Niels J. Diepeveen
May 29 '18 at 9:45
$begingroup$
@NielsJ.Diepeveen: I don't quite understand your restated version. If it is true, i.e., $1$ is a global minimum, then $E = emptyset$ and it is certainly true $bar{E} neq F$. But I am assuming $E$ is not empty (this can be done by choosing appropriate $M,N$).
$endgroup$
– user1101010
May 29 '18 at 16:29
$begingroup$
@NielsJ.Diepeveen: Thinking about the second part of your comment, I can see $X mapsto M-NX$ is open if $N$ has full rank, but could not follow why the local minimum of $f$ must be a local minimum of $rho(cdot)$. Sorry, I am having too many questions.
$endgroup$
– user1101010
May 29 '18 at 16:36
|
show 5 more comments
$begingroup$
Suppose $M in mathcal M(n times n; mathbb R)$ and $N in mathcal M(n times m; mathbb R)$ are fixed with $N neq 0$. Let
begin{align*}
E = {X in mathcal{M}(m times n; mathbb R) : rho(M-NX) < 1},
end{align*}
where $rho(cdot)$ denotes the spectral radius of a matrix.
I want to know whether the closure $bar{E}$ of $E$ is equal to
$$F = {X in mathcal{M}(m times n; mathbb R) : rho(M-NX) le 1}.$$
We will assume $E$ is not empty.
If we define following composition of continuous maps
begin{align*}
f: X mapsto M-NX mapsto (lambda_1(M-NX), dots, lambda_n(M-NX)) mapsto (|lambda_1(M-NX)|, dots, |lambda_n(M-NX)|) \
mapsto max( |lambda_1(M-NX)|, dots, |lambda_n(M-NX)|).
end{align*}
Then $f$ is continuous into $[0, infty)$. We note $E = {X: f^{-1}([0,1)}$ and $F = {X: f^{-1}([0,1])}$. So $E$ is open and $F$ is closed. Clearly $bar{E} subset F$. But I could not show the other direction (or possibly $bar{E}$ is a proper subset of $F$). I tried to construct a sequence ${X_n}$ converging to $X in F setminus E$ by multiplying a factor $1-varepsilon$ to $X$ but apparently to conclude $bar{E} = F$ we need some kind of sublinearatily of spectral radius which is not true in general.
linear-algebra general-topology spectral-radius
asked May 25 '18 at 23:48
user1101010user1101010
7391730
$endgroup$
Suppose $M in mathcal M(n times n; mathbb R)$ and $N in mathcal M(n times m; mathbb R)$ are fixed with $N neq 0$. Let
begin{align*}
E = {X in mathcal{M}(m times n; mathbb R) : rho(M-NX) < 1},
end{align*}
where $rho(cdot)$ denotes the spectral radius of a matrix.
I want to know whether the closure $bar{E}$ of $E$ is equal to
$$F = {X in mathcal{M}(m times n; mathbb R) : rho(M-NX) le 1}.$$
We will assume $E$ is not empty.
If we define following composition of continuous maps
begin{align*}
f: X mapsto M-NX mapsto (lambda_1(M-NX), dots, lambda_n(M-NX)) mapsto (|lambda_1(M-NX)|, dots, |lambda_n(M-NX)|) \
mapsto max( |lambda_1(M-NX)|, dots, |lambda_n(M-NX)|).
end{align*}
Then $f$ is continuous into $[0, infty)$. We note $E = {X: f^{-1}([0,1)}$ and $F = {X: f^{-1}([0,1])}$. So $E$ is open and $F$ is closed. Clearly $bar{E} subset F$. But I could not show the other direction (or possibly $bar{E}$ is a proper subset of $F$). I tried to construct a sequence ${X_n}$ converging to $X in F setminus E$ by multiplying a factor $1-varepsilon$ to $X$ but apparently to conclude $bar{E} = F$ we need some kind of sublinearatily of spectral radius which is not true in general.
linear-algebra general-topology spectral-radius
linear-algebra general-topology spectral-radius
asked May 25 '18 at 23:48
user1101010user1101010
7391730
asked May 25 '18 at 23:48
user1101010user1101010
7391730
edited Jun 5 '18 at 4:54
user1101010
asked May 25 '18 at 23:48
user1101010user1101010
7391730
asked May 25 '18 at 23:48
user1101010user1101010
7391730
asked May 25 '18 at 23:48
user1101010user1101010
7391730
7391730
1
$begingroup$
If $N=0$ and $rho(M)=1$, then the LHS is empty but the RHS is the whole matrix space.
$endgroup$
– user1551
May 26 '18 at 10:17
$begingroup$
Actually I made the assumption $E$ is not empty. But sorry, the assumption was in the middle which made it not clear enough.
$endgroup$
– user1101010
May 26 '18 at 14:08
$begingroup$
The problem can be restated as follows: "if $1$ is a local minimum of $f$, must it also be a global minimum?" That is certainly true if $N$ has rank $m$, as then the mapping $X mapsto M - NX$ is open and any local minimum of $f$ must be a local minimum of $rho$, which can only be $0$. For lower ranks the answer is less obvious.
$endgroup$
– Niels J. Diepeveen
May 29 '18 at 9:45
$begingroup$
@NielsJ.Diepeveen: I don't quite understand your restated version. If it is true, i.e., $1$ is a global minimum, then $E = emptyset$ and it is certainly true $bar{E} neq F$. But I am assuming $E$ is not empty (this can be done by choosing appropriate $M,N$).
$endgroup$
– user1101010
May 29 '18 at 16:29
$begingroup$
@NielsJ.Diepeveen: Thinking about the second part of your comment, I can see $X mapsto M-NX$ is open if $N$ has full rank, but could not follow why the local minimum of $f$ must be a local minimum of $rho(cdot)$. Sorry, I am having too many questions.
$endgroup$
– user1101010
May 29 '18 at 16:36
|
show 5 more comments
1
$begingroup$
If $N=0$ and $rho(M)=1$, then the LHS is empty but the RHS is the whole matrix space.
$endgroup$
– user1551
May 26 '18 at 10:17
$begingroup$
Actually I made the assumption $E$ is not empty. But sorry, the assumption was in the middle which made it not clear enough.
$endgroup$
– user1101010
May 26 '18 at 14:08
$begingroup$
The problem can be restated as follows: "if $1$ is a local minimum of $f$, must it also be a global minimum?" That is certainly true if $N$ has rank $m$, as then the mapping $X mapsto M - NX$ is open and any local minimum of $f$ must be a local minimum of $rho$, which can only be $0$. For lower ranks the answer is less obvious.
$endgroup$
– Niels J. Diepeveen
May 29 '18 at 9:45
$begingroup$
@NielsJ.Diepeveen: I don't quite understand your restated version. If it is true, i.e., $1$ is a global minimum, then $E = emptyset$ and it is certainly true $bar{E} neq F$. But I am assuming $E$ is not empty (this can be done by choosing appropriate $M,N$).
$endgroup$
– user1101010
May 29 '18 at 16:29
$begingroup$
@NielsJ.Diepeveen: Thinking about the second part of your comment, I can see $X mapsto M-NX$ is open if $N$ has full rank, but could not follow why the local minimum of $f$ must be a local minimum of $rho(cdot)$. Sorry, I am having too many questions.
$endgroup$
– user1101010
May 29 '18 at 16:36
1
1
$begingroup$
If $N=0$ and $rho(M)=1$, then the LHS is empty but the RHS is the whole matrix space.
$endgroup$
– user1551
May 26 '18 at 10:17
$begingroup$
If $N=0$ and $rho(M)=1$, then the LHS is empty but the RHS is the whole matrix space.
$endgroup$
– user1551
May 26 '18 at 10:17
$begingroup$
Actually I made the assumption $E$ is not empty. But sorry, the assumption was in the middle which made it not clear enough.
$endgroup$
– user1101010
May 26 '18 at 14:08
$begingroup$
Actually I made the assumption $E$ is not empty. But sorry, the assumption was in the middle which made it not clear enough.
$endgroup$
– user1101010
May 26 '18 at 14:08
$begingroup$
The problem can be restated as follows: "if $1$ is a local minimum of $f$, must it also be a global minimum?" That is certainly true if $N$ has rank $m$, as then the mapping $X mapsto M - NX$ is open and any local minimum of $f$ must be a local minimum of $rho$, which can only be $0$. For lower ranks the answer is less obvious.
$endgroup$
– Niels J. Diepeveen
May 29 '18 at 9:45
$begingroup$
The problem can be restated as follows: "if $1$ is a local minimum of $f$, must it also be a global minimum?" That is certainly true if $N$ has rank $m$, as then the mapping $X mapsto M - NX$ is open and any local minimum of $f$ must be a local minimum of $rho$, which can only be $0$. For lower ranks the answer is less obvious.
$endgroup$
– Niels J. Diepeveen
May 29 '18 at 9:45
$begingroup$
@NielsJ.Diepeveen: I don't quite understand your restated version. If it is true, i.e., $1$ is a global minimum, then $E = emptyset$ and it is certainly true $bar{E} neq F$. But I am assuming $E$ is not empty (this can be done by choosing appropriate $M,N$).
$endgroup$
– user1101010
May 29 '18 at 16:29
$begingroup$
@NielsJ.Diepeveen: I don't quite understand your restated version. If it is true, i.e., $1$ is a global minimum, then $E = emptyset$ and it is certainly true $bar{E} neq F$. But I am assuming $E$ is not empty (this can be done by choosing appropriate $M,N$).
$endgroup$
– user1101010
May 29 '18 at 16:29
$begingroup$
@NielsJ.Diepeveen: Thinking about the second part of your comment, I can see $X mapsto M-NX$ is open if $N$ has full rank, but could not follow why the local minimum of $f$ must be a local minimum of $rho(cdot)$. Sorry, I am having too many questions.
$endgroup$
– user1101010
May 29 '18 at 16:36
$begingroup$
@NielsJ.Diepeveen: Thinking about the second part of your comment, I can see $X mapsto M-NX$ is open if $N$ has full rank, but could not follow why the local minimum of $f$ must be a local minimum of $rho(cdot)$. Sorry, I am having too many questions.
$endgroup$
– user1101010
May 29 '18 at 16:36
|
show 5 more comments
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1
$begingroup$
If $N=0$ and $rho(M)=1$, then the LHS is empty but the RHS is the whole matrix space.
$endgroup$
– user1551
May 26 '18 at 10:17
$begingroup$
Actually I made the assumption $E$ is not empty. But sorry, the assumption was in the middle which made it not clear enough.
$endgroup$
– user1101010
May 26 '18 at 14:08
$begingroup$
The problem can be restated as follows: "if $1$ is a local minimum of $f$, must it also be a global minimum?" That is certainly true if $N$ has rank $m$, as then the mapping $X mapsto M - NX$ is open and any local minimum of $f$ must be a local minimum of $rho$, which can only be $0$. For lower ranks the answer is less obvious.
$endgroup$
– Niels J. Diepeveen
May 29 '18 at 9:45
$begingroup$
@NielsJ.Diepeveen: I don't quite understand your restated version. If it is true, i.e., $1$ is a global minimum, then $E = emptyset$ and it is certainly true $bar{E} neq F$. But I am assuming $E$ is not empty (this can be done by choosing appropriate $M,N$).
$endgroup$
– user1101010
May 29 '18 at 16:29
$begingroup$
@NielsJ.Diepeveen: Thinking about the second part of your comment, I can see $X mapsto M-NX$ is open if $N$ has full rank, but could not follow why the local minimum of $f$ must be a local minimum of $rho(cdot)$. Sorry, I am having too many questions.
$endgroup$
– user1101010
May 29 '18 at 16:36