For all matrices A exists unique matrix B with $B^2=A$ in a sufficient neighborhood.
$begingroup$
Show existence of $epsilon > 0, delta > 0$ such that for all $Ain U_epsilon(Id_n)$ exists unique $B in U_delta(Id_n)$ with $B^2=A$.
$U_epsilon(Id_n):={X in M_{ntimes n}| Vert X-Id_n Vert < epsilon; Vert cdot Vert text{ Operator norm}}$
Implicit function method: Define $F:(A,B) mapsto B^2-A$ and noticing that $(Id_n,Id_n)$ is a solution, and also $partial_{B}F|_{(Id_n,Id_n)}=2Id_n in GL(n)$. Impl.f.th. guarantees an existence of unique function $$g:U_epsilon(Id_n)to U_delta(Id_n)$$ such that $g(A)^2-A=0$. Which is equivalent to the statement if we set $B:=g(A)$.
Now I would like to show same using inverse function theorem.
Define $f(B)=B^2$, then its differential is $df_{Id_n}=2Id_n$ invertible.
Therefore we get open neighborhoods $U,V$ of $Id_n$ and $f(Id_n)=Id_n$ respectively with $f|_{U}:Uto V$ diffeomorphism.
In particular, bijectivity implies $forall Ain V: exists! B in U: f(B)=B^2=A$
Question: How to shrink $U,V$ to the form of open balls?
matrices multivariable-calculus implicit-function-theorem diffeomorphism
asked Jan 18 at 18:22
user3342072user3342072
400213
$endgroup$
add a comment |
$begingroup$
Show existence of $epsilon > 0, delta > 0$ such that for all $Ain U_epsilon(Id_n)$ exists unique $B in U_delta(Id_n)$ with $B^2=A$.
$U_epsilon(Id_n):={X in M_{ntimes n}| Vert X-Id_n Vert < epsilon; Vert cdot Vert text{ Operator norm}}$
Implicit function method: Define $F:(A,B) mapsto B^2-A$ and noticing that $(Id_n,Id_n)$ is a solution, and also $partial_{B}F|_{(Id_n,Id_n)}=2Id_n in GL(n)$. Impl.f.th. guarantees an existence of unique function $$g:U_epsilon(Id_n)to U_delta(Id_n)$$ such that $g(A)^2-A=0$. Which is equivalent to the statement if we set $B:=g(A)$.
Now I would like to show same using inverse function theorem.
Define $f(B)=B^2$, then its differential is $df_{Id_n}=2Id_n$ invertible.
Therefore we get open neighborhoods $U,V$ of $Id_n$ and $f(Id_n)=Id_n$ respectively with $f|_{U}:Uto V$ diffeomorphism.
In particular, bijectivity implies $forall Ain V: exists! B in U: f(B)=B^2=A$
Question: How to shrink $U,V$ to the form of open balls?
matrices multivariable-calculus implicit-function-theorem diffeomorphism
asked Jan 18 at 18:22
user3342072user3342072
400213
$endgroup$
add a comment |
$begingroup$
Show existence of $epsilon > 0, delta > 0$ such that for all $Ain U_epsilon(Id_n)$ exists unique $B in U_delta(Id_n)$ with $B^2=A$.
$U_epsilon(Id_n):={X in M_{ntimes n}| Vert X-Id_n Vert < epsilon; Vert cdot Vert text{ Operator norm}}$
Implicit function method: Define $F:(A,B) mapsto B^2-A$ and noticing that $(Id_n,Id_n)$ is a solution, and also $partial_{B}F|_{(Id_n,Id_n)}=2Id_n in GL(n)$. Impl.f.th. guarantees an existence of unique function $$g:U_epsilon(Id_n)to U_delta(Id_n)$$ such that $g(A)^2-A=0$. Which is equivalent to the statement if we set $B:=g(A)$.
Now I would like to show same using inverse function theorem.
Define $f(B)=B^2$, then its differential is $df_{Id_n}=2Id_n$ invertible.
Therefore we get open neighborhoods $U,V$ of $Id_n$ and $f(Id_n)=Id_n$ respectively with $f|_{U}:Uto V$ diffeomorphism.
In particular, bijectivity implies $forall Ain V: exists! B in U: f(B)=B^2=A$
Question: How to shrink $U,V$ to the form of open balls?
matrices multivariable-calculus implicit-function-theorem diffeomorphism
asked Jan 18 at 18:22
user3342072user3342072
400213
$endgroup$
Show existence of $epsilon > 0, delta > 0$ such that for all $Ain U_epsilon(Id_n)$ exists unique $B in U_delta(Id_n)$ with $B^2=A$.
$U_epsilon(Id_n):={X in M_{ntimes n}| Vert X-Id_n Vert < epsilon; Vert cdot Vert text{ Operator norm}}$
Implicit function method: Define $F:(A,B) mapsto B^2-A$ and noticing that $(Id_n,Id_n)$ is a solution, and also $partial_{B}F|_{(Id_n,Id_n)}=2Id_n in GL(n)$. Impl.f.th. guarantees an existence of unique function $$g:U_epsilon(Id_n)to U_delta(Id_n)$$ such that $g(A)^2-A=0$. Which is equivalent to the statement if we set $B:=g(A)$.
Now I would like to show same using inverse function theorem.
Define $f(B)=B^2$, then its differential is $df_{Id_n}=2Id_n$ invertible.
Therefore we get open neighborhoods $U,V$ of $Id_n$ and $f(Id_n)=Id_n$ respectively with $f|_{U}:Uto V$ diffeomorphism.
In particular, bijectivity implies $forall Ain V: exists! B in U: f(B)=B^2=A$
Question: How to shrink $U,V$ to the form of open balls?
matrices multivariable-calculus implicit-function-theorem diffeomorphism
matrices multivariable-calculus implicit-function-theorem diffeomorphism
asked Jan 18 at 18:22
user3342072user3342072
400213
asked Jan 18 at 18:22
user3342072user3342072
400213
edited Jan 18 at 18:57
user3342072
asked Jan 18 at 18:22
user3342072user3342072
400213
asked Jan 18 at 18:22
user3342072user3342072
400213
asked Jan 18 at 18:22
user3342072user3342072
400213
400213
add a comment |
add a comment |
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$begingroup$
Since $V$ is open, you can find a ball $B_1$ of radius $delta$ around the identity that is contained in $V$. Now, look at the preimage $X=f^{-1}(B_1)subseteq U$. It is open, since $f$ is continuous. Therefore you can find a ball $B_2$ of radius $epsilon$ around the identity that is contained into $f^{-1}(B_1)$ therefore into $U$. These should do the trick.
PS: I've used here that the open balls form a basis of your topology around $Id_n$.
answered Jan 18 at 19:39
J.FJ.F
33812
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$begingroup$
Since $V$ is open, you can find a ball $B_1$ of radius $delta$ around the identity that is contained in $V$. Now, look at the preimage $X=f^{-1}(B_1)subseteq U$. It is open, since $f$ is continuous. Therefore you can find a ball $B_2$ of radius $epsilon$ around the identity that is contained into $f^{-1}(B_1)$ therefore into $U$. These should do the trick.
PS: I've used here that the open balls form a basis of your topology around $Id_n$.
answered Jan 18 at 19:39
J.FJ.F
33812
$endgroup$
add a comment |
$begingroup$
Since $V$ is open, you can find a ball $B_1$ of radius $delta$ around the identity that is contained in $V$. Now, look at the preimage $X=f^{-1}(B_1)subseteq U$. It is open, since $f$ is continuous. Therefore you can find a ball $B_2$ of radius $epsilon$ around the identity that is contained into $f^{-1}(B_1)$ therefore into $U$. These should do the trick.
PS: I've used here that the open balls form a basis of your topology around $Id_n$.
answered Jan 18 at 19:39
J.FJ.F
33812
$endgroup$
add a comment |
$begingroup$
Since $V$ is open, you can find a ball $B_1$ of radius $delta$ around the identity that is contained in $V$. Now, look at the preimage $X=f^{-1}(B_1)subseteq U$. It is open, since $f$ is continuous. Therefore you can find a ball $B_2$ of radius $epsilon$ around the identity that is contained into $f^{-1}(B_1)$ therefore into $U$. These should do the trick.
PS: I've used here that the open balls form a basis of your topology around $Id_n$.
answered Jan 18 at 19:39
J.FJ.F
33812
$endgroup$
Since $V$ is open, you can find a ball $B_1$ of radius $delta$ around the identity that is contained in $V$. Now, look at the preimage $X=f^{-1}(B_1)subseteq U$. It is open, since $f$ is continuous. Therefore you can find a ball $B_2$ of radius $epsilon$ around the identity that is contained into $f^{-1}(B_1)$ therefore into $U$. These should do the trick.
PS: I've used here that the open balls form a basis of your topology around $Id_n$.
answered Jan 18 at 19:39
J.FJ.F
33812
answered Jan 18 at 19:39
J.FJ.F
33812
answered Jan 18 at 19:39
J.FJ.F
33812
answered Jan 18 at 19:39
J.FJ.F
33812
33812
add a comment |
add a comment |
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