Mixing
Is every matrix conjugate to its transpose in a continuous way?
$begingroup$
It is well-known that every square matrix is conjugate to its transpose. This means (in the case of real matrices) that, for each $ntimes n$ matrix $M$ with real entries, there is a matrix $S_Min GL(n,mathbb{R})$ such that ${S_M}^{-1}MS_M=M^T$. My question is: can you choose $S_M$ in such a way that it depends continuously on $M$? In other words:
Is there a continuous map $psicolon M_{n,n}(mathbb{R})longrightarrow GL(n,mathbb{R})$ such that$$bigl(forall Min M_{ntimes n}(mathbb{R})bigr):psi(M)^{-1}.M.psi(M)=M^T?$$
My guess is that the answer is negative even for $n=2$.
Note that, for each individual matrix $M$, there are plenty of choices for $S_M$. For instance, if $n=2$ and$$M=begin{bmatrix}x&y\z&tend{bmatrix},$$then you can take$$S_M=begin{bmatrix}az&bz\bz&bt-bx+ayend{bmatrix},$$with $a$ and $b$ chosen such that $det(S_M)neq0$ but, of course, this will only work if $zneq0$. What if $z=0$? Then you can take$$S_M=begin{bmatrix}-at+ax&ay\ay&byend{bmatrix}$$and, again, $a$ and $b$ should be chosen such that $det(S_M)neq0$; the problem now is that, of course, this will only work if $yneq0$. And so on. This looks like the problem of finding a logarithm for each $zinmathbb{C}setminus{0}$: there are plenty of choices for each individual $z$, but there is no continuous way of picking one.
linear-algebra matrices continuity
asked Dec 14 '18 at 10:03
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
It is well-known that every square matrix is conjugate to its transpose. This means (in the case of real matrices) that, for each $ntimes n$ matrix $M$ with real entries, there is a matrix $S_Min GL(n,mathbb{R})$ such that ${S_M}^{-1}MS_M=M^T$. My question is: can you choose $S_M$ in such a way that it depends continuously on $M$? In other words:
Is there a continuous map $psicolon M_{n,n}(mathbb{R})longrightarrow GL(n,mathbb{R})$ such that$$bigl(forall Min M_{ntimes n}(mathbb{R})bigr):psi(M)^{-1}.M.psi(M)=M^T?$$
My guess is that the answer is negative even for $n=2$.
Note that, for each individual matrix $M$, there are plenty of choices for $S_M$. For instance, if $n=2$ and$$M=begin{bmatrix}x&y\z&tend{bmatrix},$$then you can take$$S_M=begin{bmatrix}az&bz\bz&bt-bx+ayend{bmatrix},$$with $a$ and $b$ chosen such that $det(S_M)neq0$ but, of course, this will only work if $zneq0$. What if $z=0$? Then you can take$$S_M=begin{bmatrix}-at+ax&ay\ay&byend{bmatrix}$$and, again, $a$ and $b$ should be chosen such that $det(S_M)neq0$; the problem now is that, of course, this will only work if $yneq0$. And so on. This looks like the problem of finding a logarithm for each $zinmathbb{C}setminus{0}$: there are plenty of choices for each individual $z$, but there is no continuous way of picking one.
linear-algebra matrices continuity
asked Dec 14 '18 at 10:03
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
It is well-known that every square matrix is conjugate to its transpose. This means (in the case of real matrices) that, for each $ntimes n$ matrix $M$ with real entries, there is a matrix $S_Min GL(n,mathbb{R})$ such that ${S_M}^{-1}MS_M=M^T$. My question is: can you choose $S_M$ in such a way that it depends continuously on $M$? In other words:
Is there a continuous map $psicolon M_{n,n}(mathbb{R})longrightarrow GL(n,mathbb{R})$ such that$$bigl(forall Min M_{ntimes n}(mathbb{R})bigr):psi(M)^{-1}.M.psi(M)=M^T?$$
My guess is that the answer is negative even for $n=2$.
Note that, for each individual matrix $M$, there are plenty of choices for $S_M$. For instance, if $n=2$ and$$M=begin{bmatrix}x&y\z&tend{bmatrix},$$then you can take$$S_M=begin{bmatrix}az&bz\bz&bt-bx+ayend{bmatrix},$$with $a$ and $b$ chosen such that $det(S_M)neq0$ but, of course, this will only work if $zneq0$. What if $z=0$? Then you can take$$S_M=begin{bmatrix}-at+ax&ay\ay&byend{bmatrix}$$and, again, $a$ and $b$ should be chosen such that $det(S_M)neq0$; the problem now is that, of course, this will only work if $yneq0$. And so on. This looks like the problem of finding a logarithm for each $zinmathbb{C}setminus{0}$: there are plenty of choices for each individual $z$, but there is no continuous way of picking one.
linear-algebra matrices continuity
asked Dec 14 '18 at 10:03
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
It is well-known that every square matrix is conjugate to its transpose. This means (in the case of real matrices) that, for each $ntimes n$ matrix $M$ with real entries, there is a matrix $S_Min GL(n,mathbb{R})$ such that ${S_M}^{-1}MS_M=M^T$. My question is: can you choose $S_M$ in such a way that it depends continuously on $M$? In other words:
Is there a continuous map $psicolon M_{n,n}(mathbb{R})longrightarrow GL(n,mathbb{R})$ such that$$bigl(forall Min M_{ntimes n}(mathbb{R})bigr):psi(M)^{-1}.M.psi(M)=M^T?$$
My guess is that the answer is negative even for $n=2$.
Note that, for each individual matrix $M$, there are plenty of choices for $S_M$. For instance, if $n=2$ and$$M=begin{bmatrix}x&y\z&tend{bmatrix},$$then you can take$$S_M=begin{bmatrix}az&bz\bz&bt-bx+ayend{bmatrix},$$with $a$ and $b$ chosen such that $det(S_M)neq0$ but, of course, this will only work if $zneq0$. What if $z=0$? Then you can take$$S_M=begin{bmatrix}-at+ax&ay\ay&byend{bmatrix}$$and, again, $a$ and $b$ should be chosen such that $det(S_M)neq0$; the problem now is that, of course, this will only work if $yneq0$. And so on. This looks like the problem of finding a logarithm for each $zinmathbb{C}setminus{0}$: there are plenty of choices for each individual $z$, but there is no continuous way of picking one.
linear-algebra matrices continuity
linear-algebra matrices continuity
asked Dec 14 '18 at 10:03
José Carlos SantosJosé Carlos Santos
155k22124227
asked Dec 14 '18 at 10:03
José Carlos SantosJosé Carlos Santos
155k22124227
edited Jan 3 at 23:23
José Carlos Santos
asked Dec 14 '18 at 10:03
José Carlos SantosJosé Carlos Santos
155k22124227
asked Dec 14 '18 at 10:03
José Carlos SantosJosé Carlos Santos
155k22124227
asked Dec 14 '18 at 10:03
José Carlos SantosJosé Carlos Santos
155k22124227
155k22124227
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
My first thought is to look at a simple example - $2times 2$ rotation matrices.
Oops - rotation by $theta$ and rotation by $-theta$ are conjugate by any (real) reflection. No help there.
Second thought - OK, powers of something all work with the same $S$. So then, what happens at the identity? Which one do we choose? Actually, I can make those powers continuous by bringing in the matrix exponential.
Now we're ready. Consider a matrix $A$ with some eigenvalue $lambda$ of multiplicity $1$ and associated eigenvector $v$. $A^T$ has $lambda$ as an eigenvalue with multiplicity $1$ and associated eigenvector $w$. While we can't pin down $S(A)$ completely, we do know that $S(A)w=av$ for some nonzero $a$. This will also be true for any nonzero power of $A$, and for $exp(tA)$ for any nonzero $t$. As $tto 0$, we then have $S(I)w = lim_{tto 0}S(exp(tA))w=lim_{tto 0}a(exp(tA))v=cv$ for some $c$, possibly zero.
Almost there; all we need now are some concrete examples of what $v$ and $w$ can be. As it turns out, any pair of non-orthogonal nonzero real vectors are possible. Let $A$ be the rank-1 matrix $vw^T$, so $A^T=wv^T$. Then $Av=langle v,wrangle v$ and $A^Tw = langle v,wrangle w$, so these are the lone eigenvectors for the nonzero eigenvalue of $A$.
Combining these, $S(I)$ takes an arbitrary nonzero vector $w$ to something that's simultaneously a multiple of almost every nonzero vector $v$, which must be zero. That gives $S(I)=0$, an impossibility. By this contradiction, there is no way to choose $S$ continuously.
OK, I actually proved specifically that $S$ can't be continuous at the identity. Continuity elsewhere isn't ruled out yet.
answered Dec 14 '18 at 11:05
jmerryjmerry
3,998514
$endgroup$
add a comment |
$begingroup$
Consider the continuous function
$$
M_t=begin{cases}
tpmatrix{1&1\ 0&2}&text{ when } tge0,\
tpmatrix{1&0\ -1&2}&text{ when } t<0.
end{cases}
$$
It can be shown that all solutions to the equation $M_tS_t=S_tM_t^T$ are given by matrices of the form
$$
S_t=begin{cases}
pmatrix{a&b\ b&b}&text{ when } t>0,\
pmatrix{b&b\ b&a}&text{ when } t<0.
end{cases}
$$
It follows that if $S_t$ is chosen continuously, it must be in the form of $pmatrix{b&b\ b&b}$ at $t=0$ and hence it cannot remain non-singular.
answered Dec 14 '18 at 11:20
user1551user1551
72.3k566127
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039173%2fis-every-matrix-conjugate-to-its-transpose-in-a-continuous-way%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
My first thought is to look at a simple example - $2times 2$ rotation matrices.
Oops - rotation by $theta$ and rotation by $-theta$ are conjugate by any (real) reflection. No help there.
Second thought - OK, powers of something all work with the same $S$. So then, what happens at the identity? Which one do we choose? Actually, I can make those powers continuous by bringing in the matrix exponential.
Now we're ready. Consider a matrix $A$ with some eigenvalue $lambda$ of multiplicity $1$ and associated eigenvector $v$. $A^T$ has $lambda$ as an eigenvalue with multiplicity $1$ and associated eigenvector $w$. While we can't pin down $S(A)$ completely, we do know that $S(A)w=av$ for some nonzero $a$. This will also be true for any nonzero power of $A$, and for $exp(tA)$ for any nonzero $t$. As $tto 0$, we then have $S(I)w = lim_{tto 0}S(exp(tA))w=lim_{tto 0}a(exp(tA))v=cv$ for some $c$, possibly zero.
Almost there; all we need now are some concrete examples of what $v$ and $w$ can be. As it turns out, any pair of non-orthogonal nonzero real vectors are possible. Let $A$ be the rank-1 matrix $vw^T$, so $A^T=wv^T$. Then $Av=langle v,wrangle v$ and $A^Tw = langle v,wrangle w$, so these are the lone eigenvectors for the nonzero eigenvalue of $A$.
Combining these, $S(I)$ takes an arbitrary nonzero vector $w$ to something that's simultaneously a multiple of almost every nonzero vector $v$, which must be zero. That gives $S(I)=0$, an impossibility. By this contradiction, there is no way to choose $S$ continuously.
OK, I actually proved specifically that $S$ can't be continuous at the identity. Continuity elsewhere isn't ruled out yet.
answered Dec 14 '18 at 11:05
jmerryjmerry
3,998514
$endgroup$
add a comment |
$begingroup$
My first thought is to look at a simple example - $2times 2$ rotation matrices.
Oops - rotation by $theta$ and rotation by $-theta$ are conjugate by any (real) reflection. No help there.
Second thought - OK, powers of something all work with the same $S$. So then, what happens at the identity? Which one do we choose? Actually, I can make those powers continuous by bringing in the matrix exponential.
Now we're ready. Consider a matrix $A$ with some eigenvalue $lambda$ of multiplicity $1$ and associated eigenvector $v$. $A^T$ has $lambda$ as an eigenvalue with multiplicity $1$ and associated eigenvector $w$. While we can't pin down $S(A)$ completely, we do know that $S(A)w=av$ for some nonzero $a$. This will also be true for any nonzero power of $A$, and for $exp(tA)$ for any nonzero $t$. As $tto 0$, we then have $S(I)w = lim_{tto 0}S(exp(tA))w=lim_{tto 0}a(exp(tA))v=cv$ for some $c$, possibly zero.
Almost there; all we need now are some concrete examples of what $v$ and $w$ can be. As it turns out, any pair of non-orthogonal nonzero real vectors are possible. Let $A$ be the rank-1 matrix $vw^T$, so $A^T=wv^T$. Then $Av=langle v,wrangle v$ and $A^Tw = langle v,wrangle w$, so these are the lone eigenvectors for the nonzero eigenvalue of $A$.
Combining these, $S(I)$ takes an arbitrary nonzero vector $w$ to something that's simultaneously a multiple of almost every nonzero vector $v$, which must be zero. That gives $S(I)=0$, an impossibility. By this contradiction, there is no way to choose $S$ continuously.
OK, I actually proved specifically that $S$ can't be continuous at the identity. Continuity elsewhere isn't ruled out yet.
answered Dec 14 '18 at 11:05
jmerryjmerry
3,998514
$endgroup$
add a comment |
$begingroup$
My first thought is to look at a simple example - $2times 2$ rotation matrices.
Oops - rotation by $theta$ and rotation by $-theta$ are conjugate by any (real) reflection. No help there.
Second thought - OK, powers of something all work with the same $S$. So then, what happens at the identity? Which one do we choose? Actually, I can make those powers continuous by bringing in the matrix exponential.
Now we're ready. Consider a matrix $A$ with some eigenvalue $lambda$ of multiplicity $1$ and associated eigenvector $v$. $A^T$ has $lambda$ as an eigenvalue with multiplicity $1$ and associated eigenvector $w$. While we can't pin down $S(A)$ completely, we do know that $S(A)w=av$ for some nonzero $a$. This will also be true for any nonzero power of $A$, and for $exp(tA)$ for any nonzero $t$. As $tto 0$, we then have $S(I)w = lim_{tto 0}S(exp(tA))w=lim_{tto 0}a(exp(tA))v=cv$ for some $c$, possibly zero.
Almost there; all we need now are some concrete examples of what $v$ and $w$ can be. As it turns out, any pair of non-orthogonal nonzero real vectors are possible. Let $A$ be the rank-1 matrix $vw^T$, so $A^T=wv^T$. Then $Av=langle v,wrangle v$ and $A^Tw = langle v,wrangle w$, so these are the lone eigenvectors for the nonzero eigenvalue of $A$.
Combining these, $S(I)$ takes an arbitrary nonzero vector $w$ to something that's simultaneously a multiple of almost every nonzero vector $v$, which must be zero. That gives $S(I)=0$, an impossibility. By this contradiction, there is no way to choose $S$ continuously.
OK, I actually proved specifically that $S$ can't be continuous at the identity. Continuity elsewhere isn't ruled out yet.
answered Dec 14 '18 at 11:05
jmerryjmerry
3,998514
$endgroup$
My first thought is to look at a simple example - $2times 2$ rotation matrices.
Oops - rotation by $theta$ and rotation by $-theta$ are conjugate by any (real) reflection. No help there.
Second thought - OK, powers of something all work with the same $S$. So then, what happens at the identity? Which one do we choose? Actually, I can make those powers continuous by bringing in the matrix exponential.
Now we're ready. Consider a matrix $A$ with some eigenvalue $lambda$ of multiplicity $1$ and associated eigenvector $v$. $A^T$ has $lambda$ as an eigenvalue with multiplicity $1$ and associated eigenvector $w$. While we can't pin down $S(A)$ completely, we do know that $S(A)w=av$ for some nonzero $a$. This will also be true for any nonzero power of $A$, and for $exp(tA)$ for any nonzero $t$. As $tto 0$, we then have $S(I)w = lim_{tto 0}S(exp(tA))w=lim_{tto 0}a(exp(tA))v=cv$ for some $c$, possibly zero.
Almost there; all we need now are some concrete examples of what $v$ and $w$ can be. As it turns out, any pair of non-orthogonal nonzero real vectors are possible. Let $A$ be the rank-1 matrix $vw^T$, so $A^T=wv^T$. Then $Av=langle v,wrangle v$ and $A^Tw = langle v,wrangle w$, so these are the lone eigenvectors for the nonzero eigenvalue of $A$.
Combining these, $S(I)$ takes an arbitrary nonzero vector $w$ to something that's simultaneously a multiple of almost every nonzero vector $v$, which must be zero. That gives $S(I)=0$, an impossibility. By this contradiction, there is no way to choose $S$ continuously.
OK, I actually proved specifically that $S$ can't be continuous at the identity. Continuity elsewhere isn't ruled out yet.
answered Dec 14 '18 at 11:05
jmerryjmerry
3,998514
edited Dec 14 '18 at 11:21
answered Dec 14 '18 at 11:05
jmerryjmerry
3,998514
answered Dec 14 '18 at 11:05
jmerryjmerry
3,998514
answered Dec 14 '18 at 11:05
jmerryjmerry
3,998514
3,998514
add a comment |
add a comment |
$begingroup$
Consider the continuous function
$$
M_t=begin{cases}
tpmatrix{1&1\ 0&2}&text{ when } tge0,\
tpmatrix{1&0\ -1&2}&text{ when } t<0.
end{cases}
$$
It can be shown that all solutions to the equation $M_tS_t=S_tM_t^T$ are given by matrices of the form
$$
S_t=begin{cases}
pmatrix{a&b\ b&b}&text{ when } t>0,\
pmatrix{b&b\ b&a}&text{ when } t<0.
end{cases}
$$
It follows that if $S_t$ is chosen continuously, it must be in the form of $pmatrix{b&b\ b&b}$ at $t=0$ and hence it cannot remain non-singular.
answered Dec 14 '18 at 11:20
user1551user1551
72.3k566127
$endgroup$
add a comment |
$begingroup$
Consider the continuous function
$$
M_t=begin{cases}
tpmatrix{1&1\ 0&2}&text{ when } tge0,\
tpmatrix{1&0\ -1&2}&text{ when } t<0.
end{cases}
$$
It can be shown that all solutions to the equation $M_tS_t=S_tM_t^T$ are given by matrices of the form
$$
S_t=begin{cases}
pmatrix{a&b\ b&b}&text{ when } t>0,\
pmatrix{b&b\ b&a}&text{ when } t<0.
end{cases}
$$
It follows that if $S_t$ is chosen continuously, it must be in the form of $pmatrix{b&b\ b&b}$ at $t=0$ and hence it cannot remain non-singular.
answered Dec 14 '18 at 11:20
user1551user1551
72.3k566127
$endgroup$
add a comment |
$begingroup$
Consider the continuous function
$$
M_t=begin{cases}
tpmatrix{1&1\ 0&2}&text{ when } tge0,\
tpmatrix{1&0\ -1&2}&text{ when } t<0.
end{cases}
$$
It can be shown that all solutions to the equation $M_tS_t=S_tM_t^T$ are given by matrices of the form
$$
S_t=begin{cases}
pmatrix{a&b\ b&b}&text{ when } t>0,\
pmatrix{b&b\ b&a}&text{ when } t<0.
end{cases}
$$
It follows that if $S_t$ is chosen continuously, it must be in the form of $pmatrix{b&b\ b&b}$ at $t=0$ and hence it cannot remain non-singular.
answered Dec 14 '18 at 11:20
user1551user1551
72.3k566127
$endgroup$
Consider the continuous function
$$
M_t=begin{cases}
tpmatrix{1&1\ 0&2}&text{ when } tge0,\
tpmatrix{1&0\ -1&2}&text{ when } t<0.
end{cases}
$$
It can be shown that all solutions to the equation $M_tS_t=S_tM_t^T$ are given by matrices of the form
$$
S_t=begin{cases}
pmatrix{a&b\ b&b}&text{ when } t>0,\
pmatrix{b&b\ b&a}&text{ when } t<0.
end{cases}
$$
It follows that if $S_t$ is chosen continuously, it must be in the form of $pmatrix{b&b\ b&b}$ at $t=0$ and hence it cannot remain non-singular.
answered Dec 14 '18 at 11:20
user1551user1551
72.3k566127
edited Dec 14 '18 at 14:52
answered Dec 14 '18 at 11:20
user1551user1551
72.3k566127
answered Dec 14 '18 at 11:20
user1551user1551
72.3k566127
answered Dec 14 '18 at 11:20
user1551user1551
72.3k566127
72.3k566127
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039173%2fis-every-matrix-conjugate-to-its-transpose-in-a-continuous-way%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
