Whether the set $A$ is a compact subset of $M_3(Bbb R)$.
$begingroup$
Consider $$X=Big{A in M_3(Bbb R): rho_A(x)=x^3-3x^2+2x-1Big}$$ where $rho_A(x)$ is the characteristic polynomial of $A$ and $M_3(Bbb R)$ is the space of all $3 times 3$ matrices over $Bbb R$.
Is $X$ compact in $M_3(Bbb R)$ ?
My try: I confused with only the $2x$ term in $rho_A(x)$. Because , if the $2x$ term not appear , then $X$ becomes $$X={A in M_3(Bbb R): text{trace}(A)=3,det A=1} $$ which is unbounded, since $$(forall n in Bbb N):begin{pmatrix} 1&0 & n\0&1&0\0&0&1 end{pmatrix} in X$$
But here the problem is, the appearance of $2$. I Know $$2=A_{11}+A_{22}+A_{33}$$
so I think in this case the set becomes bounded and closed
Any help?
general-topology
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,5072928
$endgroup$
add a comment |
$begingroup$
Consider $$X=Big{A in M_3(Bbb R): rho_A(x)=x^3-3x^2+2x-1Big}$$ where $rho_A(x)$ is the characteristic polynomial of $A$ and $M_3(Bbb R)$ is the space of all $3 times 3$ matrices over $Bbb R$.
Is $X$ compact in $M_3(Bbb R)$ ?
My try: I confused with only the $2x$ term in $rho_A(x)$. Because , if the $2x$ term not appear , then $X$ becomes $$X={A in M_3(Bbb R): text{trace}(A)=3,det A=1} $$ which is unbounded, since $$(forall n in Bbb N):begin{pmatrix} 1&0 & n\0&1&0\0&0&1 end{pmatrix} in X$$
But here the problem is, the appearance of $2$. I Know $$2=A_{11}+A_{22}+A_{33}$$
so I think in this case the set becomes bounded and closed
Any help?
general-topology
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,5072928
$endgroup$
$begingroup$
I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
$endgroup$
– copper.hat
Dec 10 '18 at 3:00
$begingroup$
If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
$endgroup$
– copper.hat
Dec 10 '18 at 3:13
add a comment |
$begingroup$
Consider $$X=Big{A in M_3(Bbb R): rho_A(x)=x^3-3x^2+2x-1Big}$$ where $rho_A(x)$ is the characteristic polynomial of $A$ and $M_3(Bbb R)$ is the space of all $3 times 3$ matrices over $Bbb R$.
Is $X$ compact in $M_3(Bbb R)$ ?
My try: I confused with only the $2x$ term in $rho_A(x)$. Because , if the $2x$ term not appear , then $X$ becomes $$X={A in M_3(Bbb R): text{trace}(A)=3,det A=1} $$ which is unbounded, since $$(forall n in Bbb N):begin{pmatrix} 1&0 & n\0&1&0\0&0&1 end{pmatrix} in X$$
But here the problem is, the appearance of $2$. I Know $$2=A_{11}+A_{22}+A_{33}$$
so I think in this case the set becomes bounded and closed
Any help?
general-topology
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,5072928
$endgroup$
Consider $$X=Big{A in M_3(Bbb R): rho_A(x)=x^3-3x^2+2x-1Big}$$ where $rho_A(x)$ is the characteristic polynomial of $A$ and $M_3(Bbb R)$ is the space of all $3 times 3$ matrices over $Bbb R$.
Is $X$ compact in $M_3(Bbb R)$ ?
My try: I confused with only the $2x$ term in $rho_A(x)$. Because , if the $2x$ term not appear , then $X$ becomes $$X={A in M_3(Bbb R): text{trace}(A)=3,det A=1} $$ which is unbounded, since $$(forall n in Bbb N):begin{pmatrix} 1&0 & n\0&1&0\0&0&1 end{pmatrix} in X$$
But here the problem is, the appearance of $2$. I Know $$2=A_{11}+A_{22}+A_{33}$$
so I think in this case the set becomes bounded and closed
Any help?
general-topology
general-topology
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,5072928
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,5072928
edited Dec 10 '18 at 2:53
Chinnapparaj R
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,5072928
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,5072928
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,5072928
5,5072928
$begingroup$
I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
$endgroup$
– copper.hat
Dec 10 '18 at 3:00
$begingroup$
If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
$endgroup$
– copper.hat
Dec 10 '18 at 3:13
add a comment |
$begingroup$
I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
$endgroup$
– copper.hat
Dec 10 '18 at 3:00
$begingroup$
If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
$endgroup$
– copper.hat
Dec 10 '18 at 3:13
$begingroup$
I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
$endgroup$
– copper.hat
Dec 10 '18 at 3:00
$begingroup$
I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
$endgroup$
– copper.hat
Dec 10 '18 at 3:00
$begingroup$
If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
$endgroup$
– copper.hat
Dec 10 '18 at 3:13
$begingroup$
If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
$endgroup$
– copper.hat
Dec 10 '18 at 3:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For example,
$$ pmatrix{t & 0 & 1cr 1-2t & 0 & -2cr -t^2+3t & 1 & 3-tcr} $$
has that characteristic polynomial. This is $S^{-1} A S$ where
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
$endgroup$
– user25959
Dec 10 '18 at 3:19
$begingroup$
@Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:21
1
$begingroup$
Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
$endgroup$
– Robert Israel
Dec 10 '18 at 3:22
$begingroup$
Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
$endgroup$
– Robert Israel
Dec 10 '18 at 3:25
$begingroup$
Thank you sir! :-)
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:28
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%2f3033372%2fwhether-the-set-a-is-a-compact-subset-of-m-3-bbb-r%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For example,
$$ pmatrix{t & 0 & 1cr 1-2t & 0 & -2cr -t^2+3t & 1 & 3-tcr} $$
has that characteristic polynomial. This is $S^{-1} A S$ where
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
$endgroup$
– user25959
Dec 10 '18 at 3:19
$begingroup$
@Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:21
1
$begingroup$
Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
$endgroup$
– Robert Israel
Dec 10 '18 at 3:22
$begingroup$
Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
$endgroup$
– Robert Israel
Dec 10 '18 at 3:25
$begingroup$
Thank you sir! :-)
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:28
add a comment |
$begingroup$
For example,
$$ pmatrix{t & 0 & 1cr 1-2t & 0 & -2cr -t^2+3t & 1 & 3-tcr} $$
has that characteristic polynomial. This is $S^{-1} A S$ where
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
$endgroup$
– user25959
Dec 10 '18 at 3:19
$begingroup$
@Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:21
1
$begingroup$
Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
$endgroup$
– Robert Israel
Dec 10 '18 at 3:22
$begingroup$
Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
$endgroup$
– Robert Israel
Dec 10 '18 at 3:25
$begingroup$
Thank you sir! :-)
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:28
add a comment |
$begingroup$
For example,
$$ pmatrix{t & 0 & 1cr 1-2t & 0 & -2cr -t^2+3t & 1 & 3-tcr} $$
has that characteristic polynomial. This is $S^{-1} A S$ where
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
324k23214468
$endgroup$
For example,
$$ pmatrix{t & 0 & 1cr 1-2t & 0 & -2cr -t^2+3t & 1 & 3-tcr} $$
has that characteristic polynomial. This is $S^{-1} A S$ where
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
324k23214468
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
324k23214468
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
324k23214468
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
324k23214468
324k23214468
$begingroup$
Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
$endgroup$
– user25959
Dec 10 '18 at 3:19
$begingroup$
@Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:21
1
$begingroup$
Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
$endgroup$
– Robert Israel
Dec 10 '18 at 3:22
$begingroup$
Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
$endgroup$
– Robert Israel
Dec 10 '18 at 3:25
$begingroup$
Thank you sir! :-)
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:28
add a comment |
$begingroup$
Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
$endgroup$
– user25959
Dec 10 '18 at 3:19
$begingroup$
@Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:21
1
$begingroup$
Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
$endgroup$
– Robert Israel
Dec 10 '18 at 3:22
$begingroup$
Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
$endgroup$
– Robert Israel
Dec 10 '18 at 3:25
$begingroup$
Thank you sir! :-)
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:28
$begingroup$
Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
$endgroup$
– user25959
Dec 10 '18 at 3:19
$begingroup$
Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
$endgroup$
– user25959
Dec 10 '18 at 3:19
$begingroup$
@Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:21
$begingroup$
@Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:21
1
1
$begingroup$
Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
$endgroup$
– Robert Israel
Dec 10 '18 at 3:22
$begingroup$
Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
$endgroup$
– Robert Israel
Dec 10 '18 at 3:22
$begingroup$
Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
$endgroup$
– Robert Israel
Dec 10 '18 at 3:25
$begingroup$
Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
$endgroup$
– Robert Israel
Dec 10 '18 at 3:25
$begingroup$
Thank you sir! :-)
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:28
$begingroup$
Thank you sir! :-)
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:28
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%2f3033372%2fwhether-the-set-a-is-a-compact-subset-of-m-3-bbb-r%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
$begingroup$
I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
$endgroup$
– copper.hat
Dec 10 '18 at 3:00
$begingroup$
If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
$endgroup$
– copper.hat
Dec 10 '18 at 3:13