Mixing
Prove that $operatorname{SL}(n,R)$ is connected.
$begingroup$
Prove that $operatorname{SL}(n, R)$ is connected.
The problem is I know only topological groups from Munkres only. Again Just started fundamental groups. So if anyone can explain to me how it is true in a lucid language and in an easy way such that it remains in my boundary of knowledge then it would be a great help. I have mentioned what I know.
Again if tag this in the wrong field. Please forgive me.
linear-algebra general-topology topological-groups
edited Jan 23 at 13:31
user549397
1,5081418
asked Nov 7 '14 at 16:38
user152715user152715
1
$endgroup$
add a comment |
$begingroup$
Prove that $operatorname{SL}(n, R)$ is connected.
The problem is I know only topological groups from Munkres only. Again Just started fundamental groups. So if anyone can explain to me how it is true in a lucid language and in an easy way such that it remains in my boundary of knowledge then it would be a great help. I have mentioned what I know.
Again if tag this in the wrong field. Please forgive me.
linear-algebra general-topology topological-groups
edited Jan 23 at 13:31
user549397
1,5081418
asked Nov 7 '14 at 16:38
user152715user152715
1
$endgroup$
2
$begingroup$
Since you asked, my opinion is that you don't need the algebraic topology tag.
$endgroup$
– Matt Samuel
Nov 7 '14 at 16:40
$begingroup$
On the other hand a linear-algebra tag would be very appropriate since, as the answer of @mookid shows, this is a consequence of basic facts in basic linear algebra.
$endgroup$
– Lee Mosher
Nov 7 '14 at 20:56
add a comment |
$begingroup$
Prove that $operatorname{SL}(n, R)$ is connected.
The problem is I know only topological groups from Munkres only. Again Just started fundamental groups. So if anyone can explain to me how it is true in a lucid language and in an easy way such that it remains in my boundary of knowledge then it would be a great help. I have mentioned what I know.
Again if tag this in the wrong field. Please forgive me.
linear-algebra general-topology topological-groups
edited Jan 23 at 13:31
user549397
1,5081418
asked Nov 7 '14 at 16:38
user152715user152715
1
$endgroup$
Prove that $operatorname{SL}(n, R)$ is connected.
The problem is I know only topological groups from Munkres only. Again Just started fundamental groups. So if anyone can explain to me how it is true in a lucid language and in an easy way such that it remains in my boundary of knowledge then it would be a great help. I have mentioned what I know.
Again if tag this in the wrong field. Please forgive me.
linear-algebra general-topology topological-groups
linear-algebra general-topology topological-groups
edited Jan 23 at 13:31
user549397
1,5081418
asked Nov 7 '14 at 16:38
user152715user152715
1
edited Jan 23 at 13:31
user549397
1,5081418
asked Nov 7 '14 at 16:38
user152715user152715
1
edited Jan 23 at 13:31
user549397
1,5081418
edited Jan 23 at 13:31
user549397
1,5081418
edited Jan 23 at 13:31
user549397
1,5081418
1,5081418
asked Nov 7 '14 at 16:38
user152715user152715
1
asked Nov 7 '14 at 16:38
user152715user152715
1
asked Nov 7 '14 at 16:38
user152715user152715
1
1
2
$begingroup$
Since you asked, my opinion is that you don't need the algebraic topology tag.
$endgroup$
– Matt Samuel
Nov 7 '14 at 16:40
$begingroup$
On the other hand a linear-algebra tag would be very appropriate since, as the answer of @mookid shows, this is a consequence of basic facts in basic linear algebra.
$endgroup$
– Lee Mosher
Nov 7 '14 at 20:56
add a comment |
2
$begingroup$
Since you asked, my opinion is that you don't need the algebraic topology tag.
$endgroup$
– Matt Samuel
Nov 7 '14 at 16:40
$begingroup$
On the other hand a linear-algebra tag would be very appropriate since, as the answer of @mookid shows, this is a consequence of basic facts in basic linear algebra.
$endgroup$
– Lee Mosher
Nov 7 '14 at 20:56
2
2
$begingroup$
Since you asked, my opinion is that you don't need the algebraic topology tag.
$endgroup$
– Matt Samuel
Nov 7 '14 at 16:40
$begingroup$
Since you asked, my opinion is that you don't need the algebraic topology tag.
$endgroup$
– Matt Samuel
Nov 7 '14 at 16:40
$begingroup$
On the other hand a linear-algebra tag would be very appropriate since, as the answer of @mookid shows, this is a consequence of basic facts in basic linear algebra.
$endgroup$
– Lee Mosher
Nov 7 '14 at 20:56
$begingroup$
On the other hand a linear-algebra tag would be very appropriate since, as the answer of @mookid shows, this is a consequence of basic facts in basic linear algebra.
$endgroup$
– Lee Mosher
Nov 7 '14 at 20:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: prove that if two matrices can be transformed one into another using row-echelon transformation, then they are connected.
as we focus on elements of $SL_n$, we only need to prove that transvections $L_i to l_i + aL_j$ connect elements.
let $Ain SL_n$, $B$ is the image of $A$ under the transvection $L_i to L_i + aL_j$.
Then
$$
gamma: [0,1]to SL_n
$$defined by
"$gamma(t)$ is the image of $A$ under the transvection $L_i to L_i + taL_j$
" is continuous, and such as $gamma(0) = A$, $gamma(1) = B$ (also, check that for every $t$, $gamma (t)in SL_n$). Hence $A,B$ are path connected.
edited Apr 30 '17 at 3:32
tommy xu3
9791621
answered Nov 7 '14 at 16:40
mookidmookid
25.6k52547
$endgroup$
$begingroup$
How do I do that? Using topological group definition?
$endgroup$
– user152715
Nov 7 '14 at 16:48
1
$begingroup$
the transformations have explicit forms. Use the 'path-connected implies connected' implication.
$endgroup$
– mookid
Nov 7 '14 at 16:49
$begingroup$
can you please write this answer ilaborately then atleast I get an idea for this kind of questions.
$endgroup$
– user152715
Nov 7 '14 at 16:49
$begingroup$
Because I am new to deal with these questions. Thats why I am requesting you. Please it will be a great help.
$endgroup$
– user152715
Nov 7 '14 at 16:52
$begingroup$
@user152715 I wrote the hint part. Now it is your turn to finish the proof. Tell me if you have other issues.
$endgroup$
– mookid
Nov 7 '14 at 16:55
|
show 4 more comments
$begingroup$
Exercise 2.M.8(a) (Artin's Algebra, 2nd edition). The group $SL_n(mathbb{R})$ is generated by elementary matrices of the first type (see Exercise 2.4.8). Use this fact to prove that $SL_n(mathbb{R})$ is path-connected.
Let $sim$ be the binary operation corresponding to path-connectivity in $SL_n(mathbb{R})$; by my answer here, $sim$ is an equivalence relation.
In order to show $SL_n(mathbb{R})$ is path-connected, it suffices to show $Asim I_n$ for all $Ain SL_n(mathbb{R})$. But by my answer here, $A$ can be written as a (possibly empty) product of elementary matrices of the first type, so it in fact suffices to prove that$$E_{uv}(a)Msim M$$for all $Min SL_n(mathbb{R})$ and Type 1 elementary matrices $E_{uv}(a)$ ($1le u,,vle n$) of the form$$I_n + [a[(i,,j) = (u,,v)]]_{i,,j,=,1}^n.$$Yet$$Mto E_{uv}(b)M$$simply adds $b$ times row $j$ to row $i$, i.e. takes $r_i$ to $r_i+br_j$. For fixed $u$, $v$, $M$, this map is continuous in $b$ (and preserves the determinant), so the continuous function$$X(t) = E_{uv}(ta)M$$over $[0,1]$ takes$$X(0) = M to X(1) = E_{uv}(a)M$$while remaining inside $SL_n(mathbb{R})$, as desired.
edited Apr 13 '17 at 12:19
Community♦
1
$endgroup$
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%2f1010768%2fprove-that-operatornamesln-r-is-connected%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
$begingroup$
Hint: prove that if two matrices can be transformed one into another using row-echelon transformation, then they are connected.
as we focus on elements of $SL_n$, we only need to prove that transvections $L_i to l_i + aL_j$ connect elements.
let $Ain SL_n$, $B$ is the image of $A$ under the transvection $L_i to L_i + aL_j$.
Then
$$
gamma: [0,1]to SL_n
$$defined by
"$gamma(t)$ is the image of $A$ under the transvection $L_i to L_i + taL_j$
" is continuous, and such as $gamma(0) = A$, $gamma(1) = B$ (also, check that for every $t$, $gamma (t)in SL_n$). Hence $A,B$ are path connected.
edited Apr 30 '17 at 3:32
tommy xu3
9791621
answered Nov 7 '14 at 16:40
mookidmookid
25.6k52547
$endgroup$
$begingroup$
How do I do that? Using topological group definition?
$endgroup$
– user152715
Nov 7 '14 at 16:48
1
$begingroup$
the transformations have explicit forms. Use the 'path-connected implies connected' implication.
$endgroup$
– mookid
Nov 7 '14 at 16:49
$begingroup$
can you please write this answer ilaborately then atleast I get an idea for this kind of questions.
$endgroup$
– user152715
Nov 7 '14 at 16:49
$begingroup$
Because I am new to deal with these questions. Thats why I am requesting you. Please it will be a great help.
$endgroup$
– user152715
Nov 7 '14 at 16:52
$begingroup$
@user152715 I wrote the hint part. Now it is your turn to finish the proof. Tell me if you have other issues.
$endgroup$
– mookid
Nov 7 '14 at 16:55
|
show 4 more comments
$begingroup$
Hint: prove that if two matrices can be transformed one into another using row-echelon transformation, then they are connected.
as we focus on elements of $SL_n$, we only need to prove that transvections $L_i to l_i + aL_j$ connect elements.
let $Ain SL_n$, $B$ is the image of $A$ under the transvection $L_i to L_i + aL_j$.
Then
$$
gamma: [0,1]to SL_n
$$defined by
"$gamma(t)$ is the image of $A$ under the transvection $L_i to L_i + taL_j$
" is continuous, and such as $gamma(0) = A$, $gamma(1) = B$ (also, check that for every $t$, $gamma (t)in SL_n$). Hence $A,B$ are path connected.
edited Apr 30 '17 at 3:32
tommy xu3
9791621
answered Nov 7 '14 at 16:40
mookidmookid
25.6k52547
$endgroup$
$begingroup$
How do I do that? Using topological group definition?
$endgroup$
– user152715
Nov 7 '14 at 16:48
1
$begingroup$
the transformations have explicit forms. Use the 'path-connected implies connected' implication.
$endgroup$
– mookid
Nov 7 '14 at 16:49
$begingroup$
can you please write this answer ilaborately then atleast I get an idea for this kind of questions.
$endgroup$
– user152715
Nov 7 '14 at 16:49
$begingroup$
Because I am new to deal with these questions. Thats why I am requesting you. Please it will be a great help.
$endgroup$
– user152715
Nov 7 '14 at 16:52
$begingroup$
@user152715 I wrote the hint part. Now it is your turn to finish the proof. Tell me if you have other issues.
$endgroup$
– mookid
Nov 7 '14 at 16:55
|
show 4 more comments
$begingroup$
Hint: prove that if two matrices can be transformed one into another using row-echelon transformation, then they are connected.
as we focus on elements of $SL_n$, we only need to prove that transvections $L_i to l_i + aL_j$ connect elements.
let $Ain SL_n$, $B$ is the image of $A$ under the transvection $L_i to L_i + aL_j$.
Then
$$
gamma: [0,1]to SL_n
$$defined by
"$gamma(t)$ is the image of $A$ under the transvection $L_i to L_i + taL_j$
" is continuous, and such as $gamma(0) = A$, $gamma(1) = B$ (also, check that for every $t$, $gamma (t)in SL_n$). Hence $A,B$ are path connected.
edited Apr 30 '17 at 3:32
tommy xu3
9791621
answered Nov 7 '14 at 16:40
mookidmookid
25.6k52547
$endgroup$
Hint: prove that if two matrices can be transformed one into another using row-echelon transformation, then they are connected.
as we focus on elements of $SL_n$, we only need to prove that transvections $L_i to l_i + aL_j$ connect elements.
let $Ain SL_n$, $B$ is the image of $A$ under the transvection $L_i to L_i + aL_j$.
Then
$$
gamma: [0,1]to SL_n
$$defined by
"$gamma(t)$ is the image of $A$ under the transvection $L_i to L_i + taL_j$
" is continuous, and such as $gamma(0) = A$, $gamma(1) = B$ (also, check that for every $t$, $gamma (t)in SL_n$). Hence $A,B$ are path connected.
edited Apr 30 '17 at 3:32
tommy xu3
9791621
answered Nov 7 '14 at 16:40
mookidmookid
25.6k52547
edited Apr 30 '17 at 3:32
tommy xu3
9791621
edited Apr 30 '17 at 3:32
tommy xu3
9791621
edited Apr 30 '17 at 3:32
tommy xu3
9791621
9791621
answered Nov 7 '14 at 16:40
mookidmookid
25.6k52547
answered Nov 7 '14 at 16:40
mookidmookid
25.6k52547
answered Nov 7 '14 at 16:40
mookidmookid
25.6k52547
25.6k52547
$begingroup$
How do I do that? Using topological group definition?
$endgroup$
– user152715
Nov 7 '14 at 16:48
1
$begingroup$
the transformations have explicit forms. Use the 'path-connected implies connected' implication.
$endgroup$
– mookid
Nov 7 '14 at 16:49
$begingroup$
can you please write this answer ilaborately then atleast I get an idea for this kind of questions.
$endgroup$
– user152715
Nov 7 '14 at 16:49
$begingroup$
Because I am new to deal with these questions. Thats why I am requesting you. Please it will be a great help.
$endgroup$
– user152715
Nov 7 '14 at 16:52
$begingroup$
@user152715 I wrote the hint part. Now it is your turn to finish the proof. Tell me if you have other issues.
$endgroup$
– mookid
Nov 7 '14 at 16:55
|
show 4 more comments
$begingroup$
How do I do that? Using topological group definition?
$endgroup$
– user152715
Nov 7 '14 at 16:48
1
$begingroup$
the transformations have explicit forms. Use the 'path-connected implies connected' implication.
$endgroup$
– mookid
Nov 7 '14 at 16:49
$begingroup$
can you please write this answer ilaborately then atleast I get an idea for this kind of questions.
$endgroup$
– user152715
Nov 7 '14 at 16:49
$begingroup$
Because I am new to deal with these questions. Thats why I am requesting you. Please it will be a great help.
$endgroup$
– user152715
Nov 7 '14 at 16:52
$begingroup$
@user152715 I wrote the hint part. Now it is your turn to finish the proof. Tell me if you have other issues.
$endgroup$
– mookid
Nov 7 '14 at 16:55
$begingroup$
How do I do that? Using topological group definition?
$endgroup$
– user152715
Nov 7 '14 at 16:48
$begingroup$
How do I do that? Using topological group definition?
$endgroup$
– user152715
Nov 7 '14 at 16:48
1
1
$begingroup$
the transformations have explicit forms. Use the 'path-connected implies connected' implication.
$endgroup$
– mookid
Nov 7 '14 at 16:49
$begingroup$
the transformations have explicit forms. Use the 'path-connected implies connected' implication.
$endgroup$
– mookid
Nov 7 '14 at 16:49
$begingroup$
can you please write this answer ilaborately then atleast I get an idea for this kind of questions.
$endgroup$
– user152715
Nov 7 '14 at 16:49
$begingroup$
can you please write this answer ilaborately then atleast I get an idea for this kind of questions.
$endgroup$
– user152715
Nov 7 '14 at 16:49
$begingroup$
Because I am new to deal with these questions. Thats why I am requesting you. Please it will be a great help.
$endgroup$
– user152715
Nov 7 '14 at 16:52
$begingroup$
Because I am new to deal with these questions. Thats why I am requesting you. Please it will be a great help.
$endgroup$
– user152715
Nov 7 '14 at 16:52
$begingroup$
@user152715 I wrote the hint part. Now it is your turn to finish the proof. Tell me if you have other issues.
$endgroup$
– mookid
Nov 7 '14 at 16:55
$begingroup$
@user152715 I wrote the hint part. Now it is your turn to finish the proof. Tell me if you have other issues.
$endgroup$
– mookid
Nov 7 '14 at 16:55
|
show 4 more comments
$begingroup$
Exercise 2.M.8(a) (Artin's Algebra, 2nd edition). The group $SL_n(mathbb{R})$ is generated by elementary matrices of the first type (see Exercise 2.4.8). Use this fact to prove that $SL_n(mathbb{R})$ is path-connected.
Let $sim$ be the binary operation corresponding to path-connectivity in $SL_n(mathbb{R})$; by my answer here, $sim$ is an equivalence relation.
In order to show $SL_n(mathbb{R})$ is path-connected, it suffices to show $Asim I_n$ for all $Ain SL_n(mathbb{R})$. But by my answer here, $A$ can be written as a (possibly empty) product of elementary matrices of the first type, so it in fact suffices to prove that$$E_{uv}(a)Msim M$$for all $Min SL_n(mathbb{R})$ and Type 1 elementary matrices $E_{uv}(a)$ ($1le u,,vle n$) of the form$$I_n + [a[(i,,j) = (u,,v)]]_{i,,j,=,1}^n.$$Yet$$Mto E_{uv}(b)M$$simply adds $b$ times row $j$ to row $i$, i.e. takes $r_i$ to $r_i+br_j$. For fixed $u$, $v$, $M$, this map is continuous in $b$ (and preserves the determinant), so the continuous function$$X(t) = E_{uv}(ta)M$$over $[0,1]$ takes$$X(0) = M to X(1) = E_{uv}(a)M$$while remaining inside $SL_n(mathbb{R})$, as desired.
edited Apr 13 '17 at 12:19
Community♦
1
$endgroup$
add a comment |
$begingroup$
Exercise 2.M.8(a) (Artin's Algebra, 2nd edition). The group $SL_n(mathbb{R})$ is generated by elementary matrices of the first type (see Exercise 2.4.8). Use this fact to prove that $SL_n(mathbb{R})$ is path-connected.
Let $sim$ be the binary operation corresponding to path-connectivity in $SL_n(mathbb{R})$; by my answer here, $sim$ is an equivalence relation.
In order to show $SL_n(mathbb{R})$ is path-connected, it suffices to show $Asim I_n$ for all $Ain SL_n(mathbb{R})$. But by my answer here, $A$ can be written as a (possibly empty) product of elementary matrices of the first type, so it in fact suffices to prove that$$E_{uv}(a)Msim M$$for all $Min SL_n(mathbb{R})$ and Type 1 elementary matrices $E_{uv}(a)$ ($1le u,,vle n$) of the form$$I_n + [a[(i,,j) = (u,,v)]]_{i,,j,=,1}^n.$$Yet$$Mto E_{uv}(b)M$$simply adds $b$ times row $j$ to row $i$, i.e. takes $r_i$ to $r_i+br_j$. For fixed $u$, $v$, $M$, this map is continuous in $b$ (and preserves the determinant), so the continuous function$$X(t) = E_{uv}(ta)M$$over $[0,1]$ takes$$X(0) = M to X(1) = E_{uv}(a)M$$while remaining inside $SL_n(mathbb{R})$, as desired.
edited Apr 13 '17 at 12:19
Community♦
1
$endgroup$
add a comment |
$begingroup$
Exercise 2.M.8(a) (Artin's Algebra, 2nd edition). The group $SL_n(mathbb{R})$ is generated by elementary matrices of the first type (see Exercise 2.4.8). Use this fact to prove that $SL_n(mathbb{R})$ is path-connected.
Let $sim$ be the binary operation corresponding to path-connectivity in $SL_n(mathbb{R})$; by my answer here, $sim$ is an equivalence relation.
In order to show $SL_n(mathbb{R})$ is path-connected, it suffices to show $Asim I_n$ for all $Ain SL_n(mathbb{R})$. But by my answer here, $A$ can be written as a (possibly empty) product of elementary matrices of the first type, so it in fact suffices to prove that$$E_{uv}(a)Msim M$$for all $Min SL_n(mathbb{R})$ and Type 1 elementary matrices $E_{uv}(a)$ ($1le u,,vle n$) of the form$$I_n + [a[(i,,j) = (u,,v)]]_{i,,j,=,1}^n.$$Yet$$Mto E_{uv}(b)M$$simply adds $b$ times row $j$ to row $i$, i.e. takes $r_i$ to $r_i+br_j$. For fixed $u$, $v$, $M$, this map is continuous in $b$ (and preserves the determinant), so the continuous function$$X(t) = E_{uv}(ta)M$$over $[0,1]$ takes$$X(0) = M to X(1) = E_{uv}(a)M$$while remaining inside $SL_n(mathbb{R})$, as desired.
edited Apr 13 '17 at 12:19
Community♦
1
$endgroup$
Exercise 2.M.8(a) (Artin's Algebra, 2nd edition). The group $SL_n(mathbb{R})$ is generated by elementary matrices of the first type (see Exercise 2.4.8). Use this fact to prove that $SL_n(mathbb{R})$ is path-connected.
Let $sim$ be the binary operation corresponding to path-connectivity in $SL_n(mathbb{R})$; by my answer here, $sim$ is an equivalence relation.
In order to show $SL_n(mathbb{R})$ is path-connected, it suffices to show $Asim I_n$ for all $Ain SL_n(mathbb{R})$. But by my answer here, $A$ can be written as a (possibly empty) product of elementary matrices of the first type, so it in fact suffices to prove that$$E_{uv}(a)Msim M$$for all $Min SL_n(mathbb{R})$ and Type 1 elementary matrices $E_{uv}(a)$ ($1le u,,vle n$) of the form$$I_n + [a[(i,,j) = (u,,v)]]_{i,,j,=,1}^n.$$Yet$$Mto E_{uv}(b)M$$simply adds $b$ times row $j$ to row $i$, i.e. takes $r_i$ to $r_i+br_j$. For fixed $u$, $v$, $M$, this map is continuous in $b$ (and preserves the determinant), so the continuous function$$X(t) = E_{uv}(ta)M$$over $[0,1]$ takes$$X(0) = M to X(1) = E_{uv}(a)M$$while remaining inside $SL_n(mathbb{R})$, as desired.
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
answered Sep 24 '15 at 15:57
user203482
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%2f1010768%2fprove-that-operatornamesln-r-is-connected%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
2
$begingroup$
Since you asked, my opinion is that you don't need the algebraic topology tag.
$endgroup$
– Matt Samuel
Nov 7 '14 at 16:40
$begingroup$
On the other hand a linear-algebra tag would be very appropriate since, as the answer of @mookid shows, this is a consequence of basic facts in basic linear algebra.
$endgroup$
– Lee Mosher
Nov 7 '14 at 20:56