Is the row space of a matrix A a subspace A? If so, what are the objects of A?
$begingroup$
I have problems with an assertion that I read in a definition of the row space.
I hope somebody can help me :)
This part is clear:
Let A be a mxn-matrix, with rows $ r_{1},...,r_{m} in K^{n} $ The set of all possible linear combinations of $ r_{1},...,r_{m} $ is the row space of A.
Also the row space is a subspace of $K^{n}$.
My Problem:
In one definition that I have read it is stated that the row space of A is also a subspace of A. So the concrete Matrix A has to be a vector space. What are the objects of A?
Can somebody give me an interpretation of what it exactly means that the row space of A is a subspace of A?
PS: I'm not used to write about math in english, please ask if something doesn't makes sense to you.
matrices vector-spaces terminology
edited Jan 9 at 14:16
José Carlos Santos
155k22124227
asked Jan 9 at 14:00
CherryBlossom1878CherryBlossom1878
83
$endgroup$
add a comment |
$begingroup$
I have problems with an assertion that I read in a definition of the row space.
I hope somebody can help me :)
This part is clear:
Let A be a mxn-matrix, with rows $ r_{1},...,r_{m} in K^{n} $ The set of all possible linear combinations of $ r_{1},...,r_{m} $ is the row space of A.
Also the row space is a subspace of $K^{n}$.
My Problem:
In one definition that I have read it is stated that the row space of A is also a subspace of A. So the concrete Matrix A has to be a vector space. What are the objects of A?
Can somebody give me an interpretation of what it exactly means that the row space of A is a subspace of A?
PS: I'm not used to write about math in english, please ask if something doesn't makes sense to you.
matrices vector-spaces terminology
edited Jan 9 at 14:16
José Carlos Santos
155k22124227
asked Jan 9 at 14:00
CherryBlossom1878CherryBlossom1878
83
$endgroup$
$begingroup$
The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
$endgroup$
– David C. Ullrich
Jan 9 at 15:24
$begingroup$
sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
$endgroup$
– CherryBlossom1878
Jan 9 at 19:38
add a comment |
$begingroup$
I have problems with an assertion that I read in a definition of the row space.
I hope somebody can help me :)
This part is clear:
Let A be a mxn-matrix, with rows $ r_{1},...,r_{m} in K^{n} $ The set of all possible linear combinations of $ r_{1},...,r_{m} $ is the row space of A.
Also the row space is a subspace of $K^{n}$.
My Problem:
In one definition that I have read it is stated that the row space of A is also a subspace of A. So the concrete Matrix A has to be a vector space. What are the objects of A?
Can somebody give me an interpretation of what it exactly means that the row space of A is a subspace of A?
PS: I'm not used to write about math in english, please ask if something doesn't makes sense to you.
matrices vector-spaces terminology
edited Jan 9 at 14:16
José Carlos Santos
155k22124227
asked Jan 9 at 14:00
CherryBlossom1878CherryBlossom1878
83
$endgroup$
I have problems with an assertion that I read in a definition of the row space.
I hope somebody can help me :)
This part is clear:
Let A be a mxn-matrix, with rows $ r_{1},...,r_{m} in K^{n} $ The set of all possible linear combinations of $ r_{1},...,r_{m} $ is the row space of A.
Also the row space is a subspace of $K^{n}$.
My Problem:
In one definition that I have read it is stated that the row space of A is also a subspace of A. So the concrete Matrix A has to be a vector space. What are the objects of A?
Can somebody give me an interpretation of what it exactly means that the row space of A is a subspace of A?
PS: I'm not used to write about math in english, please ask if something doesn't makes sense to you.
matrices vector-spaces terminology
matrices vector-spaces terminology
edited Jan 9 at 14:16
José Carlos Santos
155k22124227
asked Jan 9 at 14:00
CherryBlossom1878CherryBlossom1878
83
edited Jan 9 at 14:16
José Carlos Santos
155k22124227
asked Jan 9 at 14:00
CherryBlossom1878CherryBlossom1878
83
edited Jan 9 at 14:16
José Carlos Santos
155k22124227
edited Jan 9 at 14:16
José Carlos Santos
155k22124227
edited Jan 9 at 14:16
José Carlos Santos
155k22124227
155k22124227
asked Jan 9 at 14:00
CherryBlossom1878CherryBlossom1878
83
asked Jan 9 at 14:00
CherryBlossom1878CherryBlossom1878
83
asked Jan 9 at 14:00
CherryBlossom1878CherryBlossom1878
83
83
$begingroup$
The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
$endgroup$
– David C. Ullrich
Jan 9 at 15:24
$begingroup$
sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
$endgroup$
– CherryBlossom1878
Jan 9 at 19:38
add a comment |
$begingroup$
The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
$endgroup$
– David C. Ullrich
Jan 9 at 15:24
$begingroup$
sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
$endgroup$
– CherryBlossom1878
Jan 9 at 19:38
$begingroup$
The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
$endgroup$
– David C. Ullrich
Jan 9 at 15:24
$begingroup$
The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
$endgroup$
– David C. Ullrich
Jan 9 at 15:24
$begingroup$
sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
$endgroup$
– CherryBlossom1878
Jan 9 at 19:38
$begingroup$
sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
$endgroup$
– CherryBlossom1878
Jan 9 at 19:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A matrix is not a vector space. Generally, it represents a linear transformation, from one vector space to another.
Now, since the number of columns always equals the dimension of the domain space, we can view the row space as a subspace of $V$ (where $T:Vto W$).
answered Jan 9 at 14:42
Chris CusterChris Custer
11.4k3824
$endgroup$
$begingroup$
Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
$endgroup$
– CherryBlossom1878
Jan 9 at 20:37
$begingroup$
It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
$endgroup$
– Chris Custer
Jan 9 at 20:41
$begingroup$
Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
$endgroup$
– CherryBlossom1878
Jan 9 at 20:54
$begingroup$
I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
$endgroup$
– Chris Custer
Jan 9 at 21:07
$begingroup$
I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
$endgroup$
– CherryBlossom1878
Jan 9 at 21:45
|
show 1 more comment
$begingroup$
No, a matrix is not a vector space. Asserting that the row space of a matrix is a subspace of that matrix is just an abuse of language.
answered Jan 9 at 14:15
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
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%2f3067475%2fis-the-row-space-of-a-matrix-a-a-subspace-a-if-so-what-are-the-objects-of-a%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
$begingroup$
A matrix is not a vector space. Generally, it represents a linear transformation, from one vector space to another.
Now, since the number of columns always equals the dimension of the domain space, we can view the row space as a subspace of $V$ (where $T:Vto W$).
answered Jan 9 at 14:42
Chris CusterChris Custer
11.4k3824
$endgroup$
$begingroup$
Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
$endgroup$
– CherryBlossom1878
Jan 9 at 20:37
$begingroup$
It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
$endgroup$
– Chris Custer
Jan 9 at 20:41
$begingroup$
Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
$endgroup$
– CherryBlossom1878
Jan 9 at 20:54
$begingroup$
I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
$endgroup$
– Chris Custer
Jan 9 at 21:07
$begingroup$
I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
$endgroup$
– CherryBlossom1878
Jan 9 at 21:45
|
show 1 more comment
$begingroup$
A matrix is not a vector space. Generally, it represents a linear transformation, from one vector space to another.
Now, since the number of columns always equals the dimension of the domain space, we can view the row space as a subspace of $V$ (where $T:Vto W$).
answered Jan 9 at 14:42
Chris CusterChris Custer
11.4k3824
$endgroup$
$begingroup$
Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
$endgroup$
– CherryBlossom1878
Jan 9 at 20:37
$begingroup$
It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
$endgroup$
– Chris Custer
Jan 9 at 20:41
$begingroup$
Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
$endgroup$
– CherryBlossom1878
Jan 9 at 20:54
$begingroup$
I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
$endgroup$
– Chris Custer
Jan 9 at 21:07
$begingroup$
I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
$endgroup$
– CherryBlossom1878
Jan 9 at 21:45
|
show 1 more comment
$begingroup$
A matrix is not a vector space. Generally, it represents a linear transformation, from one vector space to another.
Now, since the number of columns always equals the dimension of the domain space, we can view the row space as a subspace of $V$ (where $T:Vto W$).
answered Jan 9 at 14:42
Chris CusterChris Custer
11.4k3824
$endgroup$
A matrix is not a vector space. Generally, it represents a linear transformation, from one vector space to another.
Now, since the number of columns always equals the dimension of the domain space, we can view the row space as a subspace of $V$ (where $T:Vto W$).
answered Jan 9 at 14:42
Chris CusterChris Custer
11.4k3824
answered Jan 9 at 14:42
Chris CusterChris Custer
11.4k3824
answered Jan 9 at 14:42
Chris CusterChris Custer
11.4k3824
answered Jan 9 at 14:42
Chris CusterChris Custer
11.4k3824
11.4k3824
$begingroup$
Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
$endgroup$
– CherryBlossom1878
Jan 9 at 20:37
$begingroup$
It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
$endgroup$
– Chris Custer
Jan 9 at 20:41
$begingroup$
Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
$endgroup$
– CherryBlossom1878
Jan 9 at 20:54
$begingroup$
I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
$endgroup$
– Chris Custer
Jan 9 at 21:07
$begingroup$
I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
$endgroup$
– CherryBlossom1878
Jan 9 at 21:45
|
show 1 more comment
$begingroup$
Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
$endgroup$
– CherryBlossom1878
Jan 9 at 20:37
$begingroup$
It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
$endgroup$
– Chris Custer
Jan 9 at 20:41
$begingroup$
Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
$endgroup$
– CherryBlossom1878
Jan 9 at 20:54
$begingroup$
I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
$endgroup$
– Chris Custer
Jan 9 at 21:07
$begingroup$
I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
$endgroup$
– CherryBlossom1878
Jan 9 at 21:45
$begingroup$
Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
$endgroup$
– CherryBlossom1878
Jan 9 at 20:37
$begingroup$
Thank you for your answer :) If $f:Vrightarrow W $ is a linear transformation and I search for the transformation matrix A of f. Let $ B_{1}={v_{1},...,v_{n}}$ be the basis of V and $ B_{2}={w_{1},...,w_{m}} $ of W, then $f(v_{1})=w=a_{11}w_{1}+...+a_{m1}w_{m}$ with $a_{11},..,a_{m1}$ representing the elements of the first column of A. So if we continue we get n columns which is the dimension of the domain space V, but why should (for example) always $(a_{11},...,a_{1n})^{T}in V$ be true ?
$endgroup$
– CherryBlossom1878
Jan 9 at 20:37
$begingroup$
It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
$endgroup$
– Chris Custer
Jan 9 at 20:41
$begingroup$
It's always an $n$-tuple, of elements from the field, so it can be viewed as such.
$endgroup$
– Chris Custer
Jan 9 at 20:41
$begingroup$
Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
$endgroup$
– CherryBlossom1878
Jan 9 at 20:54
$begingroup$
Why is it always an n-tuple of elements from the field? If you aren't in the mood to explain it, I would also be happy about a link to a good explanation :)
$endgroup$
– CherryBlossom1878
Jan 9 at 20:54
$begingroup$
I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
$endgroup$
– Chris Custer
Jan 9 at 21:07
$begingroup$
I was just thinking I might refer you to Paul Halmos' book Finite Dimensional Vector Spaces. The reason though, that it's an $n$-tuple, where $n=operatorname{dim}V$, is simply that matrix multiplication won't work otherwise. Try a few examples.
$endgroup$
– Chris Custer
Jan 9 at 21:07
$begingroup$
I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
$endgroup$
– CherryBlossom1878
Jan 9 at 21:45
$begingroup$
I think that I understand it now, thank you again :) I will check if the book is availible in the library from my university. I think it would be good for my development to also read english literature about math.
$endgroup$
– CherryBlossom1878
Jan 9 at 21:45
|
show 1 more comment
$begingroup$
No, a matrix is not a vector space. Asserting that the row space of a matrix is a subspace of that matrix is just an abuse of language.
answered Jan 9 at 14:15
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
No, a matrix is not a vector space. Asserting that the row space of a matrix is a subspace of that matrix is just an abuse of language.
answered Jan 9 at 14:15
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
No, a matrix is not a vector space. Asserting that the row space of a matrix is a subspace of that matrix is just an abuse of language.
answered Jan 9 at 14:15
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
No, a matrix is not a vector space. Asserting that the row space of a matrix is a subspace of that matrix is just an abuse of language.
answered Jan 9 at 14:15
José Carlos SantosJosé Carlos Santos
155k22124227
answered Jan 9 at 14:15
José Carlos SantosJosé Carlos Santos
155k22124227
answered Jan 9 at 14:15
José Carlos SantosJosé Carlos Santos
155k22124227
answered Jan 9 at 14:15
José Carlos SantosJosé Carlos Santos
155k22124227
155k22124227
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%2f3067475%2fis-the-row-space-of-a-matrix-a-a-subspace-a-if-so-what-are-the-objects-of-a%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
$begingroup$
The phrase "subspace of $A$" is simply nonsense. Does the reference use exactly those words? If not you should tell us exactly what it does say - possibly you're misinterpreting or mistranslating sommething.
$endgroup$
– David C. Ullrich
Jan 9 at 15:24
$begingroup$
sry, I pressed "enter" too soon. The comment I wanted to post: I'm pretty sure that I translated "subspace of A" correctly but it seems like they(my source) made a mistake. I'm already satisfied with the answers (and your comment). Thank you :)
$endgroup$
– CherryBlossom1878
Jan 9 at 19:38