Clarification about the axioms of a subspaceIs $S=(a+b,a+c,2c)mid a,b,cin mathbbR$ a subspace of $mathbbR^3$?Prove that one of the following sets is a subspace and the other isn't?Is $U=(r,0,s)mid r^2+s^2=0, r,sin mathbbR$ a subspace of $mathbbR^3$?necessary condition for subspace of a vector spaceDetermine whether the set S (below) is a subspace of $M_2(mathbbR)$ (The space f all 2x2 matrices with real entities)How to verify that $Ax =3x $ is a subspace of $R^n$Proof that something is a subspace given it's a subset of a vector space.If a vector subspace contains the zero vector does it follow that there is an additive inverse as well?zero vector subspaceProve that $W_1$ is a subspace of $mathbbR^n$.
Calculate the frequency of characters in a string
What can I do if I am asked to learn different programming languages very frequently?
Relation between independence and correlation of uniform random variables
Does the attack bonus from a Masterwork weapon stack with the attack bonus from Masterwork ammunition?
Is this an example of a Neapolitan chord?
Can you move over difficult terrain with only 5 feet of movement?
What favor did Moody owe Dumbledore?
Help rendering a complicated sum/product formula
What is the significance behind "40 days" that often appears in the Bible?
Light propagating through a sound wave
How to define limit operations in general topological spaces? Are nets able to do this?
Are dual Irish/British citizens bound by the 90/180 day rule when travelling in the EU after Brexit?
PTIJ What is the inyan of the Konami code in Uncle Moishy's song?
What (if any) is the reason to buy in small local stores?
Do I need to be arrogant to get ahead?
Deletion of copy-ctor & copy-assignment - public, private or protected?
Geography in 3D perspective
How does one measure the Fourier components of a signal?
Optimising a list searching algorithm
What is the plural TO / OF something
How is the partial sum of a geometric sequence calculated?
Practical application of matrices and determinants
In the 1924 version of The Thief of Bagdad, no character is named, right?
Is honey really a supersaturated solution? Does heating to un-crystalize redissolve it or melt it?
Clarification about the axioms of a subspace
Is $S=(a+b,a+c,2c)mid a,b,cin mathbbR$ a subspace of $mathbbR^3$?Prove that one of the following sets is a subspace and the other isn't?Is $U=(r,0,s)mid r^2+s^2=0, r,sin mathbbR$ a subspace of $mathbbR^3$?necessary condition for subspace of a vector spaceDetermine whether the set S (below) is a subspace of $M_2(mathbbR)$ (The space f all 2x2 matrices with real entities)How to verify that $Ax =3x $ is a subspace of $R^n$Proof that something is a subspace given it's a subset of a vector space.If a vector subspace contains the zero vector does it follow that there is an additive inverse as well?zero vector subspaceProve that $W_1$ is a subspace of $mathbbR^n$.
$begingroup$
I'm asked to verify if:
$B=(x,y,z) in mathbbR^3 : ||(x,y,z)|| leq 1$
is a subspace. I know I have to check for the zero vector, addition and scalar multiplication. Here lies my question however.
If I use $(0,0,0)$ to check if the zero vector exists, it clearly does. However, if I multiply $(0,0,1)$ by some scalar, say, $50$, scalar multiplication does not hold so this isn't a subspace.
Here lies my problem. I thought because the $0$ vector exists, this set MUST be a subspace and hence, scalar multiplication and addition should hold.
Am I wrong in assuming that if $0$ vector exists, all the other axioms must hold? Or is it possible that some axioms work and some fail?
linear-algebra vector-spaces
asked Mar 12 at 19:23
Future Math personFuture Math person
993817
$endgroup$
add a comment |
$begingroup$
I'm asked to verify if:
$B=(x,y,z) in mathbbR^3 : ||(x,y,z)|| leq 1$
is a subspace. I know I have to check for the zero vector, addition and scalar multiplication. Here lies my question however.
If I use $(0,0,0)$ to check if the zero vector exists, it clearly does. However, if I multiply $(0,0,1)$ by some scalar, say, $50$, scalar multiplication does not hold so this isn't a subspace.
Here lies my problem. I thought because the $0$ vector exists, this set MUST be a subspace and hence, scalar multiplication and addition should hold.
Am I wrong in assuming that if $0$ vector exists, all the other axioms must hold? Or is it possible that some axioms work and some fail?
linear-algebra vector-spaces
asked Mar 12 at 19:23
Future Math personFuture Math person
993817
$endgroup$
$begingroup$
Just because a subset contains $0$ doesn’t mean it is a subspace
$endgroup$
– J. W. Tanner
Mar 12 at 19:27
$begingroup$
@J.W.Tanner But why? I thought the zero vector automatically implied a subspace since addition amd scalar multiplication must hold true as a result.
$endgroup$
– Future Math person
Mar 12 at 19:28
$begingroup$
see the answers below
$endgroup$
– J. W. Tanner
Mar 12 at 19:51
add a comment |
$begingroup$
I'm asked to verify if:
$B=(x,y,z) in mathbbR^3 : ||(x,y,z)|| leq 1$
is a subspace. I know I have to check for the zero vector, addition and scalar multiplication. Here lies my question however.
If I use $(0,0,0)$ to check if the zero vector exists, it clearly does. However, if I multiply $(0,0,1)$ by some scalar, say, $50$, scalar multiplication does not hold so this isn't a subspace.
Here lies my problem. I thought because the $0$ vector exists, this set MUST be a subspace and hence, scalar multiplication and addition should hold.
Am I wrong in assuming that if $0$ vector exists, all the other axioms must hold? Or is it possible that some axioms work and some fail?
linear-algebra vector-spaces
asked Mar 12 at 19:23
Future Math personFuture Math person
993817
$endgroup$
I'm asked to verify if:
$B=(x,y,z) in mathbbR^3 : ||(x,y,z)|| leq 1$
is a subspace. I know I have to check for the zero vector, addition and scalar multiplication. Here lies my question however.
If I use $(0,0,0)$ to check if the zero vector exists, it clearly does. However, if I multiply $(0,0,1)$ by some scalar, say, $50$, scalar multiplication does not hold so this isn't a subspace.
Here lies my problem. I thought because the $0$ vector exists, this set MUST be a subspace and hence, scalar multiplication and addition should hold.
Am I wrong in assuming that if $0$ vector exists, all the other axioms must hold? Or is it possible that some axioms work and some fail?
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Mar 12 at 19:23
Future Math personFuture Math person
993817
asked Mar 12 at 19:23
Future Math personFuture Math person
993817
asked Mar 12 at 19:23
Future Math personFuture Math person
993817
asked Mar 12 at 19:23
Future Math personFuture Math person
993817
asked Mar 12 at 19:23
Future Math personFuture Math person
993817
993817
$begingroup$
Just because a subset contains $0$ doesn’t mean it is a subspace
$endgroup$
– J. W. Tanner
Mar 12 at 19:27
$begingroup$
@J.W.Tanner But why? I thought the zero vector automatically implied a subspace since addition amd scalar multiplication must hold true as a result.
$endgroup$
– Future Math person
Mar 12 at 19:28
$begingroup$
see the answers below
$endgroup$
– J. W. Tanner
Mar 12 at 19:51
add a comment |
$begingroup$
Just because a subset contains $0$ doesn’t mean it is a subspace
$endgroup$
– J. W. Tanner
Mar 12 at 19:27
$begingroup$
@J.W.Tanner But why? I thought the zero vector automatically implied a subspace since addition amd scalar multiplication must hold true as a result.
$endgroup$
– Future Math person
Mar 12 at 19:28
$begingroup$
see the answers below
$endgroup$
– J. W. Tanner
Mar 12 at 19:51
$begingroup$
Just because a subset contains $0$ doesn’t mean it is a subspace
$endgroup$
– J. W. Tanner
Mar 12 at 19:27
$begingroup$
Just because a subset contains $0$ doesn’t mean it is a subspace
$endgroup$
– J. W. Tanner
Mar 12 at 19:27
$begingroup$
@J.W.Tanner But why? I thought the zero vector automatically implied a subspace since addition amd scalar multiplication must hold true as a result.
$endgroup$
– Future Math person
Mar 12 at 19:28
$begingroup$
@J.W.Tanner But why? I thought the zero vector automatically implied a subspace since addition amd scalar multiplication must hold true as a result.
$endgroup$
– Future Math person
Mar 12 at 19:28
$begingroup$
see the answers below
$endgroup$
– J. W. Tanner
Mar 12 at 19:51
$begingroup$
see the answers below
$endgroup$
– J. W. Tanner
Mar 12 at 19:51
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You have the idea somewhat reversed. If the space is closed under addition and scalar multiplication, then you need to have the zero vector since $$0cdot (x,y,z)=(0,0,0)$$ and $$(x,y,z)-(x,y,z)=(0,0,0)$$
So in fact the crucial axioms are addition and scalar multiplication. The zero vector comes for free
Edit thanks to comment: The zero vector comes for free after checking the set is non-empty. Just so happens that it is often easy to see that the zero vector is in the space, hence non-empty.So this whole discussion kind of chicken and egg
answered Mar 12 at 19:30
NazimJNazimJ
45218
$endgroup$
2
$begingroup$
Yes, but note that if you don't want to check that the zero vector is there you need at least to check that the subset is not empty. (otherwise there is no meaning to addition and scalar multiplication)
$endgroup$
– Mark
Mar 12 at 19:32
add a comment |
$begingroup$
Of course just checking that the zero vector is there is not enough. Just think about it-how can the existence of a zero vector imply that the subset is closed under addition and scalar multiplication? For example, is $0,1$ a subspace of $mathbbR$? Clearly not. To check if $Ssubseteq V$ is a subspace you need to check the following:
$1. $ The zero vector is in $S$. (or instead of that you can check that $S$ is not empty).
$2. $ $S$ is closed under addition and scalar multiplication.
answered Mar 12 at 19:30
MarkMark
10.2k622
$endgroup$
add a comment |
$begingroup$
Am I wrong in assuming that if $0$ vector exists, all the other axioms must hold?
In short, yes. Why else would the other axioms be there?
Technically, those aren't axioms. You have axioms describing a vector space, and given a vector space there are certain subsets of that vector space which are themselves vector spaces, using the same operations. Those are what we call subspaces.
So subspaces actually need to fulfill all the same axioms that all other vector spaces do. The checklist you're using (which you call axioms) isn't as long as the full list of vector space axioms, however. The reason is that inheriting the operations from something we know is a vector space gives us a lot for free.
answered Mar 12 at 19:32
ArthurArthur
118k7118201
$endgroup$
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%2f3145554%2fclarification-about-the-axioms-of-a-subspace%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have the idea somewhat reversed. If the space is closed under addition and scalar multiplication, then you need to have the zero vector since $$0cdot (x,y,z)=(0,0,0)$$ and $$(x,y,z)-(x,y,z)=(0,0,0)$$
So in fact the crucial axioms are addition and scalar multiplication. The zero vector comes for free
Edit thanks to comment: The zero vector comes for free after checking the set is non-empty. Just so happens that it is often easy to see that the zero vector is in the space, hence non-empty.So this whole discussion kind of chicken and egg
answered Mar 12 at 19:30
NazimJNazimJ
45218
$endgroup$
2
$begingroup$
Yes, but note that if you don't want to check that the zero vector is there you need at least to check that the subset is not empty. (otherwise there is no meaning to addition and scalar multiplication)
$endgroup$
– Mark
Mar 12 at 19:32
add a comment |
$begingroup$
You have the idea somewhat reversed. If the space is closed under addition and scalar multiplication, then you need to have the zero vector since $$0cdot (x,y,z)=(0,0,0)$$ and $$(x,y,z)-(x,y,z)=(0,0,0)$$
So in fact the crucial axioms are addition and scalar multiplication. The zero vector comes for free
Edit thanks to comment: The zero vector comes for free after checking the set is non-empty. Just so happens that it is often easy to see that the zero vector is in the space, hence non-empty.So this whole discussion kind of chicken and egg
answered Mar 12 at 19:30
NazimJNazimJ
45218
$endgroup$
2
$begingroup$
Yes, but note that if you don't want to check that the zero vector is there you need at least to check that the subset is not empty. (otherwise there is no meaning to addition and scalar multiplication)
$endgroup$
– Mark
Mar 12 at 19:32
add a comment |
$begingroup$
You have the idea somewhat reversed. If the space is closed under addition and scalar multiplication, then you need to have the zero vector since $$0cdot (x,y,z)=(0,0,0)$$ and $$(x,y,z)-(x,y,z)=(0,0,0)$$
So in fact the crucial axioms are addition and scalar multiplication. The zero vector comes for free
Edit thanks to comment: The zero vector comes for free after checking the set is non-empty. Just so happens that it is often easy to see that the zero vector is in the space, hence non-empty.So this whole discussion kind of chicken and egg
answered Mar 12 at 19:30
NazimJNazimJ
45218
$endgroup$
You have the idea somewhat reversed. If the space is closed under addition and scalar multiplication, then you need to have the zero vector since $$0cdot (x,y,z)=(0,0,0)$$ and $$(x,y,z)-(x,y,z)=(0,0,0)$$
So in fact the crucial axioms are addition and scalar multiplication. The zero vector comes for free
Edit thanks to comment: The zero vector comes for free after checking the set is non-empty. Just so happens that it is often easy to see that the zero vector is in the space, hence non-empty.So this whole discussion kind of chicken and egg
answered Mar 12 at 19:30
NazimJNazimJ
45218
edited Mar 12 at 19:39
answered Mar 12 at 19:30
NazimJNazimJ
45218
answered Mar 12 at 19:30
NazimJNazimJ
45218
answered Mar 12 at 19:30
NazimJNazimJ
45218
45218
2
$begingroup$
Yes, but note that if you don't want to check that the zero vector is there you need at least to check that the subset is not empty. (otherwise there is no meaning to addition and scalar multiplication)
$endgroup$
– Mark
Mar 12 at 19:32
add a comment |
2
$begingroup$
Yes, but note that if you don't want to check that the zero vector is there you need at least to check that the subset is not empty. (otherwise there is no meaning to addition and scalar multiplication)
$endgroup$
– Mark
Mar 12 at 19:32
2
2
$begingroup$
Yes, but note that if you don't want to check that the zero vector is there you need at least to check that the subset is not empty. (otherwise there is no meaning to addition and scalar multiplication)
$endgroup$
– Mark
Mar 12 at 19:32
$begingroup$
Yes, but note that if you don't want to check that the zero vector is there you need at least to check that the subset is not empty. (otherwise there is no meaning to addition and scalar multiplication)
$endgroup$
– Mark
Mar 12 at 19:32
add a comment |
$begingroup$
Of course just checking that the zero vector is there is not enough. Just think about it-how can the existence of a zero vector imply that the subset is closed under addition and scalar multiplication? For example, is $0,1$ a subspace of $mathbbR$? Clearly not. To check if $Ssubseteq V$ is a subspace you need to check the following:
$1. $ The zero vector is in $S$. (or instead of that you can check that $S$ is not empty).
$2. $ $S$ is closed under addition and scalar multiplication.
answered Mar 12 at 19:30
MarkMark
10.2k622
$endgroup$
add a comment |
$begingroup$
Of course just checking that the zero vector is there is not enough. Just think about it-how can the existence of a zero vector imply that the subset is closed under addition and scalar multiplication? For example, is $0,1$ a subspace of $mathbbR$? Clearly not. To check if $Ssubseteq V$ is a subspace you need to check the following:
$1. $ The zero vector is in $S$. (or instead of that you can check that $S$ is not empty).
$2. $ $S$ is closed under addition and scalar multiplication.
answered Mar 12 at 19:30
MarkMark
10.2k622
$endgroup$
add a comment |
$begingroup$
Of course just checking that the zero vector is there is not enough. Just think about it-how can the existence of a zero vector imply that the subset is closed under addition and scalar multiplication? For example, is $0,1$ a subspace of $mathbbR$? Clearly not. To check if $Ssubseteq V$ is a subspace you need to check the following:
$1. $ The zero vector is in $S$. (or instead of that you can check that $S$ is not empty).
$2. $ $S$ is closed under addition and scalar multiplication.
answered Mar 12 at 19:30
MarkMark
10.2k622
$endgroup$
Of course just checking that the zero vector is there is not enough. Just think about it-how can the existence of a zero vector imply that the subset is closed under addition and scalar multiplication? For example, is $0,1$ a subspace of $mathbbR$? Clearly not. To check if $Ssubseteq V$ is a subspace you need to check the following:
$1. $ The zero vector is in $S$. (or instead of that you can check that $S$ is not empty).
$2. $ $S$ is closed under addition and scalar multiplication.
answered Mar 12 at 19:30
MarkMark
10.2k622
answered Mar 12 at 19:30
MarkMark
10.2k622
answered Mar 12 at 19:30
MarkMark
10.2k622
answered Mar 12 at 19:30
MarkMark
10.2k622
10.2k622
add a comment |
add a comment |
$begingroup$
Am I wrong in assuming that if $0$ vector exists, all the other axioms must hold?
In short, yes. Why else would the other axioms be there?
Technically, those aren't axioms. You have axioms describing a vector space, and given a vector space there are certain subsets of that vector space which are themselves vector spaces, using the same operations. Those are what we call subspaces.
So subspaces actually need to fulfill all the same axioms that all other vector spaces do. The checklist you're using (which you call axioms) isn't as long as the full list of vector space axioms, however. The reason is that inheriting the operations from something we know is a vector space gives us a lot for free.
answered Mar 12 at 19:32
ArthurArthur
118k7118201
$endgroup$
add a comment |
$begingroup$
Am I wrong in assuming that if $0$ vector exists, all the other axioms must hold?
In short, yes. Why else would the other axioms be there?
Technically, those aren't axioms. You have axioms describing a vector space, and given a vector space there are certain subsets of that vector space which are themselves vector spaces, using the same operations. Those are what we call subspaces.
So subspaces actually need to fulfill all the same axioms that all other vector spaces do. The checklist you're using (which you call axioms) isn't as long as the full list of vector space axioms, however. The reason is that inheriting the operations from something we know is a vector space gives us a lot for free.
answered Mar 12 at 19:32
ArthurArthur
118k7118201
$endgroup$
add a comment |
$begingroup$
Am I wrong in assuming that if $0$ vector exists, all the other axioms must hold?
In short, yes. Why else would the other axioms be there?
Technically, those aren't axioms. You have axioms describing a vector space, and given a vector space there are certain subsets of that vector space which are themselves vector spaces, using the same operations. Those are what we call subspaces.
So subspaces actually need to fulfill all the same axioms that all other vector spaces do. The checklist you're using (which you call axioms) isn't as long as the full list of vector space axioms, however. The reason is that inheriting the operations from something we know is a vector space gives us a lot for free.
answered Mar 12 at 19:32
ArthurArthur
118k7118201
$endgroup$
Am I wrong in assuming that if $0$ vector exists, all the other axioms must hold?
In short, yes. Why else would the other axioms be there?
Technically, those aren't axioms. You have axioms describing a vector space, and given a vector space there are certain subsets of that vector space which are themselves vector spaces, using the same operations. Those are what we call subspaces.
So subspaces actually need to fulfill all the same axioms that all other vector spaces do. The checklist you're using (which you call axioms) isn't as long as the full list of vector space axioms, however. The reason is that inheriting the operations from something we know is a vector space gives us a lot for free.
answered Mar 12 at 19:32
ArthurArthur
118k7118201
answered Mar 12 at 19:32
ArthurArthur
118k7118201
answered Mar 12 at 19:32
ArthurArthur
118k7118201
answered Mar 12 at 19:32
ArthurArthur
118k7118201
118k7118201
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%2f3145554%2fclarification-about-the-axioms-of-a-subspace%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
$begingroup$
Just because a subset contains $0$ doesn’t mean it is a subspace
$endgroup$
– J. W. Tanner
Mar 12 at 19:27
$begingroup$
@J.W.Tanner But why? I thought the zero vector automatically implied a subspace since addition amd scalar multiplication must hold true as a result.
$endgroup$
– Future Math person
Mar 12 at 19:28
$begingroup$
see the answers below
$endgroup$
– J. W. Tanner
Mar 12 at 19:51