Vjylku

Are objects structures and/or vice versa?

"My colleague's body is amazing"

map list to bin numbers

Prime joint compound before latex paint?

Copycat chess is back

Landing in very high winds

How to move the player while also allowing forces to affect it

Landlord wants to switch my lease to a "Land contract" to "get back at the city"

Is Social Media Science Fiction?

What is the offset in a seaplane's hull?

Why did the Germans forbid the possession of pet pigeons in Rostov-on-Don in 1941?

Is it legal to have the "// (c) 2019 John Smith" header in all files when there are hundreds of contributors?

aging parents with no investments

What causes the sudden spool-up sound from an F-16 when enabling afterburner?

I’m planning on buying a laser printer but concerned about the life cycle of toner in the machine

Why do we use polarized capacitors?

Was there ever an axiom rendered a theorem?

What is GPS' 19 year rollover and does it present a cybersecurity issue?

Is "plugging out" electronic devices an American expression?

What to wear for invited talk in Canada

Patience, young "Padovan"

"listening to me about as much as you're listening to this pole here"

Unbreakable Formation vs. Cry of the Carnarium

Email Account under attack (really) - anything I can do?

Is a vector space a subspace of itself?

vector space and its subspaceProve: The set of all polynomials p with p(2) = p(3) is a vector spaceProving a vector space over itself have no subspacesVector Space vs SubspaceProving a subset is a subspace of a Vector SpaceAre all vector spaces also a subspace?Understand the definition of a vector subspaceProve that, with vector addition and scalar multiplication well-defined, $V/W$ becomes a vector space over $k$.Is the set of all exponential functions a subspace of the vector space of all continuous functions?I've seen two definitions of subspace; one involving vector spaces and one requiring linear combinations

1

1

$begingroup$

We know that a subspace is a vector space that follows the same addition and multiplication rules as $Bbb V$, but is a vector space a subspace of itself?
Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10.

linear-algebra vector-spaces

share|cite|improve this question

edited 29 mins ago

Eric Wofsey

192k14220352

asked 3 hours ago

mingming

4456

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How do you define a subspace of a vector space?
  $endgroup$
  – Brian
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Is a set a subset of itself?? What’s V1-V10?
  $endgroup$
  – J. W. Tanner
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
  $endgroup$
  – Theo Bendit
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
  $endgroup$
  – Xander Henderson
  1 hour ago
  

  
  
  
  

add a comment | 

1

1

$begingroup$

We know that a subspace is a vector space that follows the same addition and multiplication rules as $Bbb V$, but is a vector space a subspace of itself?
Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10.

linear-algebra vector-spaces

share|cite|improve this question

edited 29 mins ago

Eric Wofsey

192k14220352

asked 3 hours ago

mingming

4456

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How do you define a subspace of a vector space?
  $endgroup$
  – Brian
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Is a set a subset of itself?? What’s V1-V10?
  $endgroup$
  – J. W. Tanner
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
  $endgroup$
  – Theo Bendit
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
  $endgroup$
  – Xander Henderson
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

We know that a subspace is a vector space that follows the same addition and multiplication rules as $Bbb V$, but is a vector space a subspace of itself?
Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10.

linear-algebra vector-spaces

share|cite|improve this question

edited 29 mins ago

Eric Wofsey

192k14220352

asked 3 hours ago

mingming

4456

$endgroup$

share|cite|improve this question

edited 29 mins ago

Eric Wofsey

192k14220352

asked 3 hours ago

mingming

4456

share|cite|improve this question

edited 29 mins ago

Eric Wofsey

192k14220352

asked 3 hours ago

mingming

4456

edited 29 mins ago

Eric Wofsey

192k14220352

edited 29 mins ago

Eric Wofsey

192k14220352

asked 3 hours ago

mingming

4456

asked 3 hours ago

mingming

4456

2 Answers
2

active

oldest

votes

7

$begingroup$

Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.

share|cite|improve this answer

answered 3 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

2

$begingroup$

I'm guessing that V1 - V10 are the axioms for proving vector spaces.

To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.

There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.

Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.

I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.

You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.

share|cite|improve this answer

answered 2 hours ago

Theo BenditTheo Bendit

20.7k12354

$endgroup$

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3180447%2fis-a-vector-space-a-subspace-of-itself%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

7

$begingroup$

Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.

share|cite|improve this answer

answered 3 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

7

$begingroup$

Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.

share|cite|improve this answer

answered 3 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

$begingroup$

Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.

share|cite|improve this answer

answered 3 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

share|cite|improve this answer

answered 3 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

answered 3 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

answered 3 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

2

$begingroup$

I'm guessing that V1 - V10 are the axioms for proving vector spaces.

To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.

There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.

Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.

I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.

You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.

share|cite|improve this answer

answered 2 hours ago

Theo BenditTheo Bendit

20.7k12354

$endgroup$

add a comment | 

2

$begingroup$

I'm guessing that V1 - V10 are the axioms for proving vector spaces.

To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.

There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.

Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.

I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.

You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.

share|cite|improve this answer

answered 2 hours ago

Theo BenditTheo Bendit

20.7k12354

$endgroup$

add a comment | 

$begingroup$

I'm guessing that V1 - V10 are the axioms for proving vector spaces.

To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.

There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.

Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.

I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.

You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.

share|cite|improve this answer

answered 2 hours ago

Theo BenditTheo Bendit

20.7k12354

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Theo BenditTheo Bendit

20.7k12354

answered 2 hours ago

Theo BenditTheo Bendit

20.7k12354

answered 2 hours ago

Theo BenditTheo Bendit

20.7k12354