Sfrgttk

New order #4: World

A poker game description that does not feel gimmicky

How to make particles emit from certain parts of a 3D object?

Why do we use polarized capacitors?

If a centaur druid Wild Shapes into a Giant Elk, do their Charge features stack?

Calculate Levenshtein distance between two strings in Python

Information to fellow intern about hiring?

Can a planet have a different gravitational pull depending on its location in orbit around its sun?

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

Prime joint compound before latex paint?

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

Patience, young "Padovan"

How to make payment on the internet without leaving a money trail?

Does a dangling wire really electrocute me if I'm standing in water?

What does "enim et" mean?

Why do UK politicians seemingly ignore opinion polls on Brexit?

What do the Banks children have against barley water?

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

Is there a way to make member function NOT callable from constructor?

Is this food a bread or a loaf?

Could Giant Ground Sloths have been a good pack animal for the ancient Mayans?

Ideas for 3rd eye abilities

Extreme, but not acceptable situation and I can't start the work tomorrow morning

Where else does the Shulchan Aruch quote an authority by name?

Is a vector space a subspace?

vector space and its subspaceProve: The set of all polynomials p with p(2) = p(3) is a vector spaceVector Space vs SubspaceProving a subset is a subspace of a Vector SpaceAre all vector spaces also a subspace?Understand the definition of a vector subspaceProving 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?Linear Algebra - Zero subspace vs empty subspace

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 confuse 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 1 hour ago

José Carlos Santos

173k23133241

asked 1 hour ago

mingming

4456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is a set a subset of itself?? What’s V1-V10?
  $endgroup$
  – J. W. Tanner
  1 hour 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
  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 confuse 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 1 hour ago

José Carlos Santos

173k23133241

asked 1 hour ago

mingming

4456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is a set a subset of itself?? What’s V1-V10?
  $endgroup$
  – J. W. Tanner
  1 hour 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
  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 confuse 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 1 hour ago

José Carlos Santos

173k23133241

asked 1 hour ago

mingming

4456

$endgroup$

share|cite|improve this question

edited 1 hour ago

José Carlos Santos

173k23133241

asked 1 hour ago

mingming

4456

share|cite|improve this question

edited 1 hour ago

José Carlos Santos

173k23133241

asked 1 hour ago

mingming

4456

edited 1 hour ago

José Carlos Santos

173k23133241

edited 1 hour ago

José Carlos Santos

173k23133241

asked 1 hour ago

mingming

4456

asked 1 hour ago

mingming

4456

2 Answers
2

active

oldest

votes

6

$begingroup$

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 1 hour ago

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

1

$begingroup$

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 29 mins 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%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

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 1 hour ago

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

6

$begingroup$

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 1 hour 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 1 hour ago

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

share|cite|improve this answer

answered 1 hour ago

José Carlos SantosJosé Carlos Santos

173k23133241

answered 1 hour ago

José Carlos SantosJosé Carlos Santos

173k23133241

answered 1 hour ago

José Carlos SantosJosé Carlos Santos

173k23133241

1

$begingroup$

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 29 mins ago

Theo BenditTheo Bendit

20.7k12354

$endgroup$

add a comment | 

1

$begingroup$

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 29 mins 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 29 mins ago

Theo BenditTheo Bendit

20.7k12354

$endgroup$

share|cite|improve this answer

answered 29 mins ago

Theo BenditTheo Bendit

20.7k12354

answered 29 mins ago

Theo BenditTheo Bendit

20.7k12354

answered 29 mins ago

Theo BenditTheo Bendit

20.7k12354