Linearly independent set definition Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Showing that a one-to-one linear transformation maps a linearly independent set onto a linearly independent setProve that any finite subset of a linearly independent set is linearly independentLinearly independentdefinition of linearly dependent setDoes there exist a linearly independent subset in any linearly dependent set of vectors?If every subset of $S$ is linearly independent, then $S$ is independentImpossibility of constructing a continuum-size linearly independent set in $Bbb R$Maximal linearly independent subsets problem.About definition in Linear Algebra concerning empty subset of vector spaceWhat is the significance of the fact that a set in a vector space is linearly dependent if it contains finitely many linearly dependent vectors?
How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time
Single author papers against my advisor's will?
Jazz greats knew nothing of modes. Why are they used to improvise on standards?
What was the last x86 CPU that did not have the x87 floating-point unit built in?
How to rotate it perfectly?
Do working physicists consider Newtonian mechanics to be "falsified"?
When communicating altitude with a '9' in it, should it be pronounced "nine hundred" or "niner hundred"?
Estimated State payment too big --> money back; + 2018 Tax Reform
What did Darwin mean by 'squib' here?
Biased dice probability question
Is 1 ppb equal to 1 μg/kg?
Why does this iterative way of solving of equation work?
How to market an anarchic city as a tourism spot to people living in civilized areas?
What is the largest species of polychaete?
Who can trigger ship-wide alerts in Star Trek?
Stopping real property loss from eroding embankment
Limit for e and 1/e
What is the order of Mitzvot in Rambam's Sefer Hamitzvot?
How can I make names more distinctive without making them longer?
How does modal jazz use chord progressions?
Can a zero nonce be safely used with AES-GCM if the key is random and never used again?
Slither Like a Snake
Working around an AWS network ACL rule limit
Problem when applying foreach loop
Linearly independent set definition
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Showing that a one-to-one linear transformation maps a linearly independent set onto a linearly independent setProve that any finite subset of a linearly independent set is linearly independentLinearly independentdefinition of linearly dependent setDoes there exist a linearly independent subset in any linearly dependent set of vectors?If every subset of $S$ is linearly independent, then $S$ is independentImpossibility of constructing a continuum-size linearly independent set in $Bbb R$Maximal linearly independent subsets problem.About definition in Linear Algebra concerning empty subset of vector spaceWhat is the significance of the fact that a set in a vector space is linearly dependent if it contains finitely many linearly dependent vectors?
$begingroup$
The definition of a linearly independent set I have been given is:
A set S is linearly independent if every finite subset of S is linearly independent.
Do there exist any sets S such that all finite subsets are linearly independent, but contain an infinite subset that is linearly dependent? If not, why include the 'finite' in the definition?
linear-algebra definition
asked Apr 8 at 20:10
JosephJoseph
205
New contributor
Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The definition of a linearly independent set I have been given is:
A set S is linearly independent if every finite subset of S is linearly independent.
Do there exist any sets S such that all finite subsets are linearly independent, but contain an infinite subset that is linearly dependent? If not, why include the 'finite' in the definition?
linear-algebra definition
asked Apr 8 at 20:10
JosephJoseph
205
New contributor
Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
No, because the definition says that said infinite subset should contain a finite linearly dependent subset, which is a fortiori a linearly dependent subset of S.
$endgroup$
– Saucy O'Path
Apr 8 at 20:13
1
$begingroup$
The definition you were given seems wrong... Linear independence is used to define linear independence... is there some way you can clarify the definition?
$endgroup$
– NazimJ
Apr 8 at 20:15
$begingroup$
Before this the book defines a collection of vectors as linearly independent if they are not linearly dependent. But then for a set to be linearly independent, the property must not just hold for the set as a whole, but for every finite subset of the set.
$endgroup$
– Joseph
Apr 8 at 20:25
add a comment |
$begingroup$
The definition of a linearly independent set I have been given is:
A set S is linearly independent if every finite subset of S is linearly independent.
Do there exist any sets S such that all finite subsets are linearly independent, but contain an infinite subset that is linearly dependent? If not, why include the 'finite' in the definition?
linear-algebra definition
asked Apr 8 at 20:10
JosephJoseph
205
New contributor
Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The definition of a linearly independent set I have been given is:
A set S is linearly independent if every finite subset of S is linearly independent.
Do there exist any sets S such that all finite subsets are linearly independent, but contain an infinite subset that is linearly dependent? If not, why include the 'finite' in the definition?
linear-algebra definition
linear-algebra definition
asked Apr 8 at 20:10
JosephJoseph
205
New contributor
Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 20:10
JosephJoseph
205
New contributor
Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 20:10
JosephJoseph
205
New contributor
Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 20:10
JosephJoseph
205
asked Apr 8 at 20:10
JosephJoseph
205
205
New contributor
Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
No, because the definition says that said infinite subset should contain a finite linearly dependent subset, which is a fortiori a linearly dependent subset of S.
$endgroup$
– Saucy O'Path
Apr 8 at 20:13
1
$begingroup$
The definition you were given seems wrong... Linear independence is used to define linear independence... is there some way you can clarify the definition?
$endgroup$
– NazimJ
Apr 8 at 20:15
$begingroup$
Before this the book defines a collection of vectors as linearly independent if they are not linearly dependent. But then for a set to be linearly independent, the property must not just hold for the set as a whole, but for every finite subset of the set.
$endgroup$
– Joseph
Apr 8 at 20:25
add a comment |
$begingroup$
No, because the definition says that said infinite subset should contain a finite linearly dependent subset, which is a fortiori a linearly dependent subset of S.
$endgroup$
– Saucy O'Path
Apr 8 at 20:13
1
$begingroup$
The definition you were given seems wrong... Linear independence is used to define linear independence... is there some way you can clarify the definition?
$endgroup$
– NazimJ
Apr 8 at 20:15
$begingroup$
Before this the book defines a collection of vectors as linearly independent if they are not linearly dependent. But then for a set to be linearly independent, the property must not just hold for the set as a whole, but for every finite subset of the set.
$endgroup$
– Joseph
Apr 8 at 20:25
$begingroup$
No, because the definition says that said infinite subset should contain a finite linearly dependent subset, which is a fortiori a linearly dependent subset of S.
$endgroup$
– Saucy O'Path
Apr 8 at 20:13
$begingroup$
No, because the definition says that said infinite subset should contain a finite linearly dependent subset, which is a fortiori a linearly dependent subset of S.
$endgroup$
– Saucy O'Path
Apr 8 at 20:13
1
1
$begingroup$
The definition you were given seems wrong... Linear independence is used to define linear independence... is there some way you can clarify the definition?
$endgroup$
– NazimJ
Apr 8 at 20:15
$begingroup$
The definition you were given seems wrong... Linear independence is used to define linear independence... is there some way you can clarify the definition?
$endgroup$
– NazimJ
Apr 8 at 20:15
$begingroup$
Before this the book defines a collection of vectors as linearly independent if they are not linearly dependent. But then for a set to be linearly independent, the property must not just hold for the set as a whole, but for every finite subset of the set.
$endgroup$
– Joseph
Apr 8 at 20:25
$begingroup$
Before this the book defines a collection of vectors as linearly independent if they are not linearly dependent. But then for a set to be linearly independent, the property must not just hold for the set as a whole, but for every finite subset of the set.
$endgroup$
– Joseph
Apr 8 at 20:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, there is the usual definition of linear independence of a finite set of vectors: namely, $v_1,dots, v_k$ is linearly independent if $lambda_1v_1+dots +lambda_kv_k=0$ implies all $lambda_i=0$.
Then, one can extend it for infinite sets, say, by the given definition.
Note that vector addition, hence also linear combination, is defined only for finitely many vectors, so if a vector $u$ is linearly dependent on a set $S$, then it means it is a linear combination of finitely many of them.
answered Apr 8 at 21:15
BerciBerci
62k23776
$endgroup$
add a comment |
Your Answer
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
);
);
Joseph is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3180142%2flinearly-independent-set-definition%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, there is the usual definition of linear independence of a finite set of vectors: namely, $v_1,dots, v_k$ is linearly independent if $lambda_1v_1+dots +lambda_kv_k=0$ implies all $lambda_i=0$.
Then, one can extend it for infinite sets, say, by the given definition.
Note that vector addition, hence also linear combination, is defined only for finitely many vectors, so if a vector $u$ is linearly dependent on a set $S$, then it means it is a linear combination of finitely many of them.
answered Apr 8 at 21:15
BerciBerci
62k23776
$endgroup$
add a comment |
$begingroup$
First, there is the usual definition of linear independence of a finite set of vectors: namely, $v_1,dots, v_k$ is linearly independent if $lambda_1v_1+dots +lambda_kv_k=0$ implies all $lambda_i=0$.
Then, one can extend it for infinite sets, say, by the given definition.
Note that vector addition, hence also linear combination, is defined only for finitely many vectors, so if a vector $u$ is linearly dependent on a set $S$, then it means it is a linear combination of finitely many of them.
answered Apr 8 at 21:15
BerciBerci
62k23776
$endgroup$
add a comment |
$begingroup$
First, there is the usual definition of linear independence of a finite set of vectors: namely, $v_1,dots, v_k$ is linearly independent if $lambda_1v_1+dots +lambda_kv_k=0$ implies all $lambda_i=0$.
Then, one can extend it for infinite sets, say, by the given definition.
Note that vector addition, hence also linear combination, is defined only for finitely many vectors, so if a vector $u$ is linearly dependent on a set $S$, then it means it is a linear combination of finitely many of them.
answered Apr 8 at 21:15
BerciBerci
62k23776
$endgroup$
First, there is the usual definition of linear independence of a finite set of vectors: namely, $v_1,dots, v_k$ is linearly independent if $lambda_1v_1+dots +lambda_kv_k=0$ implies all $lambda_i=0$.
Then, one can extend it for infinite sets, say, by the given definition.
Note that vector addition, hence also linear combination, is defined only for finitely many vectors, so if a vector $u$ is linearly dependent on a set $S$, then it means it is a linear combination of finitely many of them.
answered Apr 8 at 21:15
BerciBerci
62k23776
answered Apr 8 at 21:15
BerciBerci
62k23776
answered Apr 8 at 21:15
BerciBerci
62k23776
answered Apr 8 at 21:15
BerciBerci
62k23776
62k23776
add a comment |
add a comment |
Joseph is a new contributor. Be nice, and check out our Code of Conduct.
Joseph is a new contributor. Be nice, and check out our Code of Conduct.
Joseph is a new contributor. Be nice, and check out our Code of Conduct.
Joseph is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3180142%2flinearly-independent-set-definition%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
$begingroup$
No, because the definition says that said infinite subset should contain a finite linearly dependent subset, which is a fortiori a linearly dependent subset of S.
$endgroup$
– Saucy O'Path
Apr 8 at 20:13
1
$begingroup$
The definition you were given seems wrong... Linear independence is used to define linear independence... is there some way you can clarify the definition?
$endgroup$
– NazimJ
Apr 8 at 20:15
$begingroup$
Before this the book defines a collection of vectors as linearly independent if they are not linearly dependent. But then for a set to be linearly independent, the property must not just hold for the set as a whole, but for every finite subset of the set.
$endgroup$
– Joseph
Apr 8 at 20:25