Vector not in span$left(Sright)$ implies $Scup{vec{X}}$ be linearly independent












0












$begingroup$



Theorem:
Let S be a linearly independent set in a vector space. If $vec{X} in V$ and $vec{X} in span left(Sright)$ then $S cup {vec{X}}$ is linearly independent.




Proof:



It suffices to show, by contra positivity, that $vec{X}$ $in$ $spanleft(Sright)$.



Let $S = {vec{v}_{i}}_{i=1}^{n}$



$vec{0} = c_{X}vec{X} + sum_{i=1}^{n}c_{i}vec{v}_{i}$ with coefficients $c_{i} = c_{X} = 0, forall i in mathbb{Z}_{1}^{n}$.



Rearranging:



$vec{X} = frac{1}{c_{X}} sum_{i=1}^{n}c_{i}vec{v}_{i}$



Here, I am unable to move forward. The reciprocal of $c_{X}$ gives an undefined value.



Any help to move this proof forward is appreciated.
Thanks in advance.






linear-algebra proof-verification







share|cite|improve this question









$endgroup$












  • $begingroup$
    You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
    $endgroup$
    – Michael
    Nov 24 '18 at 10:57








  • 1




    $begingroup$
    Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
    $endgroup$
    – Michael
    Nov 24 '18 at 11:04


















0












$begingroup$



Theorem:
Let S be a linearly independent set in a vector space. If $vec{X} in V$ and $vec{X} in span left(Sright)$ then $S cup {vec{X}}$ is linearly independent.




Proof:



It suffices to show, by contra positivity, that $vec{X}$ $in$ $spanleft(Sright)$.



Let $S = {vec{v}_{i}}_{i=1}^{n}$



$vec{0} = c_{X}vec{X} + sum_{i=1}^{n}c_{i}vec{v}_{i}$ with coefficients $c_{i} = c_{X} = 0, forall i in mathbb{Z}_{1}^{n}$.



Rearranging:



$vec{X} = frac{1}{c_{X}} sum_{i=1}^{n}c_{i}vec{v}_{i}$



Here, I am unable to move forward. The reciprocal of $c_{X}$ gives an undefined value.



Any help to move this proof forward is appreciated.
Thanks in advance.






linear-algebra proof-verification







share|cite|improve this question









$endgroup$












  • $begingroup$
    You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
    $endgroup$
    – Michael
    Nov 24 '18 at 10:57








  • 1




    $begingroup$
    Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
    $endgroup$
    – Michael
    Nov 24 '18 at 11:04
















0












0








0





$begingroup$



Theorem:
Let S be a linearly independent set in a vector space. If $vec{X} in V$ and $vec{X} in span left(Sright)$ then $S cup {vec{X}}$ is linearly independent.




Proof:



It suffices to show, by contra positivity, that $vec{X}$ $in$ $spanleft(Sright)$.



Let $S = {vec{v}_{i}}_{i=1}^{n}$



$vec{0} = c_{X}vec{X} + sum_{i=1}^{n}c_{i}vec{v}_{i}$ with coefficients $c_{i} = c_{X} = 0, forall i in mathbb{Z}_{1}^{n}$.



Rearranging:



$vec{X} = frac{1}{c_{X}} sum_{i=1}^{n}c_{i}vec{v}_{i}$



Here, I am unable to move forward. The reciprocal of $c_{X}$ gives an undefined value.



Any help to move this proof forward is appreciated.
Thanks in advance.






linear-algebra proof-verification







share|cite|improve this question









$endgroup$





Theorem:
Let S be a linearly independent set in a vector space. If $vec{X} in V$ and $vec{X} in span left(Sright)$ then $S cup {vec{X}}$ is linearly independent.




Proof:



It suffices to show, by contra positivity, that $vec{X}$ $in$ $spanleft(Sright)$.



Let $S = {vec{v}_{i}}_{i=1}^{n}$



$vec{0} = c_{X}vec{X} + sum_{i=1}^{n}c_{i}vec{v}_{i}$ with coefficients $c_{i} = c_{X} = 0, forall i in mathbb{Z}_{1}^{n}$.



Rearranging:



$vec{X} = frac{1}{c_{X}} sum_{i=1}^{n}c_{i}vec{v}_{i}$



Here, I am unable to move forward. The reciprocal of $c_{X}$ gives an undefined value.



Any help to move this proof forward is appreciated.
Thanks in advance.






linear-algebra proof-verification




linear-algebra proof-verification






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 '18 at 10:52









MathematicingMathematicing

2,44321854




2,44321854












  • $begingroup$
    You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
    $endgroup$
    – Michael
    Nov 24 '18 at 10:57








  • 1




    $begingroup$
    Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
    $endgroup$
    – Michael
    Nov 24 '18 at 11:04




















  • $begingroup$
    You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
    $endgroup$
    – Michael
    Nov 24 '18 at 10:57








  • 1




    $begingroup$
    Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
    $endgroup$
    – Michael
    Nov 24 '18 at 11:04


















$begingroup$
You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
$endgroup$
– Michael
Nov 24 '18 at 10:57






$begingroup$
You stated the theorem incorrectly: It is supposed to have $vec{X} notin Span(S)$, instead of $vec{X} in Span(S)$. The theorem as-stated is false (for example take $vec{X}=vec{0} in Span(S)$).
$endgroup$
– Michael
Nov 24 '18 at 10:57






1




1




$begingroup$
Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
$endgroup$
– Michael
Nov 24 '18 at 11:04






$begingroup$
Your sentence $vec{0} = c_Xvec{X} + sum_{i=1}^n c_i vec{v}_i$ "with coefficients $c_i=c_X=0$ for all $i$" does not make sense (particularly the part in quotes) since if you define those coefficients to be zero you get $vec{0}=vec{0}$ and the equation is useless to work with. That is not the definition of linear independence...you need to show that if the equation holds, then all coefficients must be zero (you are not supposed to assume they are zero, you must prove they are zero).
$endgroup$
– Michael
Nov 24 '18 at 11:04












1 Answer
1






active

oldest

votes


















1












$begingroup$

You want $vec{x} notin operatorname{span}(S)$, as the assumtion of course (as you do have in the title).



In that case we can show that $S cup {vec{x}}$ is lin. independent:



Suppose no, then there are vectors $vec{s_1},dots vec{s_n} in S$ and coefficients $c_1,ldots c_n, c$, not all $0$, such that



$$sum_{i=1}^n c_i vec{s_i} + cvec{x} = 0$$



If $c =0$ we have an immediate contradiction with the fact that $S$ is linearly independent, because not all the $c_i$ are $0$.



So we must have $c neq 0$ and so $frac{1}{c}$ is a well-defined scalar.



But then $$vec{x} = sum_{i=1}^n frac{1}{c}(-c_i) s_i$$ and we'd have that $vec{x} in operatorname{span}(S)$ contrary to the (corrected) assumption.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
    $endgroup$
    – Mathematicing
    Nov 24 '18 at 14:11












  • $begingroup$
    @Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
    $endgroup$
    – Henno Brandsma
    Nov 24 '18 at 14:13











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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011413%2fvector-not-in-span-lefts-right-implies-s-cup-vecx-be-linearly-indepe%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









1












$begingroup$

You want $vec{x} notin operatorname{span}(S)$, as the assumtion of course (as you do have in the title).



In that case we can show that $S cup {vec{x}}$ is lin. independent:



Suppose no, then there are vectors $vec{s_1},dots vec{s_n} in S$ and coefficients $c_1,ldots c_n, c$, not all $0$, such that



$$sum_{i=1}^n c_i vec{s_i} + cvec{x} = 0$$



If $c =0$ we have an immediate contradiction with the fact that $S$ is linearly independent, because not all the $c_i$ are $0$.



So we must have $c neq 0$ and so $frac{1}{c}$ is a well-defined scalar.



But then $$vec{x} = sum_{i=1}^n frac{1}{c}(-c_i) s_i$$ and we'd have that $vec{x} in operatorname{span}(S)$ contrary to the (corrected) assumption.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
    $endgroup$
    – Mathematicing
    Nov 24 '18 at 14:11












  • $begingroup$
    @Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
    $endgroup$
    – Henno Brandsma
    Nov 24 '18 at 14:13
















1












$begingroup$

You want $vec{x} notin operatorname{span}(S)$, as the assumtion of course (as you do have in the title).



In that case we can show that $S cup {vec{x}}$ is lin. independent:



Suppose no, then there are vectors $vec{s_1},dots vec{s_n} in S$ and coefficients $c_1,ldots c_n, c$, not all $0$, such that



$$sum_{i=1}^n c_i vec{s_i} + cvec{x} = 0$$



If $c =0$ we have an immediate contradiction with the fact that $S$ is linearly independent, because not all the $c_i$ are $0$.



So we must have $c neq 0$ and so $frac{1}{c}$ is a well-defined scalar.



But then $$vec{x} = sum_{i=1}^n frac{1}{c}(-c_i) s_i$$ and we'd have that $vec{x} in operatorname{span}(S)$ contrary to the (corrected) assumption.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
    $endgroup$
    – Mathematicing
    Nov 24 '18 at 14:11












  • $begingroup$
    @Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
    $endgroup$
    – Henno Brandsma
    Nov 24 '18 at 14:13














1












1








1





$begingroup$

You want $vec{x} notin operatorname{span}(S)$, as the assumtion of course (as you do have in the title).



In that case we can show that $S cup {vec{x}}$ is lin. independent:



Suppose no, then there are vectors $vec{s_1},dots vec{s_n} in S$ and coefficients $c_1,ldots c_n, c$, not all $0$, such that



$$sum_{i=1}^n c_i vec{s_i} + cvec{x} = 0$$



If $c =0$ we have an immediate contradiction with the fact that $S$ is linearly independent, because not all the $c_i$ are $0$.



So we must have $c neq 0$ and so $frac{1}{c}$ is a well-defined scalar.



But then $$vec{x} = sum_{i=1}^n frac{1}{c}(-c_i) s_i$$ and we'd have that $vec{x} in operatorname{span}(S)$ contrary to the (corrected) assumption.






share|cite|improve this answer









$endgroup$



You want $vec{x} notin operatorname{span}(S)$, as the assumtion of course (as you do have in the title).



In that case we can show that $S cup {vec{x}}$ is lin. independent:



Suppose no, then there are vectors $vec{s_1},dots vec{s_n} in S$ and coefficients $c_1,ldots c_n, c$, not all $0$, such that



$$sum_{i=1}^n c_i vec{s_i} + cvec{x} = 0$$



If $c =0$ we have an immediate contradiction with the fact that $S$ is linearly independent, because not all the $c_i$ are $0$.



So we must have $c neq 0$ and so $frac{1}{c}$ is a well-defined scalar.



But then $$vec{x} = sum_{i=1}^n frac{1}{c}(-c_i) s_i$$ and we'd have that $vec{x} in operatorname{span}(S)$ contrary to the (corrected) assumption.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 '18 at 11:06









Henno BrandsmaHenno Brandsma

106k347114




106k347114












  • $begingroup$
    I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
    $endgroup$
    – Mathematicing
    Nov 24 '18 at 14:11












  • $begingroup$
    @Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
    $endgroup$
    – Henno Brandsma
    Nov 24 '18 at 14:13


















  • $begingroup$
    I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
    $endgroup$
    – Mathematicing
    Nov 24 '18 at 14:11












  • $begingroup$
    @Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
    $endgroup$
    – Henno Brandsma
    Nov 24 '18 at 14:13
















$begingroup$
I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
$endgroup$
– Mathematicing
Nov 24 '18 at 14:11






$begingroup$
I made a mistake in the theorem. The vector X is not supposed to be in the span of the set S.
$endgroup$
– Mathematicing
Nov 24 '18 at 14:11














$begingroup$
@Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
$endgroup$
– Henno Brandsma
Nov 24 '18 at 14:13




$begingroup$
@Mathematicing I worked from the assumption that $vec{x}$ indeed is not in the span of $S$.
$endgroup$
– Henno Brandsma
Nov 24 '18 at 14:13


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011413%2fvector-not-in-span-lefts-right-implies-s-cup-vecx-be-linearly-indepe%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
Read more