How do I prove that 2 vector spaces are equal if they have different number of vectors that span them?
$begingroup$
For example, in 4-dimensions, vector space V is spanned by the linearly independent vectors x, y, and z
and vector space U is spanned by the linearly independent vectors r, and s. Would I try to prove that the vectors x, y, and z can span U and vectors r, and s can span V?
linear-algebra vector-spaces
edited Dec 10 '18 at 10:51
José Carlos Santos
170k23132238
asked Jul 23 '18 at 5:59
edmonda7edmonda7
394
$endgroup$
add a comment |
$begingroup$
For example, in 4-dimensions, vector space V is spanned by the linearly independent vectors x, y, and z
and vector space U is spanned by the linearly independent vectors r, and s. Would I try to prove that the vectors x, y, and z can span U and vectors r, and s can span V?
linear-algebra vector-spaces
edited Dec 10 '18 at 10:51
José Carlos Santos
170k23132238
asked Jul 23 '18 at 5:59
edmonda7edmonda7
394
$endgroup$
1
$begingroup$
Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
$endgroup$
– Jonas Lenz
Jul 23 '18 at 6:09
$begingroup$
That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
$endgroup$
– Jyrki Lahtonen
Jul 23 '18 at 6:12
$begingroup$
@JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
$endgroup$
– edmonda7
Jul 23 '18 at 6:16
2
$begingroup$
In that case, $dim V=3ne 2=dim U$, hence $Vne U$
$endgroup$
– Hagen von Eitzen
Jul 23 '18 at 6:19
add a comment |
$begingroup$
For example, in 4-dimensions, vector space V is spanned by the linearly independent vectors x, y, and z
and vector space U is spanned by the linearly independent vectors r, and s. Would I try to prove that the vectors x, y, and z can span U and vectors r, and s can span V?
linear-algebra vector-spaces
edited Dec 10 '18 at 10:51
José Carlos Santos
170k23132238
asked Jul 23 '18 at 5:59
edmonda7edmonda7
394
$endgroup$
For example, in 4-dimensions, vector space V is spanned by the linearly independent vectors x, y, and z
and vector space U is spanned by the linearly independent vectors r, and s. Would I try to prove that the vectors x, y, and z can span U and vectors r, and s can span V?
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Dec 10 '18 at 10:51
José Carlos Santos
170k23132238
asked Jul 23 '18 at 5:59
edmonda7edmonda7
394
edited Dec 10 '18 at 10:51
José Carlos Santos
170k23132238
asked Jul 23 '18 at 5:59
edmonda7edmonda7
394
edited Dec 10 '18 at 10:51
José Carlos Santos
170k23132238
edited Dec 10 '18 at 10:51
José Carlos Santos
170k23132238
edited Dec 10 '18 at 10:51
José Carlos Santos
170k23132238
170k23132238
asked Jul 23 '18 at 5:59
edmonda7edmonda7
394
asked Jul 23 '18 at 5:59
edmonda7edmonda7
394
asked Jul 23 '18 at 5:59
edmonda7edmonda7
394
394
1
$begingroup$
Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
$endgroup$
– Jonas Lenz
Jul 23 '18 at 6:09
$begingroup$
That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
$endgroup$
– Jyrki Lahtonen
Jul 23 '18 at 6:12
$begingroup$
@JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
$endgroup$
– edmonda7
Jul 23 '18 at 6:16
2
$begingroup$
In that case, $dim V=3ne 2=dim U$, hence $Vne U$
$endgroup$
– Hagen von Eitzen
Jul 23 '18 at 6:19
add a comment |
1
$begingroup$
Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
$endgroup$
– Jonas Lenz
Jul 23 '18 at 6:09
$begingroup$
That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
$endgroup$
– Jyrki Lahtonen
Jul 23 '18 at 6:12
$begingroup$
@JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
$endgroup$
– edmonda7
Jul 23 '18 at 6:16
2
$begingroup$
In that case, $dim V=3ne 2=dim U$, hence $Vne U$
$endgroup$
– Hagen von Eitzen
Jul 23 '18 at 6:19
1
1
$begingroup$
Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
$endgroup$
– Jonas Lenz
Jul 23 '18 at 6:09
$begingroup$
Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
$endgroup$
– Jonas Lenz
Jul 23 '18 at 6:09
$begingroup$
That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
$endgroup$
– Jyrki Lahtonen
Jul 23 '18 at 6:12
$begingroup$
That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
$endgroup$
– Jyrki Lahtonen
Jul 23 '18 at 6:12
$begingroup$
@JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
$endgroup$
– edmonda7
Jul 23 '18 at 6:16
$begingroup$
@JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
$endgroup$
– edmonda7
Jul 23 '18 at 6:16
2
2
$begingroup$
In that case, $dim V=3ne 2=dim U$, hence $Vne U$
$endgroup$
– Hagen von Eitzen
Jul 23 '18 at 6:19
$begingroup$
In that case, $dim V=3ne 2=dim U$, hence $Vne U$
$endgroup$
– Hagen von Eitzen
Jul 23 '18 at 6:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $S$ and $S'$ are subsets of a vector space $V$, then $operatorname{span}(S)=operatorname{span}(S')$ if and only if each vector of $S$ is a linear combination of elements of $S'$ and vice-versa.
Consider, for instance, the case in which:
- $V=mathbb{R}^3$;
- $S=bigl{(1,-1,0),(1,0,-1),(0,1,-1)bigr}$;
- $S'=bigl{(2,-1,-1),(-1,-1,2)bigr}$.
I claim that $operatorname{span}(S)=operatorname{span}(S')$. In order to prove it, all I have to do is to note that:
- $displaystyle(1,-1,0)=frac23(2,-1,-1)+frac13(-1,-1,2)$;
- $displaystyle(1,0,-1)=frac13(2,-1,-1)-frac13(-1,-1,2)$;
- $displaystyle(0,1,-1)=-frac13(2,-1,-1)-frac23(-1,-1,2)$;
- $displaystyle(2,-1,-1)=(1,-1,0)+(1,0,-1)$;
- $displaystyle(-1,-1,2)=-(1,0,-1)-(0,1,-1)$.
answered Jul 23 '18 at 6:41
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
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%2f2860051%2fhow-do-i-prove-that-2-vector-spaces-are-equal-if-they-have-different-number-of-v%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
$begingroup$
If $S$ and $S'$ are subsets of a vector space $V$, then $operatorname{span}(S)=operatorname{span}(S')$ if and only if each vector of $S$ is a linear combination of elements of $S'$ and vice-versa.
Consider, for instance, the case in which:
- $V=mathbb{R}^3$;
- $S=bigl{(1,-1,0),(1,0,-1),(0,1,-1)bigr}$;
- $S'=bigl{(2,-1,-1),(-1,-1,2)bigr}$.
I claim that $operatorname{span}(S)=operatorname{span}(S')$. In order to prove it, all I have to do is to note that:
- $displaystyle(1,-1,0)=frac23(2,-1,-1)+frac13(-1,-1,2)$;
- $displaystyle(1,0,-1)=frac13(2,-1,-1)-frac13(-1,-1,2)$;
- $displaystyle(0,1,-1)=-frac13(2,-1,-1)-frac23(-1,-1,2)$;
- $displaystyle(2,-1,-1)=(1,-1,0)+(1,0,-1)$;
- $displaystyle(-1,-1,2)=-(1,0,-1)-(0,1,-1)$.
answered Jul 23 '18 at 6:41
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
If $S$ and $S'$ are subsets of a vector space $V$, then $operatorname{span}(S)=operatorname{span}(S')$ if and only if each vector of $S$ is a linear combination of elements of $S'$ and vice-versa.
Consider, for instance, the case in which:
- $V=mathbb{R}^3$;
- $S=bigl{(1,-1,0),(1,0,-1),(0,1,-1)bigr}$;
- $S'=bigl{(2,-1,-1),(-1,-1,2)bigr}$.
I claim that $operatorname{span}(S)=operatorname{span}(S')$. In order to prove it, all I have to do is to note that:
- $displaystyle(1,-1,0)=frac23(2,-1,-1)+frac13(-1,-1,2)$;
- $displaystyle(1,0,-1)=frac13(2,-1,-1)-frac13(-1,-1,2)$;
- $displaystyle(0,1,-1)=-frac13(2,-1,-1)-frac23(-1,-1,2)$;
- $displaystyle(2,-1,-1)=(1,-1,0)+(1,0,-1)$;
- $displaystyle(-1,-1,2)=-(1,0,-1)-(0,1,-1)$.
answered Jul 23 '18 at 6:41
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
If $S$ and $S'$ are subsets of a vector space $V$, then $operatorname{span}(S)=operatorname{span}(S')$ if and only if each vector of $S$ is a linear combination of elements of $S'$ and vice-versa.
Consider, for instance, the case in which:
- $V=mathbb{R}^3$;
- $S=bigl{(1,-1,0),(1,0,-1),(0,1,-1)bigr}$;
- $S'=bigl{(2,-1,-1),(-1,-1,2)bigr}$.
I claim that $operatorname{span}(S)=operatorname{span}(S')$. In order to prove it, all I have to do is to note that:
- $displaystyle(1,-1,0)=frac23(2,-1,-1)+frac13(-1,-1,2)$;
- $displaystyle(1,0,-1)=frac13(2,-1,-1)-frac13(-1,-1,2)$;
- $displaystyle(0,1,-1)=-frac13(2,-1,-1)-frac23(-1,-1,2)$;
- $displaystyle(2,-1,-1)=(1,-1,0)+(1,0,-1)$;
- $displaystyle(-1,-1,2)=-(1,0,-1)-(0,1,-1)$.
answered Jul 23 '18 at 6:41
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
If $S$ and $S'$ are subsets of a vector space $V$, then $operatorname{span}(S)=operatorname{span}(S')$ if and only if each vector of $S$ is a linear combination of elements of $S'$ and vice-versa.
Consider, for instance, the case in which:
- $V=mathbb{R}^3$;
- $S=bigl{(1,-1,0),(1,0,-1),(0,1,-1)bigr}$;
- $S'=bigl{(2,-1,-1),(-1,-1,2)bigr}$.
I claim that $operatorname{span}(S)=operatorname{span}(S')$. In order to prove it, all I have to do is to note that:
- $displaystyle(1,-1,0)=frac23(2,-1,-1)+frac13(-1,-1,2)$;
- $displaystyle(1,0,-1)=frac13(2,-1,-1)-frac13(-1,-1,2)$;
- $displaystyle(0,1,-1)=-frac13(2,-1,-1)-frac23(-1,-1,2)$;
- $displaystyle(2,-1,-1)=(1,-1,0)+(1,0,-1)$;
- $displaystyle(-1,-1,2)=-(1,0,-1)-(0,1,-1)$.
answered Jul 23 '18 at 6:41
José Carlos SantosJosé Carlos Santos
170k23132238
edited Jul 23 '18 at 8:49
answered Jul 23 '18 at 6:41
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jul 23 '18 at 6:41
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jul 23 '18 at 6:41
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
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%2f2860051%2fhow-do-i-prove-that-2-vector-spaces-are-equal-if-they-have-different-number-of-v%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
1
$begingroup$
Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
$endgroup$
– Jonas Lenz
Jul 23 '18 at 6:09
$begingroup$
That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
$endgroup$
– Jyrki Lahtonen
Jul 23 '18 at 6:12
$begingroup$
@JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
$endgroup$
– edmonda7
Jul 23 '18 at 6:16
2
$begingroup$
In that case, $dim V=3ne 2=dim U$, hence $Vne U$
$endgroup$
– Hagen von Eitzen
Jul 23 '18 at 6:19