Trouble finding correct basis of subspace defined with a linear equation. The Next CEO of Stack OverflowSparse basis for linear subspaceFinding a basis for a given subspace of $Bbb R^4$Find a basis for the subspace of $BbbR^3$ that is spanned by the vectorsHow can we find a basis for affine subspace?Finding the basis of a subspaceHow to express the basis of a subspace?Basis of defined subspaceBasis of subspace UFinding Orthonormal Basis of a subspace spanned by $Bbb R^4$Systematically finding a basis of a subspace
Is micro rebar a better way to reinforce concrete than rebar?
Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?
Axiom Schema vs Axiom
Why, when going from special to general relativity, do we just replace partial derivatives with covariant derivatives?
I believe this to be a fraud - hired, then asked to cash check and send cash as Bitcoin
How did people program for Consoles with multiple CPUs?
Find non-case sensitive string in a mixed list of elements?
What did we know about the Kessel run before the prequels?
Where do students learn to solve polynomial equations these days?
What happened in Rome, when the western empire "fell"?
Some questions about different axiomatic systems for neighbourhoods
Would a completely good Muggle be able to use a wand?
Rotate a column
Why didn't Khan get resurrected in the Genesis Explosion?
Should I tutor a student who I know has cheated on their homework?
Solving system of ODEs with extra parameter
Is it possible to replace duplicates of a character with one character using tr
Why don't programming languages automatically manage the synchronous/asynchronous problem?
Is it ever safe to open a suspicious HTML file (e.g. email attachment)?
Why the difference in type-inference over the as-pattern in two similar function definitions?
How to count occurrences of text in a file?
A small doubt about the dominated convergence theorem
Is there always a complete, orthogonal set of unitary matrices?
Newlines in BSD sed vs gsed
Trouble finding correct basis of subspace defined with a linear equation.
The Next CEO of Stack OverflowSparse basis for linear subspaceFinding a basis for a given subspace of $Bbb R^4$Find a basis for the subspace of $BbbR^3$ that is spanned by the vectorsHow can we find a basis for affine subspace?Finding the basis of a subspaceHow to express the basis of a subspace?Basis of defined subspaceBasis of subspace UFinding Orthonormal Basis of a subspace spanned by $Bbb R^4$Systematically finding a basis of a subspace
$begingroup$
I have the following subspace in R^3 : H = (X,Y,Z) in R^3
When I try to solve,
I express z = x + 2y which implies that (X,Y,X+2Y).
I then find my basis as follows
T(e1) = T(1 0 0) = (1 0 1)
T(e2) = T(0 1 0) = (0 1 2)
So my basis is v1 = (1 0 1) and v2 = (0 1 2) but the correct answer for v2 is (2 -1 0)
I do not understand why. If basis is not unique I want to know how to find (2 -1 0).
linear-algebra
asked Mar 27 at 19:19
Dr.StoneDr.Stone
334
New contributor
Dr.Stone 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 |
$begingroup$
I have the following subspace in R^3 : H = (X,Y,Z) in R^3
When I try to solve,
I express z = x + 2y which implies that (X,Y,X+2Y).
I then find my basis as follows
T(e1) = T(1 0 0) = (1 0 1)
T(e2) = T(0 1 0) = (0 1 2)
So my basis is v1 = (1 0 1) and v2 = (0 1 2) but the correct answer for v2 is (2 -1 0)
I do not understand why. If basis is not unique I want to know how to find (2 -1 0).
linear-algebra
asked Mar 27 at 19:19
Dr.StoneDr.Stone
334
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Basis is not unique!
$endgroup$
– Zeekless
Mar 27 at 19:22
1
$begingroup$
A basis is not unique! Note that $v_3=(2,-1,0)=2v_1-v_2$, so $(v_1, v_3)$ and $(v_2, v_3)$ are also bases.
$endgroup$
– Bernard
Mar 27 at 19:23
add a comment |
$begingroup$
I have the following subspace in R^3 : H = (X,Y,Z) in R^3
When I try to solve,
I express z = x + 2y which implies that (X,Y,X+2Y).
I then find my basis as follows
T(e1) = T(1 0 0) = (1 0 1)
T(e2) = T(0 1 0) = (0 1 2)
So my basis is v1 = (1 0 1) and v2 = (0 1 2) but the correct answer for v2 is (2 -1 0)
I do not understand why. If basis is not unique I want to know how to find (2 -1 0).
linear-algebra
asked Mar 27 at 19:19
Dr.StoneDr.Stone
334
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have the following subspace in R^3 : H = (X,Y,Z) in R^3
When I try to solve,
I express z = x + 2y which implies that (X,Y,X+2Y).
I then find my basis as follows
T(e1) = T(1 0 0) = (1 0 1)
T(e2) = T(0 1 0) = (0 1 2)
So my basis is v1 = (1 0 1) and v2 = (0 1 2) but the correct answer for v2 is (2 -1 0)
I do not understand why. If basis is not unique I want to know how to find (2 -1 0).
linear-algebra
linear-algebra
asked Mar 27 at 19:19
Dr.StoneDr.Stone
334
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 27 at 19:19
Dr.StoneDr.Stone
334
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 27 at 19:44
Dr.Stone
asked Mar 27 at 19:19
Dr.StoneDr.Stone
334
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 27 at 19:19
Dr.StoneDr.Stone
334
asked Mar 27 at 19:19
Dr.StoneDr.Stone
334
334
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Basis is not unique!
$endgroup$
– Zeekless
Mar 27 at 19:22
1
$begingroup$
A basis is not unique! Note that $v_3=(2,-1,0)=2v_1-v_2$, so $(v_1, v_3)$ and $(v_2, v_3)$ are also bases.
$endgroup$
– Bernard
Mar 27 at 19:23
add a comment |
1
$begingroup$
Basis is not unique!
$endgroup$
– Zeekless
Mar 27 at 19:22
1
$begingroup$
A basis is not unique! Note that $v_3=(2,-1,0)=2v_1-v_2$, so $(v_1, v_3)$ and $(v_2, v_3)$ are also bases.
$endgroup$
– Bernard
Mar 27 at 19:23
1
1
$begingroup$
Basis is not unique!
$endgroup$
– Zeekless
Mar 27 at 19:22
$begingroup$
Basis is not unique!
$endgroup$
– Zeekless
Mar 27 at 19:22
1
1
$begingroup$
A basis is not unique! Note that $v_3=(2,-1,0)=2v_1-v_2$, so $(v_1, v_3)$ and $(v_2, v_3)$ are also bases.
$endgroup$
– Bernard
Mar 27 at 19:23
$begingroup$
A basis is not unique! Note that $v_3=(2,-1,0)=2v_1-v_2$, so $(v_1, v_3)$ and $(v_2, v_3)$ are also bases.
$endgroup$
– Bernard
Mar 27 at 19:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The space consists of all vectors orthogonal to $mathbf n = (1,2,-1)^T$, (why?) so any two linearly-independent vectors orthogonal to $mathbf n$ form a basis for it. Without further constraints there’s no good reason to call one choice more “correct” than another.
Now, as to how to come up with $(2,-1,0)^T$ specifically, I would first suggest following whatever method your text demonstrated for solving similar problems. However, for any nonzero vector $mathbf v = (a,b,c)^T$, we also have that the vectors $(0,c,-b)^T$, $(-c,0,a)^T$ and $(b,-a,0)^T$ are all orthogonal to $mathbf v$ and at least two of them are non-zero. (These are just the cross products of $mathbf v$ with the standard basis vectors.) For $mathbf v=(1,2,-1)^T$, the last one just happens to be $(2,-1,0)^T$.
answered Mar 27 at 20:12
amdamd
31.4k21052
$endgroup$
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
);
);
Dr.Stone 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%2f3164994%2ftrouble-finding-correct-basis-of-subspace-defined-with-a-linear-equation%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
$begingroup$
The space consists of all vectors orthogonal to $mathbf n = (1,2,-1)^T$, (why?) so any two linearly-independent vectors orthogonal to $mathbf n$ form a basis for it. Without further constraints there’s no good reason to call one choice more “correct” than another.
Now, as to how to come up with $(2,-1,0)^T$ specifically, I would first suggest following whatever method your text demonstrated for solving similar problems. However, for any nonzero vector $mathbf v = (a,b,c)^T$, we also have that the vectors $(0,c,-b)^T$, $(-c,0,a)^T$ and $(b,-a,0)^T$ are all orthogonal to $mathbf v$ and at least two of them are non-zero. (These are just the cross products of $mathbf v$ with the standard basis vectors.) For $mathbf v=(1,2,-1)^T$, the last one just happens to be $(2,-1,0)^T$.
answered Mar 27 at 20:12
amdamd
31.4k21052
$endgroup$
add a comment |
$begingroup$
The space consists of all vectors orthogonal to $mathbf n = (1,2,-1)^T$, (why?) so any two linearly-independent vectors orthogonal to $mathbf n$ form a basis for it. Without further constraints there’s no good reason to call one choice more “correct” than another.
Now, as to how to come up with $(2,-1,0)^T$ specifically, I would first suggest following whatever method your text demonstrated for solving similar problems. However, for any nonzero vector $mathbf v = (a,b,c)^T$, we also have that the vectors $(0,c,-b)^T$, $(-c,0,a)^T$ and $(b,-a,0)^T$ are all orthogonal to $mathbf v$ and at least two of them are non-zero. (These are just the cross products of $mathbf v$ with the standard basis vectors.) For $mathbf v=(1,2,-1)^T$, the last one just happens to be $(2,-1,0)^T$.
answered Mar 27 at 20:12
amdamd
31.4k21052
$endgroup$
add a comment |
$begingroup$
The space consists of all vectors orthogonal to $mathbf n = (1,2,-1)^T$, (why?) so any two linearly-independent vectors orthogonal to $mathbf n$ form a basis for it. Without further constraints there’s no good reason to call one choice more “correct” than another.
Now, as to how to come up with $(2,-1,0)^T$ specifically, I would first suggest following whatever method your text demonstrated for solving similar problems. However, for any nonzero vector $mathbf v = (a,b,c)^T$, we also have that the vectors $(0,c,-b)^T$, $(-c,0,a)^T$ and $(b,-a,0)^T$ are all orthogonal to $mathbf v$ and at least two of them are non-zero. (These are just the cross products of $mathbf v$ with the standard basis vectors.) For $mathbf v=(1,2,-1)^T$, the last one just happens to be $(2,-1,0)^T$.
answered Mar 27 at 20:12
amdamd
31.4k21052
$endgroup$
The space consists of all vectors orthogonal to $mathbf n = (1,2,-1)^T$, (why?) so any two linearly-independent vectors orthogonal to $mathbf n$ form a basis for it. Without further constraints there’s no good reason to call one choice more “correct” than another.
Now, as to how to come up with $(2,-1,0)^T$ specifically, I would first suggest following whatever method your text demonstrated for solving similar problems. However, for any nonzero vector $mathbf v = (a,b,c)^T$, we also have that the vectors $(0,c,-b)^T$, $(-c,0,a)^T$ and $(b,-a,0)^T$ are all orthogonal to $mathbf v$ and at least two of them are non-zero. (These are just the cross products of $mathbf v$ with the standard basis vectors.) For $mathbf v=(1,2,-1)^T$, the last one just happens to be $(2,-1,0)^T$.
answered Mar 27 at 20:12
amdamd
31.4k21052
answered Mar 27 at 20:12
amdamd
31.4k21052
answered Mar 27 at 20:12
amdamd
31.4k21052
answered Mar 27 at 20:12
amdamd
31.4k21052
31.4k21052
add a comment |
add a comment |
Dr.Stone is a new contributor. Be nice, and check out our Code of Conduct.
Dr.Stone is a new contributor. Be nice, and check out our Code of Conduct.
Dr.Stone is a new contributor. Be nice, and check out our Code of Conduct.
Dr.Stone 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%2f3164994%2ftrouble-finding-correct-basis-of-subspace-defined-with-a-linear-equation%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
$begingroup$
Basis is not unique!
$endgroup$
– Zeekless
Mar 27 at 19:22
1
$begingroup$
A basis is not unique! Note that $v_3=(2,-1,0)=2v_1-v_2$, so $(v_1, v_3)$ and $(v_2, v_3)$ are also bases.
$endgroup$
– Bernard
Mar 27 at 19:23