Mixing
How to show one set of vectors span R3 if its components do?
The question states
Show that if a, b, c are vectors in R3, then {b+c, c+a,a+b} spans R3 iff {a,b,c} spans R3.
I've started trying Gauss-Jordan with the first set in Ax = 0 form, but I haven't got anywhere.
What am I supposed to do from here?
linear-algebra vectors
asked Nov 20 '18 at 19:56
mathPhys
114
add a comment |
The question states
Show that if a, b, c are vectors in R3, then {b+c, c+a,a+b} spans R3 iff {a,b,c} spans R3.
I've started trying Gauss-Jordan with the first set in Ax = 0 form, but I haven't got anywhere.
What am I supposed to do from here?
linear-algebra vectors
asked Nov 20 '18 at 19:56
mathPhys
114
add a comment |
The question states
Show that if a, b, c are vectors in R3, then {b+c, c+a,a+b} spans R3 iff {a,b,c} spans R3.
I've started trying Gauss-Jordan with the first set in Ax = 0 form, but I haven't got anywhere.
What am I supposed to do from here?
linear-algebra vectors
asked Nov 20 '18 at 19:56
mathPhys
114
The question states
Show that if a, b, c are vectors in R3, then {b+c, c+a,a+b} spans R3 iff {a,b,c} spans R3.
I've started trying Gauss-Jordan with the first set in Ax = 0 form, but I haven't got anywhere.
What am I supposed to do from here?
linear-algebra vectors
linear-algebra vectors
asked Nov 20 '18 at 19:56
mathPhys
114
asked Nov 20 '18 at 19:56
mathPhys
114
asked Nov 20 '18 at 19:56
mathPhys
114
asked Nov 20 '18 at 19:56
mathPhys
114
asked Nov 20 '18 at 19:56
mathPhys
114
114
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
HINT
Let consider
$$k_1(b+c)+k_2(c+a)+k_3(a+b)=0$$
$$(k_2+k_3)a+(k_1+k_3)b+(k_1+k_2)c=0$$
answered Nov 20 '18 at 19:58
gimusi
1
add a comment |
Hint
$text{{a,b,c}}$ spans $mathbb R^3$ implies that $l_1a+l_2b+l_3c=0 $ for some $l_1,l_2,l_3$.
Now evaluate $l_1(b+c)+l_2(c+a)+l_3(a+b) $ and use the equality above.
answered Nov 20 '18 at 20:06
dougle
155
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 '18 at 20:01
add a comment |
Find a matrix $M$ such that $$begin{bmatrix}mathbf b+mathbf c & mathbf c+mathbf a & mathbf a+mathbf bend{bmatrix} = begin{bmatrix}mathbf a & mathbf b & mathbf cend{bmatrix} M$$ and use properties of determinants to argue that the columns of the matrix on the left-hand side are linearly independent iff $mathbf a$, $mathbf b$ and $mathbf c$ are.
answered Nov 20 '18 at 23:49
amd
29.2k21050
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 '18 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 '18 at 20:42
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%2f3006808%2fhow-to-show-one-set-of-vectors-span-r3-if-its-components-do%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
Let consider
$$k_1(b+c)+k_2(c+a)+k_3(a+b)=0$$
$$(k_2+k_3)a+(k_1+k_3)b+(k_1+k_2)c=0$$
answered Nov 20 '18 at 19:58
gimusi
1
add a comment |
HINT
Let consider
$$k_1(b+c)+k_2(c+a)+k_3(a+b)=0$$
$$(k_2+k_3)a+(k_1+k_3)b+(k_1+k_2)c=0$$
answered Nov 20 '18 at 19:58
gimusi
1
add a comment |
HINT
Let consider
$$k_1(b+c)+k_2(c+a)+k_3(a+b)=0$$
$$(k_2+k_3)a+(k_1+k_3)b+(k_1+k_2)c=0$$
answered Nov 20 '18 at 19:58
gimusi
1
HINT
Let consider
$$k_1(b+c)+k_2(c+a)+k_3(a+b)=0$$
$$(k_2+k_3)a+(k_1+k_3)b+(k_1+k_2)c=0$$
answered Nov 20 '18 at 19:58
gimusi
1
answered Nov 20 '18 at 19:58
gimusi
1
answered Nov 20 '18 at 19:58
gimusi
1
answered Nov 20 '18 at 19:58
gimusi
1
1
add a comment |
add a comment |
Hint
$text{{a,b,c}}$ spans $mathbb R^3$ implies that $l_1a+l_2b+l_3c=0 $ for some $l_1,l_2,l_3$.
Now evaluate $l_1(b+c)+l_2(c+a)+l_3(a+b) $ and use the equality above.
answered Nov 20 '18 at 20:06
dougle
155
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 '18 at 20:01
add a comment |
Hint
$text{{a,b,c}}$ spans $mathbb R^3$ implies that $l_1a+l_2b+l_3c=0 $ for some $l_1,l_2,l_3$.
Now evaluate $l_1(b+c)+l_2(c+a)+l_3(a+b) $ and use the equality above.
answered Nov 20 '18 at 20:06
dougle
155
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 '18 at 20:01
add a comment |
Hint
$text{{a,b,c}}$ spans $mathbb R^3$ implies that $l_1a+l_2b+l_3c=0 $ for some $l_1,l_2,l_3$.
Now evaluate $l_1(b+c)+l_2(c+a)+l_3(a+b) $ and use the equality above.
answered Nov 20 '18 at 20:06
dougle
155
Hint
$text{{a,b,c}}$ spans $mathbb R^3$ implies that $l_1a+l_2b+l_3c=0 $ for some $l_1,l_2,l_3$.
Now evaluate $l_1(b+c)+l_2(c+a)+l_3(a+b) $ and use the equality above.
answered Nov 20 '18 at 20:06
dougle
155
answered Nov 20 '18 at 20:06
dougle
155
answered Nov 20 '18 at 20:06
dougle
155
answered Nov 20 '18 at 20:06
dougle
155
155
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 '18 at 20:01
add a comment |
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 '18 at 20:01
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 '18 at 20:01
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 '18 at 20:01
add a comment |
Find a matrix $M$ such that $$begin{bmatrix}mathbf b+mathbf c & mathbf c+mathbf a & mathbf a+mathbf bend{bmatrix} = begin{bmatrix}mathbf a & mathbf b & mathbf cend{bmatrix} M$$ and use properties of determinants to argue that the columns of the matrix on the left-hand side are linearly independent iff $mathbf a$, $mathbf b$ and $mathbf c$ are.
answered Nov 20 '18 at 23:49
amd
29.2k21050
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 '18 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 '18 at 20:42
add a comment |
Find a matrix $M$ such that $$begin{bmatrix}mathbf b+mathbf c & mathbf c+mathbf a & mathbf a+mathbf bend{bmatrix} = begin{bmatrix}mathbf a & mathbf b & mathbf cend{bmatrix} M$$ and use properties of determinants to argue that the columns of the matrix on the left-hand side are linearly independent iff $mathbf a$, $mathbf b$ and $mathbf c$ are.
answered Nov 20 '18 at 23:49
amd
29.2k21050
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 '18 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 '18 at 20:42
add a comment |
Find a matrix $M$ such that $$begin{bmatrix}mathbf b+mathbf c & mathbf c+mathbf a & mathbf a+mathbf bend{bmatrix} = begin{bmatrix}mathbf a & mathbf b & mathbf cend{bmatrix} M$$ and use properties of determinants to argue that the columns of the matrix on the left-hand side are linearly independent iff $mathbf a$, $mathbf b$ and $mathbf c$ are.
answered Nov 20 '18 at 23:49
amd
29.2k21050
Find a matrix $M$ such that $$begin{bmatrix}mathbf b+mathbf c & mathbf c+mathbf a & mathbf a+mathbf bend{bmatrix} = begin{bmatrix}mathbf a & mathbf b & mathbf cend{bmatrix} M$$ and use properties of determinants to argue that the columns of the matrix on the left-hand side are linearly independent iff $mathbf a$, $mathbf b$ and $mathbf c$ are.
answered Nov 20 '18 at 23:49
amd
29.2k21050
edited Nov 21 '18 at 20:42
answered Nov 20 '18 at 23:49
amd
29.2k21050
answered Nov 20 '18 at 23:49
amd
29.2k21050
answered Nov 20 '18 at 23:49
amd
29.2k21050
29.2k21050
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 '18 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 '18 at 20:42
add a comment |
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 '18 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 '18 at 20:42
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 '18 at 20:03
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 '18 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 '18 at 20:42
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 '18 at 20:42
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3006808%2fhow-to-show-one-set-of-vectors-span-r3-if-its-components-do%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
