Proving a set of vectors are a basis
$begingroup$
Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.
So I have set $u = (a,b,c) in A$
such that $a = frac{4c-2b}{2}$, then using this
$u$ is now $(frac{4c-2b}{2},b,c)$
$$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$
I'm uncertain how to show the combination of both vectors span $A$.
linear-algebra
edited Dec 21 '18 at 19:04
hardmath
29.1k953101
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
566
$endgroup$
add a comment |
$begingroup$
Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.
So I have set $u = (a,b,c) in A$
such that $a = frac{4c-2b}{2}$, then using this
$u$ is now $(frac{4c-2b}{2},b,c)$
$$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$
I'm uncertain how to show the combination of both vectors span $A$.
linear-algebra
edited Dec 21 '18 at 19:04
hardmath
29.1k953101
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
566
$endgroup$
add a comment |
$begingroup$
Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.
So I have set $u = (a,b,c) in A$
such that $a = frac{4c-2b}{2}$, then using this
$u$ is now $(frac{4c-2b}{2},b,c)$
$$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$
I'm uncertain how to show the combination of both vectors span $A$.
linear-algebra
edited Dec 21 '18 at 19:04
hardmath
29.1k953101
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
566
$endgroup$
Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.
So I have set $u = (a,b,c) in A$
such that $a = frac{4c-2b}{2}$, then using this
$u$ is now $(frac{4c-2b}{2},b,c)$
$$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$
I'm uncertain how to show the combination of both vectors span $A$.
linear-algebra
linear-algebra
edited Dec 21 '18 at 19:04
hardmath
29.1k953101
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
566
edited Dec 21 '18 at 19:04
hardmath
29.1k953101
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
566
edited Dec 21 '18 at 19:04
hardmath
29.1k953101
edited Dec 21 '18 at 19:04
hardmath
29.1k953101
edited Dec 21 '18 at 19:04
hardmath
29.1k953101
29.1k953101
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
566
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
566
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
566
566
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.
To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.
answered Dec 15 '18 at 15:56
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
You mean $a = frac{4c-3b}2$.
If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.
If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.
Since they are not multiple of each other and you know the dimension is $2$, it is a basis.
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
add a comment |
$begingroup$
Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5718
$endgroup$
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%2f3041644%2fproving-a-set-of-vectors-are-a-basis%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
$begingroup$
First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.
To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.
answered Dec 15 '18 at 15:56
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.
To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.
answered Dec 15 '18 at 15:56
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.
To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.
answered Dec 15 '18 at 15:56
user3482749user3482749
4,296919
$endgroup$
First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.
To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.
answered Dec 15 '18 at 15:56
user3482749user3482749
4,296919
answered Dec 15 '18 at 15:56
user3482749user3482749
4,296919
answered Dec 15 '18 at 15:56
user3482749user3482749
4,296919
answered Dec 15 '18 at 15:56
user3482749user3482749
4,296919
4,296919
add a comment |
add a comment |
$begingroup$
You mean $a = frac{4c-3b}2$.
If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.
If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.
Since they are not multiple of each other and you know the dimension is $2$, it is a basis.
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
add a comment |
$begingroup$
You mean $a = frac{4c-3b}2$.
If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.
If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.
Since they are not multiple of each other and you know the dimension is $2$, it is a basis.
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
add a comment |
$begingroup$
You mean $a = frac{4c-3b}2$.
If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.
If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.
Since they are not multiple of each other and you know the dimension is $2$, it is a basis.
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
You mean $a = frac{4c-3b}2$.
If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.
If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.
Since they are not multiple of each other and you know the dimension is $2$, it is a basis.
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
102k1468119
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
102k1468119
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
102k1468119
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
102k1468119
102k1468119
add a comment |
add a comment |
$begingroup$
Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5718
$endgroup$
add a comment |
$begingroup$
Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5718
$endgroup$
add a comment |
$begingroup$
Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5718
$endgroup$
Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5718
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5718
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5718
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5718
5718
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%2f3041644%2fproving-a-set-of-vectors-are-a-basis%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
