If u is orthogonal to both v and w, and u not equal to 0, argue that u is not in the span of v and w. (
$begingroup$
QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.
Where I am at:
I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.
I also tried formulating the following steps to solve the problem.
Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa
Set u = av + bw = u (where a and b are constants)
- Disprove (2)
However, I could not get past step 1.
Any pointers would be greatly appreciated.
linear-algebra matrices
edited 44 mins ago
YuiTo Cheng
2,52341037
asked 56 mins ago
Dimen3ionalDimen3ional
82
New contributor
Dimen3ional 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 |
$begingroup$
QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.
Where I am at:
I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.
I also tried formulating the following steps to solve the problem.
Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa
Set u = av + bw = u (where a and b are constants)
- Disprove (2)
However, I could not get past step 1.
Any pointers would be greatly appreciated.
linear-algebra matrices
edited 44 mins ago
YuiTo Cheng
2,52341037
asked 56 mins ago
Dimen3ionalDimen3ional
82
New contributor
Dimen3ional 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 |
$begingroup$
QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.
Where I am at:
I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.
I also tried formulating the following steps to solve the problem.
Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa
Set u = av + bw = u (where a and b are constants)
- Disprove (2)
However, I could not get past step 1.
Any pointers would be greatly appreciated.
linear-algebra matrices
edited 44 mins ago
YuiTo Cheng
2,52341037
asked 56 mins ago
Dimen3ionalDimen3ional
82
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.
Where I am at:
I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.
I also tried formulating the following steps to solve the problem.
Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa
Set u = av + bw = u (where a and b are constants)
- Disprove (2)
However, I could not get past step 1.
Any pointers would be greatly appreciated.
linear-algebra matrices
linear-algebra matrices
edited 44 mins ago
YuiTo Cheng
2,52341037
asked 56 mins ago
Dimen3ionalDimen3ional
82
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 44 mins ago
YuiTo Cheng
2,52341037
asked 56 mins ago
Dimen3ionalDimen3ional
82
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 44 mins ago
YuiTo Cheng
2,52341037
edited 44 mins ago
YuiTo Cheng
2,52341037
edited 44 mins ago
YuiTo Cheng
2,52341037
2,52341037
asked 56 mins ago
Dimen3ionalDimen3ional
82
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 56 mins ago
Dimen3ionalDimen3ional
82
asked 56 mins ago
Dimen3ionalDimen3ional
82
82
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$defmyvec#1{{bf#1}}$
This is the same as saying
if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.
So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.
answered 48 mins ago
DavidDavid
70.1k668131
$endgroup$
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
41 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
40 mins ago
add a comment |
$begingroup$
If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.
answered 47 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
$endgroup$
add a comment |
Your Answer
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
});
}
});
Dimen3ional 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%2f3192025%2fif-u-is-orthogonal-to-both-v-and-w-and-u-not-equal-to-0-argue-that-u-is-not-in%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$defmyvec#1{{bf#1}}$
This is the same as saying
if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.
So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.
answered 48 mins ago
DavidDavid
70.1k668131
$endgroup$
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
41 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
40 mins ago
add a comment |
$begingroup$
$defmyvec#1{{bf#1}}$
This is the same as saying
if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.
So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.
answered 48 mins ago
DavidDavid
70.1k668131
$endgroup$
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
41 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
40 mins ago
add a comment |
$begingroup$
$defmyvec#1{{bf#1}}$
This is the same as saying
if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.
So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.
answered 48 mins ago
DavidDavid
70.1k668131
$endgroup$
$defmyvec#1{{bf#1}}$
This is the same as saying
if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.
So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.
answered 48 mins ago
DavidDavid
70.1k668131
answered 48 mins ago
DavidDavid
70.1k668131
answered 48 mins ago
DavidDavid
70.1k668131
answered 48 mins ago
DavidDavid
70.1k668131
70.1k668131
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
41 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
40 mins ago
add a comment |
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
41 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
40 mins ago
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
41 mins ago
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
41 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
40 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
40 mins ago
add a comment |
$begingroup$
If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.
answered 47 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
$endgroup$
add a comment |
$begingroup$
If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.
answered 47 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
$endgroup$
add a comment |
$begingroup$
If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.
answered 47 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
$endgroup$
If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.
answered 47 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
answered 47 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
answered 47 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
answered 47 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
75.6k53270
add a comment |
add a comment |
Dimen3ional is a new contributor. Be nice, and check out our Code of Conduct.
Dimen3ional is a new contributor. Be nice, and check out our Code of Conduct.
Dimen3ional is a new contributor. Be nice, and check out our Code of Conduct.
Dimen3ional 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%2f3192025%2fif-u-is-orthogonal-to-both-v-and-w-and-u-not-equal-to-0-argue-that-u-is-not-in%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
