Etydhv

2001: A Space Odyssey's use of the song "Daisy Bell" (Bicycle Built for Two); life imitates art or vice-versa?

Do wooden building fires get hotter than 600°C?

How does the math work when buying airline miles?

What is homebrew?

How to answer "Have you ever been terminated?"

What is the meaning of the new sigil in Game of Thrones Season 8 intro?

If a VARCHAR(MAX) column is included in an index, is the entire value always stored in the index page(s)?

What does "lightly crushed" mean for cardamon pods?

Dating a Former Employee

Trademark violation for app?

What is implied by the word 'Desika'

Fantasy story; one type of magic grows in power with use, but the more powerful they are, they more they are drawn to travel to their source

How to tell that you are a giant?

Why are there no cargo aircraft with "flying wing" design?

Around usage results

How do I make this wiring inside cabinet safer? (Pic)

How come Sam didn't become Lord of Horn Hill?

Do jazz musicians improvise on the parent scale in addition to the chord-scales?

また usage in a dictionary

When a candle burns, why does the top of wick glow if bottom of flame is hottest?

How could we fake a moon landing now?

How would a mousetrap for use in space work?

Using audio cues to encourage good posture

Is CEO the profession with the most psychopaths?

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. (

1

$begingroup$

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.

1. Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa




2. Set u = av + bw = u (where a and b are constants)



3. Disprove (2)

However, I could not get past step 1.

Any pointers would be greatly appreciated.

linear-algebra matrices

share|cite|improve this question

edited 11 mins ago

YuiTo Cheng

2,52341037

asked 23 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 | 

1

$begingroup$

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.

1. Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa




2. Set u = av + bw = u (where a and b are constants)



3. Disprove (2)

However, I could not get past step 1.

Any pointers would be greatly appreciated.

linear-algebra matrices

share|cite|improve this question

edited 11 mins ago

YuiTo Cheng

2,52341037

asked 23 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.

1. Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa




2. Set u = av + bw = u (where a and b are constants)



3. Disprove (2)

However, I could not get past step 1.

Any pointers would be greatly appreciated.

linear-algebra matrices

share|cite|improve this question

edited 11 mins ago

YuiTo Cheng

2,52341037

asked 23 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.

$endgroup$

share|cite|improve this question

edited 11 mins ago

YuiTo Cheng

2,52341037

asked 23 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.

share|cite|improve this question

edited 11 mins ago

YuiTo Cheng

2,52341037

asked 23 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 11 mins ago

YuiTo Cheng

2,52341037

edited 11 mins ago

YuiTo Cheng

2,52341037

asked 23 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 23 mins ago

Dimen3ionalDimen3ional

82

2 Answers
2

active

oldest

votes

2

$begingroup$

$defmyvec#1bf#1$
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 $myvec u=myvec 0$.

share|cite|improve this answer

answered 15 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
  8 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you studied proof by contradiction or by contrapositive? That's what this is.
  $endgroup$
  – David
  7 mins ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered 14 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.5k53270

$endgroup$

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.

draft saved

draft discarded

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

Name

Email

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

Name

Email

Required, but never shown

2

$begingroup$

$defmyvec#1bf#1$
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 $myvec u=myvec 0$.

share|cite|improve this answer

answered 15 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
  8 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you studied proof by contradiction or by contrapositive? That's what this is.
  $endgroup$
  – David
  7 mins ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$defmyvec#1bf#1$
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 $myvec u=myvec 0$.

share|cite|improve this answer

answered 15 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
  8 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you studied proof by contradiction or by contrapositive? That's what this is.
  $endgroup$
  – David
  7 mins ago
  
  

  
  
  
  

add a comment | 

$begingroup$

$defmyvec#1bf#1$
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 $myvec u=myvec 0$.

share|cite|improve this answer

answered 15 mins ago

DavidDavid

70.1k668131

$endgroup$

share|cite|improve this answer

answered 15 mins ago

DavidDavid

70.1k668131

answered 15 mins ago

DavidDavid

70.1k668131

answered 15 mins ago

DavidDavid

70.1k668131

2

$begingroup$

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.

share|cite|improve this answer

answered 14 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.5k53270

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered 14 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.5k53270

$endgroup$

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.

share|cite|improve this answer

answered 14 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.5k53270

$endgroup$

share|cite|improve this answer

answered 14 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.5k53270

answered 14 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.5k53270

answered 14 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.5k53270