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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 31 mins ago

YuiTo Cheng

2,52341037

asked 43 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 31 mins ago

YuiTo Cheng

2,52341037

asked 43 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 31 mins ago

YuiTo Cheng

2,52341037

asked 43 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 31 mins ago

YuiTo Cheng

2,52341037

asked 43 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 31 mins ago

YuiTo Cheng

2,52341037

asked 43 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 31 mins ago

YuiTo Cheng

2,52341037

edited 31 mins ago

YuiTo Cheng

2,52341037

asked 43 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 43 mins ago

Dimen3ionalDimen3ional

82

2 Answers
2

active

oldest

votes

2

$begingroup$

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

share|cite|improve this answer

answered 35 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
  28 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you studied proof by contradiction or by contrapositive? That's what this is.
  $endgroup$
  – David
  27 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 34 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270

$endgroup$

add a comment | 

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2

$begingroup$

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

share|cite|improve this answer

answered 35 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
  28 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

2

$begingroup$

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

share|cite|improve this answer

answered 35 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
  28 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you studied proof by contradiction or by contrapositive? That's what this is.
  $endgroup$
  – David
  27 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 $myvec u=myvec 0$.

share|cite|improve this answer

answered 35 mins ago

DavidDavid

70.1k668131

$endgroup$

share|cite|improve this answer

answered 35 mins ago

DavidDavid

70.1k668131

answered 35 mins ago

DavidDavid

70.1k668131

answered 35 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 34 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270

$endgroup$

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 34 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.

share|cite|improve this answer

answered 34 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270

$endgroup$

share|cite|improve this answer

answered 34 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270

answered 34 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270

answered 34 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270