Spanning Vectors - proof [duplicate]

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  Basis for a subspace Proof
  
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Prove that: If v and w span V, then v+w and v-w also span V.

my thinking behind this is that since the two vectors are linear combinations of v and w that is why they span V, but am unsure how to properly go about it.

linear-algebra

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asked Jul 30 at 9:21

J-Dorman

555

marked as duplicate by Arthur, Michael Hoppe, John Ma, Lord Shark the Unknown, Simply Beautiful Art Aug 1 at 14:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  math.stackexchange.com/questions/2866751/…
  – Fred
  Jul 30 at 9:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In general, linear combinations of spanning vectors won't span the vector space. For example, $e_1=beginpmatrix1\0endpmatrix$ and $e_2=beginpmatrix0\1endpmatrix$ span $mathbbR^2$, but $e_1 + e_2$ and $2e_1 + 2e_2$ only span a $1$-dimensional subspace.
  – Babelfish
  Jul 30 at 9:32
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

1

This question already has an answer here:

• 
  Basis for a subspace Proof
  
  3 answers
  
  

Prove that: If v and w span V, then v+w and v-w also span V.

my thinking behind this is that since the two vectors are linear combinations of v and w that is why they span V, but am unsure how to properly go about it.

linear-algebra

share|cite|improve this question

asked Jul 30 at 9:21

J-Dorman

555

marked as duplicate by Arthur, Michael Hoppe, John Ma, Lord Shark the Unknown, Simply Beautiful Art Aug 1 at 14:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  math.stackexchange.com/questions/2866751/…
  – Fred
  Jul 30 at 9:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In general, linear combinations of spanning vectors won't span the vector space. For example, $e_1=beginpmatrix1\0endpmatrix$ and $e_2=beginpmatrix0\1endpmatrix$ span $mathbbR^2$, but $e_1 + e_2$ and $2e_1 + 2e_2$ only span a $1$-dimensional subspace.
  – Babelfish
  Jul 30 at 9:32
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

1

up vote
-1
down vote

favorite

1

1

This question already has an answer here:

• 
  Basis for a subspace Proof
  
  3 answers
  
  

Prove that: If v and w span V, then v+w and v-w also span V.

my thinking behind this is that since the two vectors are linear combinations of v and w that is why they span V, but am unsure how to properly go about it.

linear-algebra

share|cite|improve this question

asked Jul 30 at 9:21

J-Dorman

555

This question already has an answer here:

• 
  Basis for a subspace Proof
  
  3 answers
  
  

Prove that: If v and w span V, then v+w and v-w also span V.

my thinking behind this is that since the two vectors are linear combinations of v and w that is why they span V, but am unsure how to properly go about it.

This question already has an answer here:

• 
  Basis for a subspace Proof
  
  3 answers
  
  

linear-algebra

share|cite|improve this question

asked Jul 30 at 9:21

J-Dorman

555

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 9:21

J-Dorman

555

asked Jul 30 at 9:21

J-Dorman

555

asked Jul 30 at 9:21

J-Dorman

555

555

marked as duplicate by Arthur, Michael Hoppe, John Ma, Lord Shark the Unknown, Simply Beautiful Art Aug 1 at 14:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Arthur, Michael Hoppe, John Ma, Lord Shark the Unknown, Simply Beautiful Art Aug 1 at 14:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  math.stackexchange.com/questions/2866751/…
  – Fred
  Jul 30 at 9:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In general, linear combinations of spanning vectors won't span the vector space. For example, $e_1=beginpmatrix1\0endpmatrix$ and $e_2=beginpmatrix0\1endpmatrix$ span $mathbbR^2$, but $e_1 + e_2$ and $2e_1 + 2e_2$ only span a $1$-dimensional subspace.
  – Babelfish
  Jul 30 at 9:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  math.stackexchange.com/questions/2866751/…
  – Fred
  Jul 30 at 9:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In general, linear combinations of spanning vectors won't span the vector space. For example, $e_1=beginpmatrix1\0endpmatrix$ and $e_2=beginpmatrix0\1endpmatrix$ span $mathbbR^2$, but $e_1 + e_2$ and $2e_1 + 2e_2$ only span a $1$-dimensional subspace.
  – Babelfish
  Jul 30 at 9:32
  

  
  
  
  

math.stackexchange.com/questions/2866751/…
– Fred
Jul 30 at 9:22

math.stackexchange.com/questions/2866751/…
– Fred
Jul 30 at 9:22

1

1

In general, linear combinations of spanning vectors won't span the vector space. For example, $e_1=beginpmatrix1\0endpmatrix$ and $e_2=beginpmatrix0\1endpmatrix$ span $mathbbR^2$, but $e_1 + e_2$ and $2e_1 + 2e_2$ only span a $1$-dimensional subspace.
– Babelfish
Jul 30 at 9:32

In general, linear combinations of spanning vectors won't span the vector space. For example, $e_1=beginpmatrix1\0endpmatrix$ and $e_2=beginpmatrix0\1endpmatrix$ span $mathbbR^2$, but $e_1 + e_2$ and $2e_1 + 2e_2$ only span a $1$-dimensional subspace.
– Babelfish
Jul 30 at 9:32

add a comment | 

2 Answers
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up vote
3
down vote

Consider any vector in the span

$$u=av+bw$$

then

$$u=c(v+w)+d(v-w)=(c+d)v+(c-d)w$$

then choose

• $c+d=a$


• $c-d=b$

that is

• $c=fraca+b2$


• $d=fraca-b2$

$$u=av+bw=fraca+b2(v+w)+fraca-b2(v-w)$$

and therefore also $v+w$ and $v-w$ span $V$.

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answered Jul 30 at 10:08

gimusi

64.5k73482

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up vote
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Note that $langle v+w,v-wranglesubseteqlangle v,wrangle$, bwcause any linear combination of $v+w$ and $v-w$ is obviously a linear combination of $v$ and $w$.

On the other hand,
$$
v=frac12(v+w)+frac12(v-w)
qquad
w=frac12(v+w)-frac12(v-w)
$$
For the same reason as before, $langle v,wranglesubseteqlangle v+w,v-wrangle$.

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answered Jul 30 at 11:37

egreg

164k1180187

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Consider any vector in the span

$$u=av+bw$$

then

$$u=c(v+w)+d(v-w)=(c+d)v+(c-d)w$$

then choose

• $c+d=a$


• $c-d=b$

that is

• $c=fraca+b2$


• $d=fraca-b2$

$$u=av+bw=fraca+b2(v+w)+fraca-b2(v-w)$$

and therefore also $v+w$ and $v-w$ span $V$.

share|cite|improve this answer

answered Jul 30 at 10:08

gimusi

64.5k73482

add a comment | 

up vote
3
down vote

Consider any vector in the span

$$u=av+bw$$

then

$$u=c(v+w)+d(v-w)=(c+d)v+(c-d)w$$

then choose

• $c+d=a$


• $c-d=b$

that is

• $c=fraca+b2$


• $d=fraca-b2$

$$u=av+bw=fraca+b2(v+w)+fraca-b2(v-w)$$

and therefore also $v+w$ and $v-w$ span $V$.

share|cite|improve this answer

answered Jul 30 at 10:08

gimusi

64.5k73482

add a comment | 

up vote
3
down vote

up vote
3
down vote

Consider any vector in the span

$$u=av+bw$$

then

$$u=c(v+w)+d(v-w)=(c+d)v+(c-d)w$$

then choose

• $c+d=a$


• $c-d=b$

that is

• $c=fraca+b2$


• $d=fraca-b2$

$$u=av+bw=fraca+b2(v+w)+fraca-b2(v-w)$$

and therefore also $v+w$ and $v-w$ span $V$.

share|cite|improve this answer

answered Jul 30 at 10:08

gimusi

64.5k73482

Consider any vector in the span

$$u=av+bw$$

then

$$u=c(v+w)+d(v-w)=(c+d)v+(c-d)w$$

then choose

• $c+d=a$


• $c-d=b$

that is

• $c=fraca+b2$


• $d=fraca-b2$

$$u=av+bw=fraca+b2(v+w)+fraca-b2(v-w)$$

and therefore also $v+w$ and $v-w$ span $V$.

share|cite|improve this answer

answered Jul 30 at 10:08

gimusi

64.5k73482

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 10:08

gimusi

64.5k73482

answered Jul 30 at 10:08

gimusi

64.5k73482

answered Jul 30 at 10:08

gimusi

64.5k73482

64.5k73482

add a comment | 

add a comment | 

up vote
3
down vote

Note that $langle v+w,v-wranglesubseteqlangle v,wrangle$, bwcause any linear combination of $v+w$ and $v-w$ is obviously a linear combination of $v$ and $w$.

On the other hand,
$$
v=frac12(v+w)+frac12(v-w)
qquad
w=frac12(v+w)-frac12(v-w)
$$
For the same reason as before, $langle v,wranglesubseteqlangle v+w,v-wrangle$.

share|cite|improve this answer

answered Jul 30 at 11:37

egreg

164k1180187

add a comment | 

up vote
3
down vote

Note that $langle v+w,v-wranglesubseteqlangle v,wrangle$, bwcause any linear combination of $v+w$ and $v-w$ is obviously a linear combination of $v$ and $w$.

On the other hand,
$$
v=frac12(v+w)+frac12(v-w)
qquad
w=frac12(v+w)-frac12(v-w)
$$
For the same reason as before, $langle v,wranglesubseteqlangle v+w,v-wrangle$.

share|cite|improve this answer

answered Jul 30 at 11:37

egreg

164k1180187

add a comment | 

up vote
3
down vote

up vote
3
down vote

Note that $langle v+w,v-wranglesubseteqlangle v,wrangle$, bwcause any linear combination of $v+w$ and $v-w$ is obviously a linear combination of $v$ and $w$.

On the other hand,
$$
v=frac12(v+w)+frac12(v-w)
qquad
w=frac12(v+w)-frac12(v-w)
$$
For the same reason as before, $langle v,wranglesubseteqlangle v+w,v-wrangle$.

share|cite|improve this answer

answered Jul 30 at 11:37

egreg

164k1180187

Note that $langle v+w,v-wranglesubseteqlangle v,wrangle$, bwcause any linear combination of $v+w$ and $v-w$ is obviously a linear combination of $v$ and $w$.

On the other hand,
$$
v=frac12(v+w)+frac12(v-w)
qquad
w=frac12(v+w)-frac12(v-w)
$$
For the same reason as before, $langle v,wranglesubseteqlangle v+w,v-wrangle$.

share|cite|improve this answer

answered Jul 30 at 11:37

egreg

164k1180187

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 11:37

egreg

164k1180187

answered Jul 30 at 11:37

egreg

164k1180187

answered Jul 30 at 11:37

egreg

164k1180187

164k1180187

add a comment | 

add a comment |