Dimension of Vector Space From Constraints

1

$begingroup$

Use the rank-nullity theorem to show that if $V =begin{bmatrix}a\b\c\dend{bmatrix}: x+y+z = 0, x+2y+3z =0$ that $dim($V$) = 2.$

I verified that the dimension is 2 by finding a basis from the RREF of the system of constraints, but I'm not sure what role the rank-nullity theorem plays here.

linear-algebra

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edited Mar 4 '16 at 2:15

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1

$begingroup$

Use the rank-nullity theorem to show that if $V =begin{bmatrix}a\b\c\dend{bmatrix}: x+y+z = 0, x+2y+3z =0$ that $dim($V$) = 2.$

I verified that the dimension is 2 by finding a basis from the RREF of the system of constraints, but I'm not sure what role the rank-nullity theorem plays here.

linear-algebra

share|cite|improve this question

edited Mar 4 '16 at 2:15

GoodDeeds

10.3k31435

asked Mar 4 '16 at 2:11

100001100001

17511

$endgroup$

add a comment | 

1

1

1

$begingroup$

Use the rank-nullity theorem to show that if $V =begin{bmatrix}a\b\c\dend{bmatrix}: x+y+z = 0, x+2y+3z =0$ that $dim($V$) = 2.$

I verified that the dimension is 2 by finding a basis from the RREF of the system of constraints, but I'm not sure what role the rank-nullity theorem plays here.

linear-algebra

share|cite|improve this question

edited Mar 4 '16 at 2:15

GoodDeeds

10.3k31435

asked Mar 4 '16 at 2:11

100001100001

17511

$endgroup$

Use the rank-nullity theorem to show that if $V =begin{bmatrix}a\b\c\dend{bmatrix}: x+y+z = 0, x+2y+3z =0$ that $dim($V$) = 2.$

I verified that the dimension is 2 by finding a basis from the RREF of the system of constraints, but I'm not sure what role the rank-nullity theorem plays here.

linear-algebra

linear-algebra

share|cite|improve this question

edited Mar 4 '16 at 2:15

GoodDeeds

10.3k31435

asked Mar 4 '16 at 2:11

100001100001

17511

share|cite|improve this question

edited Mar 4 '16 at 2:15

GoodDeeds

10.3k31435

asked Mar 4 '16 at 2:11

100001100001

17511

share|cite|improve this question

share|cite|improve this question

edited Mar 4 '16 at 2:15

GoodDeeds

10.3k31435

edited Mar 4 '16 at 2:15

GoodDeeds

10.3k31435

edited Mar 4 '16 at 2:15

GoodDeeds

10.3k31435

10.3k31435

asked Mar 4 '16 at 2:11

100001100001

17511

asked Mar 4 '16 at 2:11

100001100001

17511

asked Mar 4 '16 at 2:11

100001100001

17511

17511

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

1

$begingroup$

Hint Your space is the Nullspace of
$$A=begin{bmatrix} 1& 1 &1 & 0 \
1 & 2 &3 &0 end{bmatrix}$$
[the zeros correspond to the fourth variable which is missing from the equation].

Now the two rows of you matrix are not proportional, hence linearly independent. Thus $operatorname{rank}(A)=2$.

What does the rank-nullity theorem say?

share|cite|improve this answer

edited Jan 30 at 17:49

Martin Sleziak

44.9k10122277

answered Mar 4 '16 at 2:16

N. S.N. S.

105k7114210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So null($A$) = 0 and rank($A$) = 2? I don't see why null($A$) has to be zero though.
  $endgroup$
  – 100001
  Mar 4 '16 at 2:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MathLogic The rank-nullity theorem says that the rank plus the nullity is the number of COLUMNS of A ;)
  $endgroup$
  – N. S.
  Mar 4 '16 at 2:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So null($V$) = 2. But I still don't see how that shows that dim($V$) = 2?
  $endgroup$
  – 100001
  Mar 4 '16 at 2:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $W$ is the Null space of the matrix ;)
  $endgroup$
  – N. S.
  Mar 4 '16 at 3:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, I don't understand. Could you expound upon your explanations a bit more? I can't figure out what you are trying to get at.
  $endgroup$
  – 100001
  Mar 4 '16 at 3:41
  

  
  
  
  

 | 
show 1 more comment

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Hint Your space is the Nullspace of
$$A=begin{bmatrix} 1& 1 &1 & 0 \
1 & 2 &3 &0 end{bmatrix}$$
[the zeros correspond to the fourth variable which is missing from the equation].

Now the two rows of you matrix are not proportional, hence linearly independent. Thus $operatorname{rank}(A)=2$.

What does the rank-nullity theorem say?

share|cite|improve this answer

edited Jan 30 at 17:49

Martin Sleziak

44.9k10122277

answered Mar 4 '16 at 2:16

N. S.N. S.

105k7114210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So null($A$) = 0 and rank($A$) = 2? I don't see why null($A$) has to be zero though.
  $endgroup$
  – 100001
  Mar 4 '16 at 2:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MathLogic The rank-nullity theorem says that the rank plus the nullity is the number of COLUMNS of A ;)
  $endgroup$
  – N. S.
  Mar 4 '16 at 2:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So null($V$) = 2. But I still don't see how that shows that dim($V$) = 2?
  $endgroup$
  – 100001
  Mar 4 '16 at 2:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $W$ is the Null space of the matrix ;)
  $endgroup$
  – N. S.
  Mar 4 '16 at 3:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, I don't understand. Could you expound upon your explanations a bit more? I can't figure out what you are trying to get at.
  $endgroup$
  – 100001
  Mar 4 '16 at 3:41
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

Hint Your space is the Nullspace of
$$A=begin{bmatrix} 1& 1 &1 & 0 \
1 & 2 &3 &0 end{bmatrix}$$
[the zeros correspond to the fourth variable which is missing from the equation].

Now the two rows of you matrix are not proportional, hence linearly independent. Thus $operatorname{rank}(A)=2$.

What does the rank-nullity theorem say?

share|cite|improve this answer

edited Jan 30 at 17:49

Martin Sleziak

44.9k10122277

answered Mar 4 '16 at 2:16

N. S.N. S.

105k7114210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So null($A$) = 0 and rank($A$) = 2? I don't see why null($A$) has to be zero though.
  $endgroup$
  – 100001
  Mar 4 '16 at 2:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MathLogic The rank-nullity theorem says that the rank plus the nullity is the number of COLUMNS of A ;)
  $endgroup$
  – N. S.
  Mar 4 '16 at 2:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So null($V$) = 2. But I still don't see how that shows that dim($V$) = 2?
  $endgroup$
  – 100001
  Mar 4 '16 at 2:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $W$ is the Null space of the matrix ;)
  $endgroup$
  – N. S.
  Mar 4 '16 at 3:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, I don't understand. Could you expound upon your explanations a bit more? I can't figure out what you are trying to get at.
  $endgroup$
  – 100001
  Mar 4 '16 at 3:41
  

  
  
  
  

 | 
show 1 more comment

1

1

1

$begingroup$

Hint Your space is the Nullspace of
$$A=begin{bmatrix} 1& 1 &1 & 0 \
1 & 2 &3 &0 end{bmatrix}$$
[the zeros correspond to the fourth variable which is missing from the equation].

Now the two rows of you matrix are not proportional, hence linearly independent. Thus $operatorname{rank}(A)=2$.

What does the rank-nullity theorem say?

share|cite|improve this answer

edited Jan 30 at 17:49

Martin Sleziak

44.9k10122277

answered Mar 4 '16 at 2:16

N. S.N. S.

105k7114210

$endgroup$

Hint Your space is the Nullspace of
$$A=begin{bmatrix} 1& 1 &1 & 0 \
1 & 2 &3 &0 end{bmatrix}$$
[the zeros correspond to the fourth variable which is missing from the equation].

Now the two rows of you matrix are not proportional, hence linearly independent. Thus $operatorname{rank}(A)=2$.

What does the rank-nullity theorem say?

share|cite|improve this answer

edited Jan 30 at 17:49

Martin Sleziak

44.9k10122277

answered Mar 4 '16 at 2:16

N. S.N. S.

105k7114210

share|cite|improve this answer

share|cite|improve this answer

edited Jan 30 at 17:49

Martin Sleziak

44.9k10122277

edited Jan 30 at 17:49

Martin Sleziak

44.9k10122277

edited Jan 30 at 17:49

Martin Sleziak

44.9k10122277

44.9k10122277

answered Mar 4 '16 at 2:16

N. S.N. S.

105k7114210

answered Mar 4 '16 at 2:16

N. S.N. S.

105k7114210

answered Mar 4 '16 at 2:16

N. S.N. S.

105k7114210

105k7114210

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So null($A$) = 0 and rank($A$) = 2? I don't see why null($A$) has to be zero though.
  $endgroup$
  – 100001
  Mar 4 '16 at 2:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MathLogic The rank-nullity theorem says that the rank plus the nullity is the number of COLUMNS of A ;)
  $endgroup$
  – N. S.
  Mar 4 '16 at 2:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So null($V$) = 2. But I still don't see how that shows that dim($V$) = 2?
  $endgroup$
  – 100001
  Mar 4 '16 at 2:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $W$ is the Null space of the matrix ;)
  $endgroup$
  – N. S.
  Mar 4 '16 at 3:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, I don't understand. Could you expound upon your explanations a bit more? I can't figure out what you are trying to get at.
  $endgroup$
  – 100001
  Mar 4 '16 at 3:41
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So null($A$) = 0 and rank($A$) = 2? I don't see why null($A$) has to be zero though.
  $endgroup$
  – 100001
  Mar 4 '16 at 2:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MathLogic The rank-nullity theorem says that the rank plus the nullity is the number of COLUMNS of A ;)
  $endgroup$
  – N. S.
  Mar 4 '16 at 2:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So null($V$) = 2. But I still don't see how that shows that dim($V$) = 2?
  $endgroup$
  – 100001
  Mar 4 '16 at 2:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $W$ is the Null space of the matrix ;)
  $endgroup$
  – N. S.
  Mar 4 '16 at 3:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, I don't understand. Could you expound upon your explanations a bit more? I can't figure out what you are trying to get at.
  $endgroup$
  – 100001
  Mar 4 '16 at 3:41
  

  
  
  
  

$begingroup$
So null($A$) = 0 and rank($A$) = 2? I don't see why null($A$) has to be zero though.
$endgroup$
– 100001
Mar 4 '16 at 2:21

$begingroup$
So null($A$) = 0 and rank($A$) = 2? I don't see why null($A$) has to be zero though.
$endgroup$
– 100001
Mar 4 '16 at 2:21

1

1

$begingroup$
@MathLogic The rank-nullity theorem says that the rank plus the nullity is the number of COLUMNS of A ;)
$endgroup$
– N. S.
Mar 4 '16 at 2:48

$begingroup$
@MathLogic The rank-nullity theorem says that the rank plus the nullity is the number of COLUMNS of A ;)
$endgroup$
– N. S.
Mar 4 '16 at 2:48

$begingroup$
So null($V$) = 2. But I still don't see how that shows that dim($V$) = 2?
$endgroup$
– 100001
Mar 4 '16 at 2:55

$begingroup$
So null($V$) = 2. But I still don't see how that shows that dim($V$) = 2?
$endgroup$
– 100001
Mar 4 '16 at 2:55

$begingroup$
$W$ is the Null space of the matrix ;)
$endgroup$
– N. S.
Mar 4 '16 at 3:34

$begingroup$
$W$ is the Null space of the matrix ;)
$endgroup$
– N. S.
Mar 4 '16 at 3:34

$begingroup$
I'm sorry, I don't understand. Could you expound upon your explanations a bit more? I can't figure out what you are trying to get at.
$endgroup$
– 100001
Mar 4 '16 at 3:41

$begingroup$
I'm sorry, I don't understand. Could you expound upon your explanations a bit more? I can't figure out what you are trying to get at.
$endgroup$
– 100001
Mar 4 '16 at 3:41

 | 
show 1 more 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.

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Use MathJax to format equations. MathJax reference.

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