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If A is an m by n matrix, prove that the set of vectors b that are not in C(A) forms a subspace.

2

$begingroup$

If A is an m by n matrix, prove that the set of vectors b that are not in C(A) forms a subspace.

I would like to first understand if I am interpreting the question correctly. My understanding of this question is that I need to first prove that the set of vectors b are equal to the 0 vector, and if b1 and b2 are members of the subspace, then the sum of set b should be a member, and so should some scalar C multiplied by the set of vectors b. I just don't understand how to actually prove this

linear-algebra matrices vector-spaces

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1862937

asked yesterday

AdamAdam

345

$endgroup$

add a comment | 

2

$begingroup$

If A is an m by n matrix, prove that the set of vectors b that are not in C(A) forms a subspace.

I would like to first understand if I am interpreting the question correctly. My understanding of this question is that I need to first prove that the set of vectors b are equal to the 0 vector, and if b1 and b2 are members of the subspace, then the sum of set b should be a member, and so should some scalar C multiplied by the set of vectors b. I just don't understand how to actually prove this

linear-algebra matrices vector-spaces

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1862937

asked yesterday

AdamAdam

345

$endgroup$

add a comment | 

$begingroup$

If A is an m by n matrix, prove that the set of vectors b that are not in C(A) forms a subspace.

I would like to first understand if I am interpreting the question correctly. My understanding of this question is that I need to first prove that the set of vectors b are equal to the 0 vector, and if b1 and b2 are members of the subspace, then the sum of set b should be a member, and so should some scalar C multiplied by the set of vectors b. I just don't understand how to actually prove this

linear-algebra matrices vector-spaces

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1862937

asked yesterday

AdamAdam

345

$endgroup$

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1862937

asked yesterday

AdamAdam

345

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1862937

asked yesterday

AdamAdam

345

edited yesterday

YuiTo Cheng

2,1862937

edited yesterday

YuiTo Cheng

2,1862937

asked yesterday

AdamAdam

345

asked yesterday

AdamAdam

345

2 Answers
2

active

oldest

votes

2

$begingroup$

Your result is wrong. Maybe this picture can help you figure out.

Since every subspace must contain the zero vector, noted by $underline0$.

We know that $C(A)$ is a subspace for $mathbbR^m$, so $underline0in C(A) subseteq mathbbR^m$, that means the zero vector are inside the column space and also $mathbbR^m$.

But if we consider $mathbbR^msetminus C(A)$, by the picture, it means we are now cancelling the red circle out, so the zero vector is no longer inside this "space".

Therefore, it can not form a subspace.

share|cite|improve this answer

answered yesterday

Jade PangJade Pang

1964

$endgroup$

add a comment | 

5

$begingroup$

Your first condition should be $0$ is in a subspace.

Also, the result is not true.

Let $0_m$ be the zero vector in $mathbbR^m$. We know that $0_m in C(A)$ since $sum_i=1^n A_i cdot 0=0_m$.

$0_m$ is in $C(A)$, $0_m$ can't be in the set of vector that are not in $C(A)$. Hence the set of vectors that are not in $C(A)$ can't form a subspace.

share|cite|improve this answer

edited yesterday

answered yesterday

Siong Thye GohSiong Thye Goh

103k1468119

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do I know that 0 is in C(A)?
  $endgroup$
  – Adam
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $C(A)$ is the column space right?
  $endgroup$
  – Siong Thye Goh
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Correct. I am new to linear algebra and am trying to understand the definitions Column space, null space, subspace, etc...Perhaps I am not understanding the proper definition of column space
  $endgroup$
  – Adam
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I have included an explaination of why $0_m in C(A)$.
  $endgroup$
  – Siong Thye Goh
  yesterday
  

  
  
  
  

add a comment | 

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2

$begingroup$

Your result is wrong. Maybe this picture can help you figure out.

Since every subspace must contain the zero vector, noted by $underline0$.

We know that $C(A)$ is a subspace for $mathbbR^m$, so $underline0in C(A) subseteq mathbbR^m$, that means the zero vector are inside the column space and also $mathbbR^m$.

But if we consider $mathbbR^msetminus C(A)$, by the picture, it means we are now cancelling the red circle out, so the zero vector is no longer inside this "space".

Therefore, it can not form a subspace.

share|cite|improve this answer

answered yesterday

Jade PangJade Pang

1964

$endgroup$

add a comment | 

2

$begingroup$

Your result is wrong. Maybe this picture can help you figure out.

Since every subspace must contain the zero vector, noted by $underline0$.

We know that $C(A)$ is a subspace for $mathbbR^m$, so $underline0in C(A) subseteq mathbbR^m$, that means the zero vector are inside the column space and also $mathbbR^m$.

But if we consider $mathbbR^msetminus C(A)$, by the picture, it means we are now cancelling the red circle out, so the zero vector is no longer inside this "space".

Therefore, it can not form a subspace.

share|cite|improve this answer

answered yesterday

Jade PangJade Pang

1964

$endgroup$

add a comment | 

$begingroup$

Your result is wrong. Maybe this picture can help you figure out.

Since every subspace must contain the zero vector, noted by $underline0$.

We know that $C(A)$ is a subspace for $mathbbR^m$, so $underline0in C(A) subseteq mathbbR^m$, that means the zero vector are inside the column space and also $mathbbR^m$.

But if we consider $mathbbR^msetminus C(A)$, by the picture, it means we are now cancelling the red circle out, so the zero vector is no longer inside this "space".

Therefore, it can not form a subspace.

share|cite|improve this answer

answered yesterday

Jade PangJade Pang

1964

$endgroup$

share|cite|improve this answer

answered yesterday

Jade PangJade Pang

1964

answered yesterday

Jade PangJade Pang

1964

answered yesterday

Jade PangJade Pang

1964

5

$begingroup$

Your first condition should be $0$ is in a subspace.

Also, the result is not true.

Let $0_m$ be the zero vector in $mathbbR^m$. We know that $0_m in C(A)$ since $sum_i=1^n A_i cdot 0=0_m$.

$0_m$ is in $C(A)$, $0_m$ can't be in the set of vector that are not in $C(A)$. Hence the set of vectors that are not in $C(A)$ can't form a subspace.

share|cite|improve this answer

edited yesterday

answered yesterday

Siong Thye GohSiong Thye Goh

103k1468119

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do I know that 0 is in C(A)?
  $endgroup$
  – Adam
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $C(A)$ is the column space right?
  $endgroup$
  – Siong Thye Goh
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Correct. I am new to linear algebra and am trying to understand the definitions Column space, null space, subspace, etc...Perhaps I am not understanding the proper definition of column space
  $endgroup$
  – Adam
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I have included an explaination of why $0_m in C(A)$.
  $endgroup$
  – Siong Thye Goh
  yesterday
  

  
  
  
  

add a comment | 

5

$begingroup$

Your first condition should be $0$ is in a subspace.

Also, the result is not true.

Let $0_m$ be the zero vector in $mathbbR^m$. We know that $0_m in C(A)$ since $sum_i=1^n A_i cdot 0=0_m$.

$0_m$ is in $C(A)$, $0_m$ can't be in the set of vector that are not in $C(A)$. Hence the set of vectors that are not in $C(A)$ can't form a subspace.

share|cite|improve this answer

edited yesterday

answered yesterday

Siong Thye GohSiong Thye Goh

103k1468119

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do I know that 0 is in C(A)?
  $endgroup$
  – Adam
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $C(A)$ is the column space right?
  $endgroup$
  – Siong Thye Goh
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Correct. I am new to linear algebra and am trying to understand the definitions Column space, null space, subspace, etc...Perhaps I am not understanding the proper definition of column space
  $endgroup$
  – Adam
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I have included an explaination of why $0_m in C(A)$.
  $endgroup$
  – Siong Thye Goh
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Your first condition should be $0$ is in a subspace.

Also, the result is not true.

Let $0_m$ be the zero vector in $mathbbR^m$. We know that $0_m in C(A)$ since $sum_i=1^n A_i cdot 0=0_m$.

$0_m$ is in $C(A)$, $0_m$ can't be in the set of vector that are not in $C(A)$. Hence the set of vectors that are not in $C(A)$ can't form a subspace.

share|cite|improve this answer

edited yesterday

answered yesterday

Siong Thye GohSiong Thye Goh

103k1468119

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Siong Thye GohSiong Thye Goh

103k1468119

answered yesterday

Siong Thye GohSiong Thye Goh

103k1468119

answered yesterday

Siong Thye GohSiong Thye Goh

103k1468119