If A is an m by n matrix, prove that the set of vectors b that are not in C(A) forms a subspace. The Next CEO of Stack OverflowProve that S forms a subspace of R^3Prove that is a subspaceWhy do these vectors not belong to the same vector space?Am I correctly determining whether the vectors are in the subspace?Prove the following set of vectors is a subspaceCan a subspace S containing vectors with a finite number of nonzero components contain the zero vector?Vector subspace of two linear transformationsProve that if the following two vectors are not parallel that the following holds.Are the polynomials of form $a_0 + a_1x + a_2x^2 +a_3x^3$, with $a_i$ rational, a subspace of $P_3$?Why are all vectors with exactly one nonzero component not a subspace of $mathbbR^3$?
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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.
The Next CEO of Stack OverflowProve that S forms a subspace of R^3Prove that is a subspaceWhy do these vectors not belong to the same vector space?Am I correctly determining whether the vectors are in the subspace?Prove the following set of vectors is a subspaceCan a subspace S containing vectors with a finite number of nonzero components contain the zero vector?Vector subspace of two linear transformationsProve that if the following two vectors are not parallel that the following holds.Are the polynomials of form $a_0 + a_1x + a_2x^2 +a_3x^3$, with $a_i$ rational, a subspace of $P_3$?Why are all vectors with exactly one nonzero component not a subspace of $mathbbR^3$?
$begingroup$
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
edited 2 days ago
YuiTo Cheng
2,1863937
asked 2 days ago
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
edited 2 days ago
YuiTo Cheng
2,1863937
asked 2 days ago
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
edited 2 days ago
YuiTo Cheng
2,1863937
asked 2 days ago
AdamAdam
345
$endgroup$
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
linear-algebra matrices vector-spaces
edited 2 days ago
YuiTo Cheng
2,1863937
asked 2 days ago
AdamAdam
345
edited 2 days ago
YuiTo Cheng
2,1863937
asked 2 days ago
AdamAdam
345
edited 2 days ago
YuiTo Cheng
2,1863937
edited 2 days ago
YuiTo Cheng
2,1863937
edited 2 days ago
YuiTo Cheng
2,1863937
2,1863937
asked 2 days ago
AdamAdam
345
asked 2 days ago
AdamAdam
345
asked 2 days ago
AdamAdam
345
345
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 2 days ago
Jade PangJade Pang
1114
$endgroup$
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.
answered 2 days ago
Siong Thye GohSiong Thye Goh
103k1468120
$endgroup$
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
2 days ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
2 days ago
$begingroup$
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
2 days ago
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
2 days ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 2 days ago
Jade PangJade Pang
1114
$endgroup$
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.
answered 2 days ago
Jade PangJade Pang
1114
$endgroup$
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.
answered 2 days ago
Jade PangJade Pang
1114
$endgroup$
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.
answered 2 days ago
Jade PangJade Pang
1114
answered 2 days ago
Jade PangJade Pang
1114
answered 2 days ago
Jade PangJade Pang
1114
answered 2 days ago
Jade PangJade Pang
1114
1114
add a comment |
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.
answered 2 days ago
Siong Thye GohSiong Thye Goh
103k1468120
$endgroup$
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
2 days ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
2 days ago
$begingroup$
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
2 days ago
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
2 days ago
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.
answered 2 days ago
Siong Thye GohSiong Thye Goh
103k1468120
$endgroup$
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
2 days ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
2 days ago
$begingroup$
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
2 days ago
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
2 days ago
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.
answered 2 days ago
Siong Thye GohSiong Thye Goh
103k1468120
$endgroup$
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.
answered 2 days ago
Siong Thye GohSiong Thye Goh
103k1468120
edited 2 days ago
answered 2 days ago
Siong Thye GohSiong Thye Goh
103k1468120
answered 2 days ago
Siong Thye GohSiong Thye Goh
103k1468120
answered 2 days ago
Siong Thye GohSiong Thye Goh
103k1468120
103k1468120
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
2 days ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
2 days ago
$begingroup$
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
2 days ago
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
2 days ago
add a comment |
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
2 days ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
2 days ago
$begingroup$
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
2 days ago
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
2 days ago
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
2 days ago
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
2 days ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
2 days ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
1
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
2 days ago
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
2 days ago
add a comment |
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