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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 23 mins ago
YuiTo Cheng
2,1862937
asked 3 hours ago
AdamAdam
295
$endgroup$
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 23 mins ago
YuiTo Cheng
2,1862937
asked 3 hours ago
AdamAdam
295
$endgroup$
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 23 mins ago
YuiTo Cheng
2,1862937
asked 3 hours ago
AdamAdam
295
$endgroup$
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 23 mins ago
YuiTo Cheng
2,1862937
asked 3 hours ago
AdamAdam
295
edited 23 mins ago
YuiTo Cheng
2,1862937
asked 3 hours ago
AdamAdam
295
edited 23 mins ago
YuiTo Cheng
2,1862937
edited 23 mins ago
YuiTo Cheng
2,1862937
edited 23 mins ago
YuiTo Cheng
2,1862937
2,1862937
asked 3 hours ago
AdamAdam
295
asked 3 hours ago
AdamAdam
295
asked 3 hours ago
AdamAdam
295
295
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 hours ago
Jade PangJade Pang
1714
$endgroup$
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 3 hours ago
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
3 hours ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
3 hours 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
3 hours ago
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
3 hours ago
add a comment |
Your Answer
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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 hours ago
Jade PangJade Pang
1714
$endgroup$
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 hours ago
Jade PangJade Pang
1714
$endgroup$
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 hours ago
Jade PangJade Pang
1714
$endgroup$
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 hours ago
Jade PangJade Pang
1714
answered 2 hours ago
Jade PangJade Pang
1714
answered 2 hours ago
Jade PangJade Pang
1714
answered 2 hours ago
Jade PangJade Pang
1714
1714
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 3 hours ago
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
3 hours ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
3 hours 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
3 hours ago
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
3 hours 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 3 hours ago
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
3 hours ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
3 hours 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
3 hours ago
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
3 hours 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 3 hours ago
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
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 3 hours ago
Siong Thye GohSiong Thye Goh
103k1468119
edited 3 hours ago
answered 3 hours ago
Siong Thye GohSiong Thye Goh
103k1468119
answered 3 hours ago
Siong Thye GohSiong Thye Goh
103k1468119
answered 3 hours ago
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
3 hours ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
3 hours 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
3 hours ago
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
3 hours ago
add a comment |
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
3 hours ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
3 hours 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
3 hours ago
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
3 hours ago
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
3 hours ago
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
3 hours ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
3 hours ago
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
3 hours 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
3 hours 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
3 hours ago
1
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
3 hours ago
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
3 hours ago
add a comment |
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