If A is an m by n matrix, prove that the set of vectors b that are not in C(A) forms a subspace.
$begingroup$
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 Mar 30 at 11:05
YuiTo Cheng
2,4064937
asked Mar 30 at 7:37
AdamAdam
365
$endgroup$
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 Mar 30 at 11:05
YuiTo Cheng
2,4064937
asked Mar 30 at 7:37
AdamAdam
365
$endgroup$
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 Mar 30 at 11:05
YuiTo Cheng
2,4064937
asked Mar 30 at 7:37
AdamAdam
365
$endgroup$
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 Mar 30 at 11:05
YuiTo Cheng
2,4064937
asked Mar 30 at 7:37
AdamAdam
365
edited Mar 30 at 11:05
YuiTo Cheng
2,4064937
asked Mar 30 at 7:37
AdamAdam
365
edited Mar 30 at 11:05
YuiTo Cheng
2,4064937
edited Mar 30 at 11:05
YuiTo Cheng
2,4064937
edited Mar 30 at 11:05
YuiTo Cheng
2,4064937
2,4064937
asked Mar 30 at 7:37
AdamAdam
365
asked Mar 30 at 7:37
AdamAdam
365
asked Mar 30 at 7:37
AdamAdam
365
365
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 $underline{0}$.
We know that $C(A)$ is a subspace for $mathbb{R}^m$, so $underline{0}in C(A) subseteq mathbb{R}^m$, that means the zero vector are inside the column space and also $mathbb{R}^m$.
But if we consider $mathbb{R}^msetminus C(A)$, 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 Mar 30 at 9:27
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 $mathbb{R}^m$. 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 Mar 30 at 7:44
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
Mar 30 at 7:48
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
Mar 30 at 7:50
$begingroup$
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
Mar 30 at 7:52
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
Mar 30 at 7:55
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168021%2fif-a-is-an-m-by-n-matrix-prove-that-the-set-of-vectors-b-that-are-not-in-ca-f%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 $underline{0}$.
We know that $C(A)$ is a subspace for $mathbb{R}^m$, so $underline{0}in C(A) subseteq mathbb{R}^m$, that means the zero vector are inside the column space and also $mathbb{R}^m$.
But if we consider $mathbb{R}^msetminus C(A)$, 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 Mar 30 at 9:27
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 $underline{0}$.
We know that $C(A)$ is a subspace for $mathbb{R}^m$, so $underline{0}in C(A) subseteq mathbb{R}^m$, that means the zero vector are inside the column space and also $mathbb{R}^m$.
But if we consider $mathbb{R}^msetminus C(A)$, 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 Mar 30 at 9:27
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 $underline{0}$.
We know that $C(A)$ is a subspace for $mathbb{R}^m$, so $underline{0}in C(A) subseteq mathbb{R}^m$, that means the zero vector are inside the column space and also $mathbb{R}^m$.
But if we consider $mathbb{R}^msetminus C(A)$, 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 Mar 30 at 9:27
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 $underline{0}$.
We know that $C(A)$ is a subspace for $mathbb{R}^m$, so $underline{0}in C(A) subseteq mathbb{R}^m$, that means the zero vector are inside the column space and also $mathbb{R}^m$.
But if we consider $mathbb{R}^msetminus C(A)$, 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 Mar 30 at 9:27
Jade PangJade Pang
1114
answered Mar 30 at 9:27
Jade PangJade Pang
1114
answered Mar 30 at 9:27
Jade PangJade Pang
1114
answered Mar 30 at 9:27
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 $mathbb{R}^m$. 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 Mar 30 at 7:44
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
Mar 30 at 7:48
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
Mar 30 at 7:50
$begingroup$
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
Mar 30 at 7:52
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
Mar 30 at 7:55
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 $mathbb{R}^m$. 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 Mar 30 at 7:44
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
Mar 30 at 7:48
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
Mar 30 at 7:50
$begingroup$
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
Mar 30 at 7:52
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
Mar 30 at 7:55
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 $mathbb{R}^m$. 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 Mar 30 at 7:44
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
Your first condition should be $0$ is in a subspace.
Also, the result is not true.
Let $0_m$ be the zero vector in $mathbb{R}^m$. 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 Mar 30 at 7:44
Siong Thye GohSiong Thye Goh
104k1468120
edited Mar 30 at 7:54
answered Mar 30 at 7:44
Siong Thye GohSiong Thye Goh
104k1468120
answered Mar 30 at 7:44
Siong Thye GohSiong Thye Goh
104k1468120
answered Mar 30 at 7:44
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
Mar 30 at 7:48
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
Mar 30 at 7:50
$begingroup$
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
Mar 30 at 7:52
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
Mar 30 at 7:55
add a comment |
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
Mar 30 at 7:48
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
Mar 30 at 7:50
$begingroup$
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
Mar 30 at 7:52
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
Mar 30 at 7:55
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
Mar 30 at 7:48
$begingroup$
How do I know that 0 is in C(A)?
$endgroup$
– Adam
Mar 30 at 7:48
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
Mar 30 at 7:50
$begingroup$
$C(A)$ is the column space right?
$endgroup$
– Siong Thye Goh
Mar 30 at 7:50
$begingroup$
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
Mar 30 at 7:52
$begingroup$
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
Mar 30 at 7:52
1
1
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
Mar 30 at 7:55
$begingroup$
I have included an explaination of why $0_m in C(A)$.
$endgroup$
– Siong Thye Goh
Mar 30 at 7:55
add a 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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168021%2fif-a-is-an-m-by-n-matrix-prove-that-the-set-of-vectors-b-that-are-not-in-ca-f%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
