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 2019 Stack Overflow Developer Survey Results Are InProve 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 $mathbbR^3$?
Worn-tile Scrabble
Why can't devices on different VLANs, but on the same subnet, communicate?
The phrase "to the numbers born"?
How come people say “Would of”?
Why was M87 targeted for the Event Horizon Telescope instead of Sagittarius A*?
How can I connect public and private node through a reverse SSH tunnel?
"as much details as you can remember"
Can we generate random numbers using irrational numbers like π and e?
Cooking pasta in a water boiler
How to obtain a position of last non-zero element
Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?
Why are there uneven bright areas in this photo of black hole?
What are the motivations for publishing new editions of an existing textbook, beyond new discoveries in a field?
Output the Arecibo Message
Can withdrawing asylum be illegal?
Right tool to dig six foot holes?
Button changing its text & action. Good or terrible?
Why didn't the Event Horizon Telescope team mention Sagittarius A*?
Does a dangling wire really electrocute me if I'm standing in water?
What could be the right powersource for 15 seconds lifespan disposable giant chainsaw?
How much of the clove should I use when using big garlic heads?
What to expect from an e-bike service?
Why not take a picture of a closer black hole?
What to do when moving next to a bird sanctuary with a loosely-domesticated cat?
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 2019 Stack Overflow Developer Survey Results Are InProve 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 Mar 30 at 11:05
YuiTo Cheng
2,3694937
asked Mar 30 at 7:37
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 Mar 30 at 11:05
YuiTo Cheng
2,3694937
asked Mar 30 at 7:37
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 Mar 30 at 11:05
YuiTo Cheng
2,3694937
asked Mar 30 at 7:37
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 Mar 30 at 11:05
YuiTo Cheng
2,3694937
asked Mar 30 at 7:37
AdamAdam
345
edited Mar 30 at 11:05
YuiTo Cheng
2,3694937
asked Mar 30 at 7:37
AdamAdam
345
edited Mar 30 at 11:05
YuiTo Cheng
2,3694937
edited Mar 30 at 11:05
YuiTo Cheng
2,3694937
edited Mar 30 at 11:05
YuiTo Cheng
2,3694937
2,3694937
asked Mar 30 at 7:37
AdamAdam
345
asked Mar 30 at 7:37
AdamAdam
345
asked Mar 30 at 7:37
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 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 $mathbbR^m$. 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 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.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
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 $underline0$.
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 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 $underline0$.
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 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 $underline0$.
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 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 $underline0$.
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 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 $mathbbR^m$. 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 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 $mathbbR^m$. 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 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 $mathbbR^m$. 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 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 $mathbbR^m$. 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 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
