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 $mathbbR^3$?
Calculate the Mean mean of two numbers
Is it possible to make a 9x9 table fit within the default margins?
Find a path from s to t using as few red nodes as possible
What does this strange code stamp on my passport mean?
Man transported from Alternate World into ours by a Neutrino Detector
Why does the freezing point matter when picking cooler ice packs?
Traveling with my 5 year old daughter (as the father) without the mother from Germany to Mexico
Why can't we say "I have been having a dog"?
Why does sin(x) - sin(y) equal this?
What is a typical Mizrachi Seder like?
How dangerous is XSS
Gödel's incompleteness theorems - what are the religious implications?
Which acid/base does a strong base/acid react when added to a buffer solution?
Is it reasonable to ask other researchers to send me their previous grant applications?
Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?
Is it correct to say moon starry nights?
Can Sri Krishna be called 'a person'?
Can a PhD from a non-TU9 German university become a professor in a TU9 university?
Early programmable calculators with RS-232
Is the 21st century's idea of "freedom of speech" based on precedent?
Are British MPs missing the point, with these 'Indicative Votes'?
Why was Sir Cadogan fired?
Calculating discount not working
Why did early computer designers eschew integers?
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 |
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 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 |
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
