Find the rank of the $mathbbC[x]$-module $mathbbC^3$ given by a matrix The Next CEO of Stack OverflowQuotient of a free $mathbbZ$-moduleElementary Row MatricesDetermining a linear relation via the wedge product.Basis of free abelian group tensor $mathbb R$How do I find a dual basis given the following basis?Define $phi:mathbbR^3 rightarrow mathbbR$ by $phi(e_1) = 1$, $phi(e_2) = 2$, $phi(e_3)=-1$. Determine ker$phi$ and im$phi$Finding the rank of the matrix directly from eigenvaluesHow is the given set a basis for $K$?R-module: Prove $Wncong R^3$Basis in respect of an inner product
Easy to read palindrome checker
Make solar eclipses exceedingly rare, but still have new moons
Why doesn't UK go for the same deal Japan has with EU to resolve Brexit?
Math-accent symbol over parentheses enclosing accented symbol (amsmath)
How do I align (1) and (2)?
RigExpert AA-35 - Interpreting The Information
If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?
INSERT to a table from a database to other (same SQL Server) using Dynamic SQL
Does Germany produce more waste than the US?
Is the D&D universe the same as the Forgotten Realms universe?
How to get from Geneva Airport to Metabief, Doubs, France by public transport?
How to install OpenCV on Raspbian Stretch?
What happened in Rome, when the western empire "fell"?
How to edit “Name” property in GCI output?
Domestic-to-international connection at Orlando (MCO)
A Man With a Stainless Steel Endoskeleton (like The Terminator) Fighting Cloaked Aliens Only He Can See
How to count occurrences of text in a file?
Can MTA send mail via a relay without being told so?
Is it professional to write unrelated content in an almost-empty email?
Would be okay to drive on this tire?
Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?
What did we know about the Kessel run before the prequels?
Prepend last line of stdin to entire stdin
Is it okay to majorly distort historical facts while writing a fiction story?
Find the rank of the $mathbbC[x]$-module $mathbbC^3$ given by a matrix
The Next CEO of Stack OverflowQuotient of a free $mathbbZ$-moduleElementary Row MatricesDetermining a linear relation via the wedge product.Basis of free abelian group tensor $mathbb R$How do I find a dual basis given the following basis?Define $phi:mathbbR^3 rightarrow mathbbR$ by $phi(e_1) = 1$, $phi(e_2) = 2$, $phi(e_3)=-1$. Determine ker$phi$ and im$phi$Finding the rank of the matrix directly from eigenvaluesHow is the given set a basis for $K$?R-module: Prove $Wncong R^3$Basis in respect of an inner product
$begingroup$
As the title says, I'm trying to solve a problem which asks me to find the rank of the $mathbbC[x]$-module $N=mathbbC^3$ given by
$$
A=
beginbmatrix
0 & 0 & 1 \
1 & 0 & 0 \
0 & 1 & 0
endbmatrix.
$$
I'm struggling to see how to do this, because I can't seem to think of a basis. The vector $e_1$ clearly spans $N$ since $A(e_1)=e_2,A(e_2)=e_3$ and hence $(ax^2 +bx + c)e_1 = (a, b, c)^t$, but is obviously not linearly independent since $(x^3 - 1)e_1 = (0, 0, 0)^t $ where clearly $x^3-1 neq 0$ in $N$.
Can anyone suggest an alternative basis? Or is there another way to find the rank of $N$ as a $mathbbC[x]$-module?
abstract-algebra matrices ring-theory modules
edited Mar 27 at 22:36
user26857
39.5k124283
asked Mar 27 at 17:05
nongnerunongneru
62
New contributor
nongneru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
As the title says, I'm trying to solve a problem which asks me to find the rank of the $mathbbC[x]$-module $N=mathbbC^3$ given by
$$
A=
beginbmatrix
0 & 0 & 1 \
1 & 0 & 0 \
0 & 1 & 0
endbmatrix.
$$
I'm struggling to see how to do this, because I can't seem to think of a basis. The vector $e_1$ clearly spans $N$ since $A(e_1)=e_2,A(e_2)=e_3$ and hence $(ax^2 +bx + c)e_1 = (a, b, c)^t$, but is obviously not linearly independent since $(x^3 - 1)e_1 = (0, 0, 0)^t $ where clearly $x^3-1 neq 0$ in $N$.
Can anyone suggest an alternative basis? Or is there another way to find the rank of $N$ as a $mathbbC[x]$-module?
abstract-algebra matrices ring-theory modules
edited Mar 27 at 22:36
user26857
39.5k124283
asked Mar 27 at 17:05
nongnerunongneru
62
New contributor
nongneru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
As the title says, I'm trying to solve a problem which asks me to find the rank of the $mathbbC[x]$-module $N=mathbbC^3$ given by
$$
A=
beginbmatrix
0 & 0 & 1 \
1 & 0 & 0 \
0 & 1 & 0
endbmatrix.
$$
I'm struggling to see how to do this, because I can't seem to think of a basis. The vector $e_1$ clearly spans $N$ since $A(e_1)=e_2,A(e_2)=e_3$ and hence $(ax^2 +bx + c)e_1 = (a, b, c)^t$, but is obviously not linearly independent since $(x^3 - 1)e_1 = (0, 0, 0)^t $ where clearly $x^3-1 neq 0$ in $N$.
Can anyone suggest an alternative basis? Or is there another way to find the rank of $N$ as a $mathbbC[x]$-module?
abstract-algebra matrices ring-theory modules
edited Mar 27 at 22:36
user26857
39.5k124283
asked Mar 27 at 17:05
nongnerunongneru
62
New contributor
nongneru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
As the title says, I'm trying to solve a problem which asks me to find the rank of the $mathbbC[x]$-module $N=mathbbC^3$ given by
$$
A=
beginbmatrix
0 & 0 & 1 \
1 & 0 & 0 \
0 & 1 & 0
endbmatrix.
$$
I'm struggling to see how to do this, because I can't seem to think of a basis. The vector $e_1$ clearly spans $N$ since $A(e_1)=e_2,A(e_2)=e_3$ and hence $(ax^2 +bx + c)e_1 = (a, b, c)^t$, but is obviously not linearly independent since $(x^3 - 1)e_1 = (0, 0, 0)^t $ where clearly $x^3-1 neq 0$ in $N$.
Can anyone suggest an alternative basis? Or is there another way to find the rank of $N$ as a $mathbbC[x]$-module?
abstract-algebra matrices ring-theory modules
abstract-algebra matrices ring-theory modules
edited Mar 27 at 22:36
user26857
39.5k124283
asked Mar 27 at 17:05
nongnerunongneru
62
New contributor
nongneru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 27 at 22:36
user26857
39.5k124283
asked Mar 27 at 17:05
nongnerunongneru
62
New contributor
nongneru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 27 at 22:36
user26857
39.5k124283
edited Mar 27 at 22:36
user26857
39.5k124283
edited Mar 27 at 22:36
user26857
39.5k124283
39.5k124283
asked Mar 27 at 17:05
nongnerunongneru
62
New contributor
nongneru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 27 at 17:05
nongnerunongneru
62
asked Mar 27 at 17:05
nongnerunongneru
62
62
New contributor
nongneru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
nongneru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
nongneru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The problem you have has nothing to do with $e_1$. For any $vinmathbb C^3$, you will have $(x^3-1)cdot v=0$. So $N$ cannot have a basis in the setup you are looking at.
What one usually does is to consider, instead of $mathbb C[x]$, the quotient $mathbb C[x]/(x^3-1)$. With this new ring of coefficients, the set $e_1$ will be a basis. In fact, any one-element set will be a basis, and so your module is free of rank one.
answered Mar 27 at 19:41
Martin ArgeramiMartin Argerami
129k1184184
$endgroup$
$begingroup$
Or, if by rank the question means the dimension of $N otimes_mathbbC[x] mathbbC(x)$ as a $mathbbC(x)$-vector space - then the rank would be 0.
$endgroup$
– Daniel Schepler
Mar 27 at 19:56
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
);
);
nongneru is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3164784%2ffind-the-rank-of-the-mathbbcx-module-mathbbc3-given-by-a-matrix%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem you have has nothing to do with $e_1$. For any $vinmathbb C^3$, you will have $(x^3-1)cdot v=0$. So $N$ cannot have a basis in the setup you are looking at.
What one usually does is to consider, instead of $mathbb C[x]$, the quotient $mathbb C[x]/(x^3-1)$. With this new ring of coefficients, the set $e_1$ will be a basis. In fact, any one-element set will be a basis, and so your module is free of rank one.
answered Mar 27 at 19:41
Martin ArgeramiMartin Argerami
129k1184184
$endgroup$
$begingroup$
Or, if by rank the question means the dimension of $N otimes_mathbbC[x] mathbbC(x)$ as a $mathbbC(x)$-vector space - then the rank would be 0.
$endgroup$
– Daniel Schepler
Mar 27 at 19:56
add a comment |
$begingroup$
The problem you have has nothing to do with $e_1$. For any $vinmathbb C^3$, you will have $(x^3-1)cdot v=0$. So $N$ cannot have a basis in the setup you are looking at.
What one usually does is to consider, instead of $mathbb C[x]$, the quotient $mathbb C[x]/(x^3-1)$. With this new ring of coefficients, the set $e_1$ will be a basis. In fact, any one-element set will be a basis, and so your module is free of rank one.
answered Mar 27 at 19:41
Martin ArgeramiMartin Argerami
129k1184184
$endgroup$
$begingroup$
Or, if by rank the question means the dimension of $N otimes_mathbbC[x] mathbbC(x)$ as a $mathbbC(x)$-vector space - then the rank would be 0.
$endgroup$
– Daniel Schepler
Mar 27 at 19:56
add a comment |
$begingroup$
The problem you have has nothing to do with $e_1$. For any $vinmathbb C^3$, you will have $(x^3-1)cdot v=0$. So $N$ cannot have a basis in the setup you are looking at.
What one usually does is to consider, instead of $mathbb C[x]$, the quotient $mathbb C[x]/(x^3-1)$. With this new ring of coefficients, the set $e_1$ will be a basis. In fact, any one-element set will be a basis, and so your module is free of rank one.
answered Mar 27 at 19:41
Martin ArgeramiMartin Argerami
129k1184184
$endgroup$
The problem you have has nothing to do with $e_1$. For any $vinmathbb C^3$, you will have $(x^3-1)cdot v=0$. So $N$ cannot have a basis in the setup you are looking at.
What one usually does is to consider, instead of $mathbb C[x]$, the quotient $mathbb C[x]/(x^3-1)$. With this new ring of coefficients, the set $e_1$ will be a basis. In fact, any one-element set will be a basis, and so your module is free of rank one.
answered Mar 27 at 19:41
Martin ArgeramiMartin Argerami
129k1184184
answered Mar 27 at 19:41
Martin ArgeramiMartin Argerami
129k1184184
answered Mar 27 at 19:41
Martin ArgeramiMartin Argerami
129k1184184
answered Mar 27 at 19:41
Martin ArgeramiMartin Argerami
129k1184184
129k1184184
$begingroup$
Or, if by rank the question means the dimension of $N otimes_mathbbC[x] mathbbC(x)$ as a $mathbbC(x)$-vector space - then the rank would be 0.
$endgroup$
– Daniel Schepler
Mar 27 at 19:56
add a comment |
$begingroup$
Or, if by rank the question means the dimension of $N otimes_mathbbC[x] mathbbC(x)$ as a $mathbbC(x)$-vector space - then the rank would be 0.
$endgroup$
– Daniel Schepler
Mar 27 at 19:56
$begingroup$
Or, if by rank the question means the dimension of $N otimes_mathbbC[x] mathbbC(x)$ as a $mathbbC(x)$-vector space - then the rank would be 0.
$endgroup$
– Daniel Schepler
Mar 27 at 19:56
$begingroup$
Or, if by rank the question means the dimension of $N otimes_mathbbC[x] mathbbC(x)$ as a $mathbbC(x)$-vector space - then the rank would be 0.
$endgroup$
– Daniel Schepler
Mar 27 at 19:56
add a comment |
nongneru is a new contributor. Be nice, and check out our Code of Conduct.
nongneru is a new contributor. Be nice, and check out our Code of Conduct.
nongneru is a new contributor. Be nice, and check out our Code of Conduct.
nongneru is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3164784%2ffind-the-rank-of-the-mathbbcx-module-mathbbc3-given-by-a-matrix%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
