Linear Independence for Vectors of Cosine ValuesRemembering exact sine cosine and tangent values?A basic question on linear independence of eigen vectorsLinear Independence of a set of Complex VectorsWorking with Eigen vectors and values, How do you verify linear independence?Linear independence of standard basis vectors from Vandermonde style vectors$det(A+I)=1+texttr(A)$, if $textrank(A)=1$ proof?Finding the “Larger” of Two Cosine ValuesLinear Independence of Cosine FunctionLinear independence of row vectorsWhat is the dimension of $Xin M_n,n(F); AX=XA=0$?
Is there a reason why Turkey took the Balkan territories of the Ottoman Empire, instead of Greece or another of the Balkan states?
Magical Modulo Squares
What computer port is this?
Company stopped paying my salary. What are my options?
What should I use to get rid of some kind of weed in my onions
Capturing the entire webpage with WebExecute's CaptureImage
Steganography in Latex
Are there vaccine ingredients which may not be disclosed ("hidden", "trade secret", or similar)?
How do I politely tell my players to shut up about their backstory?
Did any early RISC OS precursor run on the BBC Micro?
Linear Independence for Vectors of Cosine Values
My parents are Afghan
Why doesn't Dany protect her dragons better?
Identity of a supposed anonymous referee revealed through "Description" of the report
Mindfulness of Watching Youtube
Can I bring back Planetary Romance as a genre?
Opposite party turned away from voting when ballot is all opposing party
why it is 2>&1 and not 2>>&1 to append to a log file
Is it possible to do moon sighting in advance for 5 years with 100% accuracy?
Is the tensor product (of vector spaces) commutative?
Was Mohammed the most popular first name for boys born in Berlin in 2018?
Visual Studio Code download existing code
Why is the episode called "The Last of the Starks"?
What's an appropriate age to involve kids in life changing decisions?
Linear Independence for Vectors of Cosine Values
Remembering exact sine cosine and tangent values?A basic question on linear independence of eigen vectorsLinear Independence of a set of Complex VectorsWorking with Eigen vectors and values, How do you verify linear independence?Linear independence of standard basis vectors from Vandermonde style vectors$det(A+I)=1+texttr(A)$, if $textrank(A)=1$ proof?Finding the “Larger” of Two Cosine ValuesLinear Independence of Cosine FunctionLinear independence of row vectorsWhat is the dimension of $Xin M_n,n(F); AX=XA=0$?
$begingroup$
For every integer $ngeq3$ define an $ntimes n$ matrix $A$ using the following row vector as the first row: $(c_0,c_1,c_2,ldots, c_n-1)$ where $c_k=cos (2pi k/n)$.
The subsequent rows are obtained as cyclic shifts. This matrix seems to be always of rank 2.
Actually I was working with some families of representations of dihedral groups, arising in another context and was computing the degrees of the representations. That calculation implied this matrix should be of rank 2.
But definitely there must be some direct way of doing it.
Can anyone tell me how to accomplish it?
matrices trigonometry
asked 2 hours ago
P VanchinathanP Vanchinathan
15.7k12237
$endgroup$
add a comment |
$begingroup$
For every integer $ngeq3$ define an $ntimes n$ matrix $A$ using the following row vector as the first row: $(c_0,c_1,c_2,ldots, c_n-1)$ where $c_k=cos (2pi k/n)$.
The subsequent rows are obtained as cyclic shifts. This matrix seems to be always of rank 2.
Actually I was working with some families of representations of dihedral groups, arising in another context and was computing the degrees of the representations. That calculation implied this matrix should be of rank 2.
But definitely there must be some direct way of doing it.
Can anyone tell me how to accomplish it?
matrices trigonometry
asked 2 hours ago
P VanchinathanP Vanchinathan
15.7k12237
$endgroup$
$begingroup$
You may want to read this page: en.wikipedia.org/wiki/Circulant_matrix
$endgroup$
– angryavian
2 hours ago
$begingroup$
@angryavian: Thanks. A quick glance tells me that the solution is more algebraic (with gcd of polynomials). This is easier to deal for me than explicit computations!
$endgroup$
– P Vanchinathan
2 hours ago
add a comment |
$begingroup$
For every integer $ngeq3$ define an $ntimes n$ matrix $A$ using the following row vector as the first row: $(c_0,c_1,c_2,ldots, c_n-1)$ where $c_k=cos (2pi k/n)$.
The subsequent rows are obtained as cyclic shifts. This matrix seems to be always of rank 2.
Actually I was working with some families of representations of dihedral groups, arising in another context and was computing the degrees of the representations. That calculation implied this matrix should be of rank 2.
But definitely there must be some direct way of doing it.
Can anyone tell me how to accomplish it?
matrices trigonometry
asked 2 hours ago
P VanchinathanP Vanchinathan
15.7k12237
$endgroup$
For every integer $ngeq3$ define an $ntimes n$ matrix $A$ using the following row vector as the first row: $(c_0,c_1,c_2,ldots, c_n-1)$ where $c_k=cos (2pi k/n)$.
The subsequent rows are obtained as cyclic shifts. This matrix seems to be always of rank 2.
Actually I was working with some families of representations of dihedral groups, arising in another context and was computing the degrees of the representations. That calculation implied this matrix should be of rank 2.
But definitely there must be some direct way of doing it.
Can anyone tell me how to accomplish it?
matrices trigonometry
matrices trigonometry
asked 2 hours ago
P VanchinathanP Vanchinathan
15.7k12237
asked 2 hours ago
P VanchinathanP Vanchinathan
15.7k12237
asked 2 hours ago
P VanchinathanP Vanchinathan
15.7k12237
asked 2 hours ago
P VanchinathanP Vanchinathan
15.7k12237
asked 2 hours ago
P VanchinathanP Vanchinathan
15.7k12237
15.7k12237
$begingroup$
You may want to read this page: en.wikipedia.org/wiki/Circulant_matrix
$endgroup$
– angryavian
2 hours ago
$begingroup$
@angryavian: Thanks. A quick glance tells me that the solution is more algebraic (with gcd of polynomials). This is easier to deal for me than explicit computations!
$endgroup$
– P Vanchinathan
2 hours ago
add a comment |
$begingroup$
You may want to read this page: en.wikipedia.org/wiki/Circulant_matrix
$endgroup$
– angryavian
2 hours ago
$begingroup$
@angryavian: Thanks. A quick glance tells me that the solution is more algebraic (with gcd of polynomials). This is easier to deal for me than explicit computations!
$endgroup$
– P Vanchinathan
2 hours ago
$begingroup$
You may want to read this page: en.wikipedia.org/wiki/Circulant_matrix
$endgroup$
– angryavian
2 hours ago
$begingroup$
You may want to read this page: en.wikipedia.org/wiki/Circulant_matrix
$endgroup$
– angryavian
2 hours ago
$begingroup$
@angryavian: Thanks. A quick glance tells me that the solution is more algebraic (with gcd of polynomials). This is easier to deal for me than explicit computations!
$endgroup$
– P Vanchinathan
2 hours ago
$begingroup$
@angryavian: Thanks. A quick glance tells me that the solution is more algebraic (with gcd of polynomials). This is easier to deal for me than explicit computations!
$endgroup$
– P Vanchinathan
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The identity
$cos((n+1)x)
=2cos(x)cos(nx)-cos((n-1)x)
$
shows that each column
is a linear combination
of the previous two columns,
so the rank is 2.
This also holds for $sin$.
answered 2 hours ago
marty cohenmarty cohen
76.5k549130
$endgroup$
$begingroup$
[+1] Very nice answer !
$endgroup$
– Jean Marie
26 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
13 mins ago
$begingroup$
That identity has stuck in my mind since reading a numerical analysis book many years ago written by Henrici. It is quite stable and used to create a relatively simple method of computing the points on a circle.
$endgroup$
– marty cohen
10 mins ago
add a comment |
$begingroup$
A typical cyclic shift is
$$(c_k,c_k+1,ldots,c_k+n-1)$$
noting that $c_n=c_0$, $c_n+1=c_1$ etc.
This is the real part of the complex vector
$$(zeta^k,zeta^k+1,ldots,zeta^k+n-1)$$
where $zeta=exp(2pi i/n)$. Thus
beginalign
(c_k,c_k+1,ldots,c_k+n-1)
&=frac12(zeta^k,zeta^k+1,ldots,zeta^k+n-1)
+frac12(zeta^-k,zeta^-k-1,ldots,zeta^-k-n+1)\
&=fraczeta^k2(1,zeta,ldots,zeta^n-1)
+fraczeta^-k2(1,zeta^-1,ldots,zeta^-n+1).
endalign
Each row is a linear combination of the two vectors
$(1,zeta,ldots,zeta^n-1)$ and $(1,zeta^-1,ldots,zeta^-n+1)$
so the matrix has rank $le2$.
answered 2 hours ago
Lord Shark the UnknownLord Shark the Unknown
110k1163137
$endgroup$
$begingroup$
+1 I like this answer too. Unfortunately I can accept only one.
$endgroup$
– P Vanchinathan
14 mins ago
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%2f3217835%2flinear-independence-for-vectors-of-cosine-values%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
$begingroup$
The identity
$cos((n+1)x)
=2cos(x)cos(nx)-cos((n-1)x)
$
shows that each column
is a linear combination
of the previous two columns,
so the rank is 2.
This also holds for $sin$.
answered 2 hours ago
marty cohenmarty cohen
76.5k549130
$endgroup$
$begingroup$
[+1] Very nice answer !
$endgroup$
– Jean Marie
26 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
13 mins ago
$begingroup$
That identity has stuck in my mind since reading a numerical analysis book many years ago written by Henrici. It is quite stable and used to create a relatively simple method of computing the points on a circle.
$endgroup$
– marty cohen
10 mins ago
add a comment |
$begingroup$
The identity
$cos((n+1)x)
=2cos(x)cos(nx)-cos((n-1)x)
$
shows that each column
is a linear combination
of the previous two columns,
so the rank is 2.
This also holds for $sin$.
answered 2 hours ago
marty cohenmarty cohen
76.5k549130
$endgroup$
$begingroup$
[+1] Very nice answer !
$endgroup$
– Jean Marie
26 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
13 mins ago
$begingroup$
That identity has stuck in my mind since reading a numerical analysis book many years ago written by Henrici. It is quite stable and used to create a relatively simple method of computing the points on a circle.
$endgroup$
– marty cohen
10 mins ago
add a comment |
$begingroup$
The identity
$cos((n+1)x)
=2cos(x)cos(nx)-cos((n-1)x)
$
shows that each column
is a linear combination
of the previous two columns,
so the rank is 2.
This also holds for $sin$.
answered 2 hours ago
marty cohenmarty cohen
76.5k549130
$endgroup$
The identity
$cos((n+1)x)
=2cos(x)cos(nx)-cos((n-1)x)
$
shows that each column
is a linear combination
of the previous two columns,
so the rank is 2.
This also holds for $sin$.
answered 2 hours ago
marty cohenmarty cohen
76.5k549130
answered 2 hours ago
marty cohenmarty cohen
76.5k549130
answered 2 hours ago
marty cohenmarty cohen
76.5k549130
answered 2 hours ago
marty cohenmarty cohen
76.5k549130
76.5k549130
$begingroup$
[+1] Very nice answer !
$endgroup$
– Jean Marie
26 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
13 mins ago
$begingroup$
That identity has stuck in my mind since reading a numerical analysis book many years ago written by Henrici. It is quite stable and used to create a relatively simple method of computing the points on a circle.
$endgroup$
– marty cohen
10 mins ago
add a comment |
$begingroup$
[+1] Very nice answer !
$endgroup$
– Jean Marie
26 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
13 mins ago
$begingroup$
That identity has stuck in my mind since reading a numerical analysis book many years ago written by Henrici. It is quite stable and used to create a relatively simple method of computing the points on a circle.
$endgroup$
– marty cohen
10 mins ago
$begingroup$
[+1] Very nice answer !
$endgroup$
– Jean Marie
26 mins ago
$begingroup$
[+1] Very nice answer !
$endgroup$
– Jean Marie
26 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
13 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
13 mins ago
$begingroup$
That identity has stuck in my mind since reading a numerical analysis book many years ago written by Henrici. It is quite stable and used to create a relatively simple method of computing the points on a circle.
$endgroup$
– marty cohen
10 mins ago
$begingroup$
That identity has stuck in my mind since reading a numerical analysis book many years ago written by Henrici. It is quite stable and used to create a relatively simple method of computing the points on a circle.
$endgroup$
– marty cohen
10 mins ago
add a comment |
$begingroup$
A typical cyclic shift is
$$(c_k,c_k+1,ldots,c_k+n-1)$$
noting that $c_n=c_0$, $c_n+1=c_1$ etc.
This is the real part of the complex vector
$$(zeta^k,zeta^k+1,ldots,zeta^k+n-1)$$
where $zeta=exp(2pi i/n)$. Thus
beginalign
(c_k,c_k+1,ldots,c_k+n-1)
&=frac12(zeta^k,zeta^k+1,ldots,zeta^k+n-1)
+frac12(zeta^-k,zeta^-k-1,ldots,zeta^-k-n+1)\
&=fraczeta^k2(1,zeta,ldots,zeta^n-1)
+fraczeta^-k2(1,zeta^-1,ldots,zeta^-n+1).
endalign
Each row is a linear combination of the two vectors
$(1,zeta,ldots,zeta^n-1)$ and $(1,zeta^-1,ldots,zeta^-n+1)$
so the matrix has rank $le2$.
answered 2 hours ago
Lord Shark the UnknownLord Shark the Unknown
110k1163137
$endgroup$
$begingroup$
+1 I like this answer too. Unfortunately I can accept only one.
$endgroup$
– P Vanchinathan
14 mins ago
add a comment |
$begingroup$
A typical cyclic shift is
$$(c_k,c_k+1,ldots,c_k+n-1)$$
noting that $c_n=c_0$, $c_n+1=c_1$ etc.
This is the real part of the complex vector
$$(zeta^k,zeta^k+1,ldots,zeta^k+n-1)$$
where $zeta=exp(2pi i/n)$. Thus
beginalign
(c_k,c_k+1,ldots,c_k+n-1)
&=frac12(zeta^k,zeta^k+1,ldots,zeta^k+n-1)
+frac12(zeta^-k,zeta^-k-1,ldots,zeta^-k-n+1)\
&=fraczeta^k2(1,zeta,ldots,zeta^n-1)
+fraczeta^-k2(1,zeta^-1,ldots,zeta^-n+1).
endalign
Each row is a linear combination of the two vectors
$(1,zeta,ldots,zeta^n-1)$ and $(1,zeta^-1,ldots,zeta^-n+1)$
so the matrix has rank $le2$.
answered 2 hours ago
Lord Shark the UnknownLord Shark the Unknown
110k1163137
$endgroup$
$begingroup$
+1 I like this answer too. Unfortunately I can accept only one.
$endgroup$
– P Vanchinathan
14 mins ago
add a comment |
$begingroup$
A typical cyclic shift is
$$(c_k,c_k+1,ldots,c_k+n-1)$$
noting that $c_n=c_0$, $c_n+1=c_1$ etc.
This is the real part of the complex vector
$$(zeta^k,zeta^k+1,ldots,zeta^k+n-1)$$
where $zeta=exp(2pi i/n)$. Thus
beginalign
(c_k,c_k+1,ldots,c_k+n-1)
&=frac12(zeta^k,zeta^k+1,ldots,zeta^k+n-1)
+frac12(zeta^-k,zeta^-k-1,ldots,zeta^-k-n+1)\
&=fraczeta^k2(1,zeta,ldots,zeta^n-1)
+fraczeta^-k2(1,zeta^-1,ldots,zeta^-n+1).
endalign
Each row is a linear combination of the two vectors
$(1,zeta,ldots,zeta^n-1)$ and $(1,zeta^-1,ldots,zeta^-n+1)$
so the matrix has rank $le2$.
answered 2 hours ago
Lord Shark the UnknownLord Shark the Unknown
110k1163137
$endgroup$
A typical cyclic shift is
$$(c_k,c_k+1,ldots,c_k+n-1)$$
noting that $c_n=c_0$, $c_n+1=c_1$ etc.
This is the real part of the complex vector
$$(zeta^k,zeta^k+1,ldots,zeta^k+n-1)$$
where $zeta=exp(2pi i/n)$. Thus
beginalign
(c_k,c_k+1,ldots,c_k+n-1)
&=frac12(zeta^k,zeta^k+1,ldots,zeta^k+n-1)
+frac12(zeta^-k,zeta^-k-1,ldots,zeta^-k-n+1)\
&=fraczeta^k2(1,zeta,ldots,zeta^n-1)
+fraczeta^-k2(1,zeta^-1,ldots,zeta^-n+1).
endalign
Each row is a linear combination of the two vectors
$(1,zeta,ldots,zeta^n-1)$ and $(1,zeta^-1,ldots,zeta^-n+1)$
so the matrix has rank $le2$.
answered 2 hours ago
Lord Shark the UnknownLord Shark the Unknown
110k1163137
answered 2 hours ago
Lord Shark the UnknownLord Shark the Unknown
110k1163137
answered 2 hours ago
Lord Shark the UnknownLord Shark the Unknown
110k1163137
answered 2 hours ago
Lord Shark the UnknownLord Shark the Unknown
110k1163137
110k1163137
$begingroup$
+1 I like this answer too. Unfortunately I can accept only one.
$endgroup$
– P Vanchinathan
14 mins ago
add a comment |
$begingroup$
+1 I like this answer too. Unfortunately I can accept only one.
$endgroup$
– P Vanchinathan
14 mins ago
$begingroup$
+1 I like this answer too. Unfortunately I can accept only one.
$endgroup$
– P Vanchinathan
14 mins ago
$begingroup$
+1 I like this answer too. Unfortunately I can accept only one.
$endgroup$
– P Vanchinathan
14 mins 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%2f3217835%2flinear-independence-for-vectors-of-cosine-values%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
$begingroup$
You may want to read this page: en.wikipedia.org/wiki/Circulant_matrix
$endgroup$
– angryavian
2 hours ago
$begingroup$
@angryavian: Thanks. A quick glance tells me that the solution is more algebraic (with gcd of polynomials). This is easier to deal for me than explicit computations!
$endgroup$
– P Vanchinathan
2 hours ago