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 $Xin M_n,n(F); AX=XA=0$?
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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
15 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
1 min 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
2 mins ago
add a comment |
Your Answer
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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
15 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
1 min 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
15 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
1 min 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
15 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
1 min ago
add a comment |
$begingroup$
[+1] Very nice answer !
$endgroup$
– Jean Marie
15 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
1 min ago
$begingroup$
[+1] Very nice answer !
$endgroup$
– Jean Marie
15 mins ago
$begingroup$
[+1] Very nice answer !
$endgroup$
– Jean Marie
15 mins ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
1 min ago
$begingroup$
Pleasantly simple answer!
$endgroup$
– P Vanchinathan
1 min 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
2 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
2 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
2 mins ago
add a comment |
$begingroup$
+1 I like this answer too. Unfortunately I can accept only one.
$endgroup$
– P Vanchinathan
2 mins ago
$begingroup$
+1 I like this answer too. Unfortunately I can accept only one.
$endgroup$
– P Vanchinathan
2 mins ago
$begingroup$
+1 I like this answer too. Unfortunately I can accept only one.
$endgroup$
– P Vanchinathan
2 mins ago
add a comment |
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$begingroup$
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