Some sort of commutativity of circulant matrices with a certain transformation
$begingroup$
Given a vector $v = (a_0, a_1, a_2, ... , a_n)$, we call a $k$-circulant matrix $C_k(v)$, where $k geq n$, the following $k*k$ matrix :
$$ C_k(v)= begin{bmatrix}
a_1 & a_2 & ... & a_n & 0 & ... & 0 \
0 & a_1 & a_2 & ... & a_n & 0 & ... \
cdot & & & & cdot & & \
& cdot & & & & cdot & \
& & cdot & & & & cdot \
a_3 & a_4 & ... & a_n & 0 & ... & 0 \
a_2 & a_3 & a_4 & ... & a_n & 0 & ...
end{bmatrix} $$
(basically, you pad the vector with zeros and then you form a circulant matrix with it.)
For simplicity, given two natural numbers $n,k$, define the block-matrix $$I_{n,k} = begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} $$
where $I_n$ repeats k times (I_n is the identity matrix of order $n$.)
I found out that, given a vector $v = (a_1,a_2,...,a_n)$ of length $n$ and a natural number $k$, we have the following relation:
$$ C_n(v) cdot I_{n,k} = I_{n,k}cdot C_{nk}(v)$$
This really isn't that hard to prove, since if for the vector $v$ we consider the following matrices:
$$ B = begin{bmatrix}
0 & 0 & 0 & 0 &... & 0 \
a_n & 0 & 0 & 0 & ... & 0 \
a_{n - 1}& a_n & 0 & 0 & ... & 0 \
a_{n - 2} & a_{n-1} & a_n & 0 & ... & 0 \
& & cdot & & & \
& & &cdot & & \
& & & & cdot & \
a_{2} & a_{3} & ... & a_{n-1} & a_n& 0 \
end{bmatrix} $$
$$A = begin{bmatrix}
a_1 & a_2 & a_3 & ... & a_n \
0 & a_1 & a_2 & ... & a_{n-1} \
& & cdot \
& & & cdot \
& & & & cdot \
0 & 0 & ... & 0 & a_1
end{bmatrix}$$
we observe that $C_n(v) = A+B$,
$$C_{n,k} = begin{bmatrix}
A & B & 0 & 0 & ... & 0\
0 & A & B & 0 & ... & 0\
&& cdot \
&&& cdot \
&&&& cdot \
B & 0 & 0 & 0 & ... & A
end{bmatrix}, $$
and so the relation to prove becomes
$$ [ A + B ] cdot begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} = begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} cdot begin{bmatrix}
A & B & 0 & 0 & ... & 0\
0 & A & B & 0 & ... & 0\
&& cdot \
&&& cdot \
&&&& cdot \
B & 0 & 0 & 0 & ... & A
end{bmatrix}, $$
which is obvious.
However, here comes the question: I have tested that given two vectors $v = (a_1, a_2, a_3, ... a_n) $ and $w = (b_1, b_2, ... , b_n)$, the following is also true:
$$ C_n(v) cdot C_n(w) cdot I_{n,k} = I_{n,k} cdot C_{nk}(v) cdot C_{nk}(w) $$
I haven't been able to prove this one; I know that the product of two circulant matrices is also a circulant matrix, but it doesn't seem like there is a vector $z = (c_1,c_2,...,c_n)$ such that $C_n(v) cdot C_n(w) = C_n(z)$ and $ C_{nk}(v) cdot C_{nk}(w) = C_{nk}(z)$.
Any ideas would be appreciated.
linear-algebra linear-transformations circulant-matrices
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
34318
$endgroup$
add a comment |
$begingroup$
Given a vector $v = (a_0, a_1, a_2, ... , a_n)$, we call a $k$-circulant matrix $C_k(v)$, where $k geq n$, the following $k*k$ matrix :
$$ C_k(v)= begin{bmatrix}
a_1 & a_2 & ... & a_n & 0 & ... & 0 \
0 & a_1 & a_2 & ... & a_n & 0 & ... \
cdot & & & & cdot & & \
& cdot & & & & cdot & \
& & cdot & & & & cdot \
a_3 & a_4 & ... & a_n & 0 & ... & 0 \
a_2 & a_3 & a_4 & ... & a_n & 0 & ...
end{bmatrix} $$
(basically, you pad the vector with zeros and then you form a circulant matrix with it.)
For simplicity, given two natural numbers $n,k$, define the block-matrix $$I_{n,k} = begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} $$
where $I_n$ repeats k times (I_n is the identity matrix of order $n$.)
I found out that, given a vector $v = (a_1,a_2,...,a_n)$ of length $n$ and a natural number $k$, we have the following relation:
$$ C_n(v) cdot I_{n,k} = I_{n,k}cdot C_{nk}(v)$$
This really isn't that hard to prove, since if for the vector $v$ we consider the following matrices:
$$ B = begin{bmatrix}
0 & 0 & 0 & 0 &... & 0 \
a_n & 0 & 0 & 0 & ... & 0 \
a_{n - 1}& a_n & 0 & 0 & ... & 0 \
a_{n - 2} & a_{n-1} & a_n & 0 & ... & 0 \
& & cdot & & & \
& & &cdot & & \
& & & & cdot & \
a_{2} & a_{3} & ... & a_{n-1} & a_n& 0 \
end{bmatrix} $$
$$A = begin{bmatrix}
a_1 & a_2 & a_3 & ... & a_n \
0 & a_1 & a_2 & ... & a_{n-1} \
& & cdot \
& & & cdot \
& & & & cdot \
0 & 0 & ... & 0 & a_1
end{bmatrix}$$
we observe that $C_n(v) = A+B$,
$$C_{n,k} = begin{bmatrix}
A & B & 0 & 0 & ... & 0\
0 & A & B & 0 & ... & 0\
&& cdot \
&&& cdot \
&&&& cdot \
B & 0 & 0 & 0 & ... & A
end{bmatrix}, $$
and so the relation to prove becomes
$$ [ A + B ] cdot begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} = begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} cdot begin{bmatrix}
A & B & 0 & 0 & ... & 0\
0 & A & B & 0 & ... & 0\
&& cdot \
&&& cdot \
&&&& cdot \
B & 0 & 0 & 0 & ... & A
end{bmatrix}, $$
which is obvious.
However, here comes the question: I have tested that given two vectors $v = (a_1, a_2, a_3, ... a_n) $ and $w = (b_1, b_2, ... , b_n)$, the following is also true:
$$ C_n(v) cdot C_n(w) cdot I_{n,k} = I_{n,k} cdot C_{nk}(v) cdot C_{nk}(w) $$
I haven't been able to prove this one; I know that the product of two circulant matrices is also a circulant matrix, but it doesn't seem like there is a vector $z = (c_1,c_2,...,c_n)$ such that $C_n(v) cdot C_n(w) = C_n(z)$ and $ C_{nk}(v) cdot C_{nk}(w) = C_{nk}(z)$.
Any ideas would be appreciated.
linear-algebra linear-transformations circulant-matrices
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
34318
$endgroup$
add a comment |
$begingroup$
Given a vector $v = (a_0, a_1, a_2, ... , a_n)$, we call a $k$-circulant matrix $C_k(v)$, where $k geq n$, the following $k*k$ matrix :
$$ C_k(v)= begin{bmatrix}
a_1 & a_2 & ... & a_n & 0 & ... & 0 \
0 & a_1 & a_2 & ... & a_n & 0 & ... \
cdot & & & & cdot & & \
& cdot & & & & cdot & \
& & cdot & & & & cdot \
a_3 & a_4 & ... & a_n & 0 & ... & 0 \
a_2 & a_3 & a_4 & ... & a_n & 0 & ...
end{bmatrix} $$
(basically, you pad the vector with zeros and then you form a circulant matrix with it.)
For simplicity, given two natural numbers $n,k$, define the block-matrix $$I_{n,k} = begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} $$
where $I_n$ repeats k times (I_n is the identity matrix of order $n$.)
I found out that, given a vector $v = (a_1,a_2,...,a_n)$ of length $n$ and a natural number $k$, we have the following relation:
$$ C_n(v) cdot I_{n,k} = I_{n,k}cdot C_{nk}(v)$$
This really isn't that hard to prove, since if for the vector $v$ we consider the following matrices:
$$ B = begin{bmatrix}
0 & 0 & 0 & 0 &... & 0 \
a_n & 0 & 0 & 0 & ... & 0 \
a_{n - 1}& a_n & 0 & 0 & ... & 0 \
a_{n - 2} & a_{n-1} & a_n & 0 & ... & 0 \
& & cdot & & & \
& & &cdot & & \
& & & & cdot & \
a_{2} & a_{3} & ... & a_{n-1} & a_n& 0 \
end{bmatrix} $$
$$A = begin{bmatrix}
a_1 & a_2 & a_3 & ... & a_n \
0 & a_1 & a_2 & ... & a_{n-1} \
& & cdot \
& & & cdot \
& & & & cdot \
0 & 0 & ... & 0 & a_1
end{bmatrix}$$
we observe that $C_n(v) = A+B$,
$$C_{n,k} = begin{bmatrix}
A & B & 0 & 0 & ... & 0\
0 & A & B & 0 & ... & 0\
&& cdot \
&&& cdot \
&&&& cdot \
B & 0 & 0 & 0 & ... & A
end{bmatrix}, $$
and so the relation to prove becomes
$$ [ A + B ] cdot begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} = begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} cdot begin{bmatrix}
A & B & 0 & 0 & ... & 0\
0 & A & B & 0 & ... & 0\
&& cdot \
&&& cdot \
&&&& cdot \
B & 0 & 0 & 0 & ... & A
end{bmatrix}, $$
which is obvious.
However, here comes the question: I have tested that given two vectors $v = (a_1, a_2, a_3, ... a_n) $ and $w = (b_1, b_2, ... , b_n)$, the following is also true:
$$ C_n(v) cdot C_n(w) cdot I_{n,k} = I_{n,k} cdot C_{nk}(v) cdot C_{nk}(w) $$
I haven't been able to prove this one; I know that the product of two circulant matrices is also a circulant matrix, but it doesn't seem like there is a vector $z = (c_1,c_2,...,c_n)$ such that $C_n(v) cdot C_n(w) = C_n(z)$ and $ C_{nk}(v) cdot C_{nk}(w) = C_{nk}(z)$.
Any ideas would be appreciated.
linear-algebra linear-transformations circulant-matrices
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
34318
$endgroup$
Given a vector $v = (a_0, a_1, a_2, ... , a_n)$, we call a $k$-circulant matrix $C_k(v)$, where $k geq n$, the following $k*k$ matrix :
$$ C_k(v)= begin{bmatrix}
a_1 & a_2 & ... & a_n & 0 & ... & 0 \
0 & a_1 & a_2 & ... & a_n & 0 & ... \
cdot & & & & cdot & & \
& cdot & & & & cdot & \
& & cdot & & & & cdot \
a_3 & a_4 & ... & a_n & 0 & ... & 0 \
a_2 & a_3 & a_4 & ... & a_n & 0 & ...
end{bmatrix} $$
(basically, you pad the vector with zeros and then you form a circulant matrix with it.)
For simplicity, given two natural numbers $n,k$, define the block-matrix $$I_{n,k} = begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} $$
where $I_n$ repeats k times (I_n is the identity matrix of order $n$.)
I found out that, given a vector $v = (a_1,a_2,...,a_n)$ of length $n$ and a natural number $k$, we have the following relation:
$$ C_n(v) cdot I_{n,k} = I_{n,k}cdot C_{nk}(v)$$
This really isn't that hard to prove, since if for the vector $v$ we consider the following matrices:
$$ B = begin{bmatrix}
0 & 0 & 0 & 0 &... & 0 \
a_n & 0 & 0 & 0 & ... & 0 \
a_{n - 1}& a_n & 0 & 0 & ... & 0 \
a_{n - 2} & a_{n-1} & a_n & 0 & ... & 0 \
& & cdot & & & \
& & &cdot & & \
& & & & cdot & \
a_{2} & a_{3} & ... & a_{n-1} & a_n& 0 \
end{bmatrix} $$
$$A = begin{bmatrix}
a_1 & a_2 & a_3 & ... & a_n \
0 & a_1 & a_2 & ... & a_{n-1} \
& & cdot \
& & & cdot \
& & & & cdot \
0 & 0 & ... & 0 & a_1
end{bmatrix}$$
we observe that $C_n(v) = A+B$,
$$C_{n,k} = begin{bmatrix}
A & B & 0 & 0 & ... & 0\
0 & A & B & 0 & ... & 0\
&& cdot \
&&& cdot \
&&&& cdot \
B & 0 & 0 & 0 & ... & A
end{bmatrix}, $$
and so the relation to prove becomes
$$ [ A + B ] cdot begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} = begin{bmatrix}
I_n & I_n & ... & I_n
end{bmatrix} cdot begin{bmatrix}
A & B & 0 & 0 & ... & 0\
0 & A & B & 0 & ... & 0\
&& cdot \
&&& cdot \
&&&& cdot \
B & 0 & 0 & 0 & ... & A
end{bmatrix}, $$
which is obvious.
However, here comes the question: I have tested that given two vectors $v = (a_1, a_2, a_3, ... a_n) $ and $w = (b_1, b_2, ... , b_n)$, the following is also true:
$$ C_n(v) cdot C_n(w) cdot I_{n,k} = I_{n,k} cdot C_{nk}(v) cdot C_{nk}(w) $$
I haven't been able to prove this one; I know that the product of two circulant matrices is also a circulant matrix, but it doesn't seem like there is a vector $z = (c_1,c_2,...,c_n)$ such that $C_n(v) cdot C_n(w) = C_n(z)$ and $ C_{nk}(v) cdot C_{nk}(w) = C_{nk}(z)$.
Any ideas would be appreciated.
linear-algebra linear-transformations circulant-matrices
linear-algebra linear-transformations circulant-matrices
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
34318
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
34318
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
34318
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
34318
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
34318
34318
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I figured it out. Basically we have
$$C_n(w) cdot begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} = begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} cdot C_{nk} (w) $$
By splitting $C_{nk} (w) $ into matrices $X$ and $Y$, we want to show that
$$begin{bmatrix} C_{n}(w)C_n(v) & C_{n}(w) C_n(v) & ... & C_{n}(w)C_n(v) end{bmatrix} = begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} cdot begin{bmatrix} X & Y & 0 & 0 & ... & 0 \ 0 & X & Y & 0 & ... & 0 \ &&cdot \ &&&cdot \ &&&&cdot \ Y & 0 & 0 & 0 & ... & X end{bmatrix} ,$$
which is obvious, since $ X+Y = C_n(w)$,
answered Dec 14 '18 at 8:02
Tanny SiebenTanny Sieben
34318
$endgroup$
add a comment |
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$begingroup$
I figured it out. Basically we have
$$C_n(w) cdot begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} = begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} cdot C_{nk} (w) $$
By splitting $C_{nk} (w) $ into matrices $X$ and $Y$, we want to show that
$$begin{bmatrix} C_{n}(w)C_n(v) & C_{n}(w) C_n(v) & ... & C_{n}(w)C_n(v) end{bmatrix} = begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} cdot begin{bmatrix} X & Y & 0 & 0 & ... & 0 \ 0 & X & Y & 0 & ... & 0 \ &&cdot \ &&&cdot \ &&&&cdot \ Y & 0 & 0 & 0 & ... & X end{bmatrix} ,$$
which is obvious, since $ X+Y = C_n(w)$,
answered Dec 14 '18 at 8:02
Tanny SiebenTanny Sieben
34318
$endgroup$
add a comment |
$begingroup$
I figured it out. Basically we have
$$C_n(w) cdot begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} = begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} cdot C_{nk} (w) $$
By splitting $C_{nk} (w) $ into matrices $X$ and $Y$, we want to show that
$$begin{bmatrix} C_{n}(w)C_n(v) & C_{n}(w) C_n(v) & ... & C_{n}(w)C_n(v) end{bmatrix} = begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} cdot begin{bmatrix} X & Y & 0 & 0 & ... & 0 \ 0 & X & Y & 0 & ... & 0 \ &&cdot \ &&&cdot \ &&&&cdot \ Y & 0 & 0 & 0 & ... & X end{bmatrix} ,$$
which is obvious, since $ X+Y = C_n(w)$,
answered Dec 14 '18 at 8:02
Tanny SiebenTanny Sieben
34318
$endgroup$
add a comment |
$begingroup$
I figured it out. Basically we have
$$C_n(w) cdot begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} = begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} cdot C_{nk} (w) $$
By splitting $C_{nk} (w) $ into matrices $X$ and $Y$, we want to show that
$$begin{bmatrix} C_{n}(w)C_n(v) & C_{n}(w) C_n(v) & ... & C_{n}(w)C_n(v) end{bmatrix} = begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} cdot begin{bmatrix} X & Y & 0 & 0 & ... & 0 \ 0 & X & Y & 0 & ... & 0 \ &&cdot \ &&&cdot \ &&&&cdot \ Y & 0 & 0 & 0 & ... & X end{bmatrix} ,$$
which is obvious, since $ X+Y = C_n(w)$,
answered Dec 14 '18 at 8:02
Tanny SiebenTanny Sieben
34318
$endgroup$
I figured it out. Basically we have
$$C_n(w) cdot begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} = begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} cdot C_{nk} (w) $$
By splitting $C_{nk} (w) $ into matrices $X$ and $Y$, we want to show that
$$begin{bmatrix} C_{n}(w)C_n(v) & C_{n}(w) C_n(v) & ... & C_{n}(w)C_n(v) end{bmatrix} = begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) end{bmatrix} cdot begin{bmatrix} X & Y & 0 & 0 & ... & 0 \ 0 & X & Y & 0 & ... & 0 \ &&cdot \ &&&cdot \ &&&&cdot \ Y & 0 & 0 & 0 & ... & X end{bmatrix} ,$$
which is obvious, since $ X+Y = C_n(w)$,
answered Dec 14 '18 at 8:02
Tanny SiebenTanny Sieben
34318
answered Dec 14 '18 at 8:02
Tanny SiebenTanny Sieben
34318
answered Dec 14 '18 at 8:02
Tanny SiebenTanny Sieben
34318
answered Dec 14 '18 at 8:02
Tanny SiebenTanny Sieben
34318
34318
add a comment |
add a comment |
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