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
36318
$endgroup$
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
36318
$endgroup$
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
36318
$endgroup$
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
36318
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
36318
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
36318
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
36318
asked Dec 14 '18 at 1:10
Tanny SiebenTanny Sieben
36318
36318
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
36318
$endgroup$
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
});
}
});
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%2f3038803%2fsome-sort-of-commutativity-of-circulant-matrices-with-a-certain-transformation%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
$begingroup$
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
36318
$endgroup$
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
36318
$endgroup$
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
36318
$endgroup$
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
36318
answered Dec 14 '18 at 8:02
Tanny SiebenTanny Sieben
36318
answered Dec 14 '18 at 8:02
Tanny SiebenTanny Sieben
36318
answered Dec 14 '18 at 8:02
Tanny SiebenTanny Sieben
36318
36318
add a comment |
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%2f3038803%2fsome-sort-of-commutativity-of-circulant-matrices-with-a-certain-transformation%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
