How to minimize the minimum mean square error of this difference












0












$begingroup$


I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem



$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$



where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.



Any hints or ideas on finding the minimizing vectors of this problem?



It is also given that



$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while



$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$



where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector



I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$



But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...



Thanks






optimization mean-square-error







share|cite|improve this question











$endgroup$












  • $begingroup$
    There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:17












  • $begingroup$
    sorry had typo thanks
    $endgroup$
    – Tyrone
    Oct 22 '15 at 0:19










  • $begingroup$
    I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:43












  • $begingroup$
    thnk you for the reference
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:41
















0












$begingroup$


I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem



$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$



where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.



Any hints or ideas on finding the minimizing vectors of this problem?



It is also given that



$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while



$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$



where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector



I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$



But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...



Thanks






optimization mean-square-error







share|cite|improve this question











$endgroup$












  • $begingroup$
    There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:17












  • $begingroup$
    sorry had typo thanks
    $endgroup$
    – Tyrone
    Oct 22 '15 at 0:19










  • $begingroup$
    I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:43












  • $begingroup$
    thnk you for the reference
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:41














0












0








0





$begingroup$


I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem



$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$



where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.



Any hints or ideas on finding the minimizing vectors of this problem?



It is also given that



$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while



$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$



where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector



I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$



But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...



Thanks






optimization mean-square-error







share|cite|improve this question











$endgroup$




I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem



$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$



where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.



Any hints or ideas on finding the minimizing vectors of this problem?



It is also given that



$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while



$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$



where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector



I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$



But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...



Thanks






optimization mean-square-error




optimization mean-square-error






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 22 '15 at 3:00







Tyrone

















asked Oct 21 '15 at 23:46









TyroneTyrone

244418




244418












  • $begingroup$
    There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:17












  • $begingroup$
    sorry had typo thanks
    $endgroup$
    – Tyrone
    Oct 22 '15 at 0:19










  • $begingroup$
    I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:43












  • $begingroup$
    thnk you for the reference
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:41


















  • $begingroup$
    There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:17












  • $begingroup$
    sorry had typo thanks
    $endgroup$
    – Tyrone
    Oct 22 '15 at 0:19










  • $begingroup$
    I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
    $endgroup$
    – callculus
    Oct 22 '15 at 0:43












  • $begingroup$
    thnk you for the reference
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:41
















$begingroup$
There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
$endgroup$
– callculus
Oct 22 '15 at 0:17






$begingroup$
There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
$endgroup$
– callculus
Oct 22 '15 at 0:17














$begingroup$
sorry had typo thanks
$endgroup$
– Tyrone
Oct 22 '15 at 0:19




$begingroup$
sorry had typo thanks
$endgroup$
– Tyrone
Oct 22 '15 at 0:19












$begingroup$
I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
$endgroup$
– callculus
Oct 22 '15 at 0:43






$begingroup$
I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
$endgroup$
– callculus
Oct 22 '15 at 0:43














$begingroup$
thnk you for the reference
$endgroup$
– Tyrone
Oct 22 '15 at 1:41




$begingroup$
thnk you for the reference
$endgroup$
– Tyrone
Oct 22 '15 at 1:41










1 Answer
1






active

oldest

votes


















0












$begingroup$

I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$

where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.

This problem can be restated as

$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.

After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.

Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as

$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.

Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    i am not sure... wht how did you decompose vector y?
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:49










  • $begingroup$
    The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
    $endgroup$
    – shaktiman
    Oct 22 '15 at 3:31













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$

where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.

This problem can be restated as

$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.

After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.

Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as

$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.

Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    i am not sure... wht how did you decompose vector y?
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:49










  • $begingroup$
    The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
    $endgroup$
    – shaktiman
    Oct 22 '15 at 3:31


















0












$begingroup$

I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$

where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.

This problem can be restated as

$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.

After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.

Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as

$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.

Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    i am not sure... wht how did you decompose vector y?
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:49










  • $begingroup$
    The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
    $endgroup$
    – shaktiman
    Oct 22 '15 at 3:31
















0












0








0





$begingroup$

I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$

where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.

This problem can be restated as

$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.

After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.

Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as

$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.

Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.






share|cite|improve this answer









$endgroup$



I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$

where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.

This problem can be restated as

$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.

After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.

Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as

$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.

Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 22 '15 at 0:29









shaktimanshaktiman

628




628












  • $begingroup$
    i am not sure... wht how did you decompose vector y?
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:49










  • $begingroup$
    The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
    $endgroup$
    – shaktiman
    Oct 22 '15 at 3:31




















  • $begingroup$
    i am not sure... wht how did you decompose vector y?
    $endgroup$
    – Tyrone
    Oct 22 '15 at 1:49










  • $begingroup$
    The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
    $endgroup$
    – shaktiman
    Oct 22 '15 at 3:31


















$begingroup$
i am not sure... wht how did you decompose vector y?
$endgroup$
– Tyrone
Oct 22 '15 at 1:49




$begingroup$
i am not sure... wht how did you decompose vector y?
$endgroup$
– Tyrone
Oct 22 '15 at 1:49












$begingroup$
The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
$endgroup$
– shaktiman
Oct 22 '15 at 3:31






$begingroup$
The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
$endgroup$
– shaktiman
Oct 22 '15 at 3:31




















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Plaza Victoria

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Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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