How to rewrite $M_1otimes M_2$ isolating $M_2$?












0












$begingroup$


I have 2 matrices $M_1, M_2$.



Is there a way to rewrite $M_1otimes M_2$ as $M cdot M_2$? i.e. $M$ is a matrix that it's being multiplied by $M_2$.



My objective here is to isolate $M_2$ as a product of matrices.



Can we do something similar with $operatorname{vec}(M_1otimes M_2)$ as $Mcdot operatorname{vec}(M_2)$?






matrix-calculus







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  • $begingroup$
    For those who are voting to close. Please give me some feedback. The reasons to vote are too general...
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    – An old man in the sea.
    Dec 21 '18 at 10:29
















0












$begingroup$


I have 2 matrices $M_1, M_2$.



Is there a way to rewrite $M_1otimes M_2$ as $M cdot M_2$? i.e. $M$ is a matrix that it's being multiplied by $M_2$.



My objective here is to isolate $M_2$ as a product of matrices.



Can we do something similar with $operatorname{vec}(M_1otimes M_2)$ as $Mcdot operatorname{vec}(M_2)$?






matrix-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    For those who are voting to close. Please give me some feedback. The reasons to vote are too general...
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 10:29














0












0








0





$begingroup$


I have 2 matrices $M_1, M_2$.



Is there a way to rewrite $M_1otimes M_2$ as $M cdot M_2$? i.e. $M$ is a matrix that it's being multiplied by $M_2$.



My objective here is to isolate $M_2$ as a product of matrices.



Can we do something similar with $operatorname{vec}(M_1otimes M_2)$ as $Mcdot operatorname{vec}(M_2)$?






matrix-calculus







share|cite|improve this question











$endgroup$




I have 2 matrices $M_1, M_2$.



Is there a way to rewrite $M_1otimes M_2$ as $M cdot M_2$? i.e. $M$ is a matrix that it's being multiplied by $M_2$.



My objective here is to isolate $M_2$ as a product of matrices.



Can we do something similar with $operatorname{vec}(M_1otimes M_2)$ as $Mcdot operatorname{vec}(M_2)$?






matrix-calculus




matrix-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 13:04









Davide Giraudo

126k16150261




126k16150261










asked Dec 21 '18 at 8:38









An old man in the sea.An old man in the sea.

1,61311134




1,61311134












  • $begingroup$
    For those who are voting to close. Please give me some feedback. The reasons to vote are too general...
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 10:29


















  • $begingroup$
    For those who are voting to close. Please give me some feedback. The reasons to vote are too general...
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 10:29
















$begingroup$
For those who are voting to close. Please give me some feedback. The reasons to vote are too general...
$endgroup$
– An old man in the sea.
Dec 21 '18 at 10:29




$begingroup$
For those who are voting to close. Please give me some feedback. The reasons to vote are too general...
$endgroup$
– An old man in the sea.
Dec 21 '18 at 10:29










3 Answers
3






active

oldest

votes


















1












$begingroup$

Assume the following dimensions for the matrices:
$$eqalign{
M_1quad&isquad(mtimes n) cr
M_2quad&isquad(ptimes q) cr
Mquad&isquad(rtimes p) cr
}$$
Then for their products:
$$eqalign{
M_1otimes M_2quad&isquad(mptimes nq) cr
Mcdot M_2quad&isquad(rtimes q) cr
}$$
Note that the final dimensions do not match, except in the trivial case $n=1$.



However, finding a matrix such that
$${rm vec}(M_1otimes M_2)=Mcdot{rm vec}(M_2)$$
is possible; something along the lines that Nadiels has suggested.



Let $c_k$ be the $k^{th}$ column of $M_1,,$ and let $,(P,Q),$ be the $(ptimes p)$ and $(qtimes q)$ identity matrices, respectively. Then
$$eqalign{
M &= pmatrix{Qotimes c_1cr Qotimes c_2cr vdotscr Qotimes c_n}otimes Pcr
}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You're completely right... I never thought about that. I'm a bit dumb... thanks for all the help. ;)
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:15










  • $begingroup$
    One thing I didn't understand from the answer of Nadiels is what's $frac{partial}{partial m_i}$...
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:17










  • $begingroup$
    Nadiels's derivative zeros out all elements of $M_1$ and $m$ except for the one that corresponds to the $m_i$ component. So it transforms $m$ into a vector from the standard basis ($e_i$), and transforms $M_1$ into a single-entry matrix.
    $endgroup$
    – greg
    Dec 21 '18 at 17:40












  • $begingroup$
    Greg I always learn with you. Thank you! ;)
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 18:40



















1












$begingroup$

Since $operatorname{vec}(M_1 otimes M_2)$ is linear in the entries of $M_2$, then letting $mathbf{m} = operatorname{vec}(M_2)$ you could write
begin{align}
operatorname{vec}(M_1 otimes M_2) &= D cdot mathbf{m}_2
end{align}
where the columns of $D$ are given by
$$
begin{align}
D_{:, i} &= frac{partial}{partial m_i}operatorname{vec}(M_1 otimes M_2) \
&= operatorname{vec}left(M_1 otimes frac{partial}{partial m_i}M_2right)
end{align}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nadiels, Thanks for the answer. However, I don't understand where the derivative comes from...
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 10:44










  • $begingroup$
    Let me reformulate my comment above... What's $m_i$?
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:16










  • $begingroup$
    Can't elaborate 'cause phone, but someone else has given you a more complete answer. Basically once you vectorise everything you can write linear transformations as a matrix/vec product using their Jacobian. I was just too lazy to keep track of the change in indices moving to the vectorised form, hence my $mathbf{m}$
    $endgroup$
    – Nadiels
    Dec 21 '18 at 17:56










  • $begingroup$
    =OOOOO I would never have thought about something like that!!!!! WOW! =D thanks ;)
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:58





















1












$begingroup$

Another way to solve the problem is to take advantage of the block structure of the Kronecker product $$P = M_1otimes M_2$$
Simply extract the $(1,1)$-block of the product and divide it by the $(1,1)$ element of the first matrix. Assuming the dimensions of the matrices are
$$M_1in{mathbb R}^{mtimes n},quad M_2in{mathbb R}^{ptimes q}$$
one can isolate the second matrix as follows
$$eqalign{
M_2 &= frac{pmatrix{I_{p} & 0}Big(M_1otimes M_2Big)pmatrix{I_{q}cr 0}}{pmatrix{1&0}M_1pmatrix{1cr 0}} cr
}$$






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Assume the following dimensions for the matrices:
    $$eqalign{
    M_1quad&isquad(mtimes n) cr
    M_2quad&isquad(ptimes q) cr
    Mquad&isquad(rtimes p) cr
    }$$
    Then for their products:
    $$eqalign{
    M_1otimes M_2quad&isquad(mptimes nq) cr
    Mcdot M_2quad&isquad(rtimes q) cr
    }$$
    Note that the final dimensions do not match, except in the trivial case $n=1$.



    However, finding a matrix such that
    $${rm vec}(M_1otimes M_2)=Mcdot{rm vec}(M_2)$$
    is possible; something along the lines that Nadiels has suggested.



    Let $c_k$ be the $k^{th}$ column of $M_1,,$ and let $,(P,Q),$ be the $(ptimes p)$ and $(qtimes q)$ identity matrices, respectively. Then
    $$eqalign{
    M &= pmatrix{Qotimes c_1cr Qotimes c_2cr vdotscr Qotimes c_n}otimes Pcr
    }$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You're completely right... I never thought about that. I'm a bit dumb... thanks for all the help. ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:15










    • $begingroup$
      One thing I didn't understand from the answer of Nadiels is what's $frac{partial}{partial m_i}$...
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:17










    • $begingroup$
      Nadiels's derivative zeros out all elements of $M_1$ and $m$ except for the one that corresponds to the $m_i$ component. So it transforms $m$ into a vector from the standard basis ($e_i$), and transforms $M_1$ into a single-entry matrix.
      $endgroup$
      – greg
      Dec 21 '18 at 17:40












    • $begingroup$
      Greg I always learn with you. Thank you! ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 18:40
















    1












    $begingroup$

    Assume the following dimensions for the matrices:
    $$eqalign{
    M_1quad&isquad(mtimes n) cr
    M_2quad&isquad(ptimes q) cr
    Mquad&isquad(rtimes p) cr
    }$$
    Then for their products:
    $$eqalign{
    M_1otimes M_2quad&isquad(mptimes nq) cr
    Mcdot M_2quad&isquad(rtimes q) cr
    }$$
    Note that the final dimensions do not match, except in the trivial case $n=1$.



    However, finding a matrix such that
    $${rm vec}(M_1otimes M_2)=Mcdot{rm vec}(M_2)$$
    is possible; something along the lines that Nadiels has suggested.



    Let $c_k$ be the $k^{th}$ column of $M_1,,$ and let $,(P,Q),$ be the $(ptimes p)$ and $(qtimes q)$ identity matrices, respectively. Then
    $$eqalign{
    M &= pmatrix{Qotimes c_1cr Qotimes c_2cr vdotscr Qotimes c_n}otimes Pcr
    }$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You're completely right... I never thought about that. I'm a bit dumb... thanks for all the help. ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:15










    • $begingroup$
      One thing I didn't understand from the answer of Nadiels is what's $frac{partial}{partial m_i}$...
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:17










    • $begingroup$
      Nadiels's derivative zeros out all elements of $M_1$ and $m$ except for the one that corresponds to the $m_i$ component. So it transforms $m$ into a vector from the standard basis ($e_i$), and transforms $M_1$ into a single-entry matrix.
      $endgroup$
      – greg
      Dec 21 '18 at 17:40












    • $begingroup$
      Greg I always learn with you. Thank you! ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 18:40














    1












    1








    1





    $begingroup$

    Assume the following dimensions for the matrices:
    $$eqalign{
    M_1quad&isquad(mtimes n) cr
    M_2quad&isquad(ptimes q) cr
    Mquad&isquad(rtimes p) cr
    }$$
    Then for their products:
    $$eqalign{
    M_1otimes M_2quad&isquad(mptimes nq) cr
    Mcdot M_2quad&isquad(rtimes q) cr
    }$$
    Note that the final dimensions do not match, except in the trivial case $n=1$.



    However, finding a matrix such that
    $${rm vec}(M_1otimes M_2)=Mcdot{rm vec}(M_2)$$
    is possible; something along the lines that Nadiels has suggested.



    Let $c_k$ be the $k^{th}$ column of $M_1,,$ and let $,(P,Q),$ be the $(ptimes p)$ and $(qtimes q)$ identity matrices, respectively. Then
    $$eqalign{
    M &= pmatrix{Qotimes c_1cr Qotimes c_2cr vdotscr Qotimes c_n}otimes Pcr
    }$$






    share|cite|improve this answer











    $endgroup$



    Assume the following dimensions for the matrices:
    $$eqalign{
    M_1quad&isquad(mtimes n) cr
    M_2quad&isquad(ptimes q) cr
    Mquad&isquad(rtimes p) cr
    }$$
    Then for their products:
    $$eqalign{
    M_1otimes M_2quad&isquad(mptimes nq) cr
    Mcdot M_2quad&isquad(rtimes q) cr
    }$$
    Note that the final dimensions do not match, except in the trivial case $n=1$.



    However, finding a matrix such that
    $${rm vec}(M_1otimes M_2)=Mcdot{rm vec}(M_2)$$
    is possible; something along the lines that Nadiels has suggested.



    Let $c_k$ be the $k^{th}$ column of $M_1,,$ and let $,(P,Q),$ be the $(ptimes p)$ and $(qtimes q)$ identity matrices, respectively. Then
    $$eqalign{
    M &= pmatrix{Qotimes c_1cr Qotimes c_2cr vdotscr Qotimes c_n}otimes Pcr
    }$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 21 '18 at 17:32

























    answered Dec 21 '18 at 17:08









    greggreg

    8,0251822




    8,0251822












    • $begingroup$
      You're completely right... I never thought about that. I'm a bit dumb... thanks for all the help. ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:15










    • $begingroup$
      One thing I didn't understand from the answer of Nadiels is what's $frac{partial}{partial m_i}$...
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:17










    • $begingroup$
      Nadiels's derivative zeros out all elements of $M_1$ and $m$ except for the one that corresponds to the $m_i$ component. So it transforms $m$ into a vector from the standard basis ($e_i$), and transforms $M_1$ into a single-entry matrix.
      $endgroup$
      – greg
      Dec 21 '18 at 17:40












    • $begingroup$
      Greg I always learn with you. Thank you! ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 18:40


















    • $begingroup$
      You're completely right... I never thought about that. I'm a bit dumb... thanks for all the help. ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:15










    • $begingroup$
      One thing I didn't understand from the answer of Nadiels is what's $frac{partial}{partial m_i}$...
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:17










    • $begingroup$
      Nadiels's derivative zeros out all elements of $M_1$ and $m$ except for the one that corresponds to the $m_i$ component. So it transforms $m$ into a vector from the standard basis ($e_i$), and transforms $M_1$ into a single-entry matrix.
      $endgroup$
      – greg
      Dec 21 '18 at 17:40












    • $begingroup$
      Greg I always learn with you. Thank you! ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 18:40
















    $begingroup$
    You're completely right... I never thought about that. I'm a bit dumb... thanks for all the help. ;)
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:15




    $begingroup$
    You're completely right... I never thought about that. I'm a bit dumb... thanks for all the help. ;)
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:15












    $begingroup$
    One thing I didn't understand from the answer of Nadiels is what's $frac{partial}{partial m_i}$...
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:17




    $begingroup$
    One thing I didn't understand from the answer of Nadiels is what's $frac{partial}{partial m_i}$...
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:17












    $begingroup$
    Nadiels's derivative zeros out all elements of $M_1$ and $m$ except for the one that corresponds to the $m_i$ component. So it transforms $m$ into a vector from the standard basis ($e_i$), and transforms $M_1$ into a single-entry matrix.
    $endgroup$
    – greg
    Dec 21 '18 at 17:40






    $begingroup$
    Nadiels's derivative zeros out all elements of $M_1$ and $m$ except for the one that corresponds to the $m_i$ component. So it transforms $m$ into a vector from the standard basis ($e_i$), and transforms $M_1$ into a single-entry matrix.
    $endgroup$
    – greg
    Dec 21 '18 at 17:40














    $begingroup$
    Greg I always learn with you. Thank you! ;)
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 18:40




    $begingroup$
    Greg I always learn with you. Thank you! ;)
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 18:40











    1












    $begingroup$

    Since $operatorname{vec}(M_1 otimes M_2)$ is linear in the entries of $M_2$, then letting $mathbf{m} = operatorname{vec}(M_2)$ you could write
    begin{align}
    operatorname{vec}(M_1 otimes M_2) &= D cdot mathbf{m}_2
    end{align}
    where the columns of $D$ are given by
    $$
    begin{align}
    D_{:, i} &= frac{partial}{partial m_i}operatorname{vec}(M_1 otimes M_2) \
    &= operatorname{vec}left(M_1 otimes frac{partial}{partial m_i}M_2right)
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Nadiels, Thanks for the answer. However, I don't understand where the derivative comes from...
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 10:44










    • $begingroup$
      Let me reformulate my comment above... What's $m_i$?
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:16










    • $begingroup$
      Can't elaborate 'cause phone, but someone else has given you a more complete answer. Basically once you vectorise everything you can write linear transformations as a matrix/vec product using their Jacobian. I was just too lazy to keep track of the change in indices moving to the vectorised form, hence my $mathbf{m}$
      $endgroup$
      – Nadiels
      Dec 21 '18 at 17:56










    • $begingroup$
      =OOOOO I would never have thought about something like that!!!!! WOW! =D thanks ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:58


















    1












    $begingroup$

    Since $operatorname{vec}(M_1 otimes M_2)$ is linear in the entries of $M_2$, then letting $mathbf{m} = operatorname{vec}(M_2)$ you could write
    begin{align}
    operatorname{vec}(M_1 otimes M_2) &= D cdot mathbf{m}_2
    end{align}
    where the columns of $D$ are given by
    $$
    begin{align}
    D_{:, i} &= frac{partial}{partial m_i}operatorname{vec}(M_1 otimes M_2) \
    &= operatorname{vec}left(M_1 otimes frac{partial}{partial m_i}M_2right)
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Nadiels, Thanks for the answer. However, I don't understand where the derivative comes from...
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 10:44










    • $begingroup$
      Let me reformulate my comment above... What's $m_i$?
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:16










    • $begingroup$
      Can't elaborate 'cause phone, but someone else has given you a more complete answer. Basically once you vectorise everything you can write linear transformations as a matrix/vec product using their Jacobian. I was just too lazy to keep track of the change in indices moving to the vectorised form, hence my $mathbf{m}$
      $endgroup$
      – Nadiels
      Dec 21 '18 at 17:56










    • $begingroup$
      =OOOOO I would never have thought about something like that!!!!! WOW! =D thanks ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:58
















    1












    1








    1





    $begingroup$

    Since $operatorname{vec}(M_1 otimes M_2)$ is linear in the entries of $M_2$, then letting $mathbf{m} = operatorname{vec}(M_2)$ you could write
    begin{align}
    operatorname{vec}(M_1 otimes M_2) &= D cdot mathbf{m}_2
    end{align}
    where the columns of $D$ are given by
    $$
    begin{align}
    D_{:, i} &= frac{partial}{partial m_i}operatorname{vec}(M_1 otimes M_2) \
    &= operatorname{vec}left(M_1 otimes frac{partial}{partial m_i}M_2right)
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$



    Since $operatorname{vec}(M_1 otimes M_2)$ is linear in the entries of $M_2$, then letting $mathbf{m} = operatorname{vec}(M_2)$ you could write
    begin{align}
    operatorname{vec}(M_1 otimes M_2) &= D cdot mathbf{m}_2
    end{align}
    where the columns of $D$ are given by
    $$
    begin{align}
    D_{:, i} &= frac{partial}{partial m_i}operatorname{vec}(M_1 otimes M_2) \
    &= operatorname{vec}left(M_1 otimes frac{partial}{partial m_i}M_2right)
    end{align}
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 21 '18 at 10:22









    NadielsNadiels

    2,385413




    2,385413












    • $begingroup$
      Nadiels, Thanks for the answer. However, I don't understand where the derivative comes from...
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 10:44










    • $begingroup$
      Let me reformulate my comment above... What's $m_i$?
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:16










    • $begingroup$
      Can't elaborate 'cause phone, but someone else has given you a more complete answer. Basically once you vectorise everything you can write linear transformations as a matrix/vec product using their Jacobian. I was just too lazy to keep track of the change in indices moving to the vectorised form, hence my $mathbf{m}$
      $endgroup$
      – Nadiels
      Dec 21 '18 at 17:56










    • $begingroup$
      =OOOOO I would never have thought about something like that!!!!! WOW! =D thanks ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:58




















    • $begingroup$
      Nadiels, Thanks for the answer. However, I don't understand where the derivative comes from...
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 10:44










    • $begingroup$
      Let me reformulate my comment above... What's $m_i$?
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:16










    • $begingroup$
      Can't elaborate 'cause phone, but someone else has given you a more complete answer. Basically once you vectorise everything you can write linear transformations as a matrix/vec product using their Jacobian. I was just too lazy to keep track of the change in indices moving to the vectorised form, hence my $mathbf{m}$
      $endgroup$
      – Nadiels
      Dec 21 '18 at 17:56










    • $begingroup$
      =OOOOO I would never have thought about something like that!!!!! WOW! =D thanks ;)
      $endgroup$
      – An old man in the sea.
      Dec 21 '18 at 17:58


















    $begingroup$
    Nadiels, Thanks for the answer. However, I don't understand where the derivative comes from...
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 10:44




    $begingroup$
    Nadiels, Thanks for the answer. However, I don't understand where the derivative comes from...
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 10:44












    $begingroup$
    Let me reformulate my comment above... What's $m_i$?
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:16




    $begingroup$
    Let me reformulate my comment above... What's $m_i$?
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:16












    $begingroup$
    Can't elaborate 'cause phone, but someone else has given you a more complete answer. Basically once you vectorise everything you can write linear transformations as a matrix/vec product using their Jacobian. I was just too lazy to keep track of the change in indices moving to the vectorised form, hence my $mathbf{m}$
    $endgroup$
    – Nadiels
    Dec 21 '18 at 17:56




    $begingroup$
    Can't elaborate 'cause phone, but someone else has given you a more complete answer. Basically once you vectorise everything you can write linear transformations as a matrix/vec product using their Jacobian. I was just too lazy to keep track of the change in indices moving to the vectorised form, hence my $mathbf{m}$
    $endgroup$
    – Nadiels
    Dec 21 '18 at 17:56












    $begingroup$
    =OOOOO I would never have thought about something like that!!!!! WOW! =D thanks ;)
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:58






    $begingroup$
    =OOOOO I would never have thought about something like that!!!!! WOW! =D thanks ;)
    $endgroup$
    – An old man in the sea.
    Dec 21 '18 at 17:58













    1












    $begingroup$

    Another way to solve the problem is to take advantage of the block structure of the Kronecker product $$P = M_1otimes M_2$$
    Simply extract the $(1,1)$-block of the product and divide it by the $(1,1)$ element of the first matrix. Assuming the dimensions of the matrices are
    $$M_1in{mathbb R}^{mtimes n},quad M_2in{mathbb R}^{ptimes q}$$
    one can isolate the second matrix as follows
    $$eqalign{
    M_2 &= frac{pmatrix{I_{p} & 0}Big(M_1otimes M_2Big)pmatrix{I_{q}cr 0}}{pmatrix{1&0}M_1pmatrix{1cr 0}} cr
    }$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Another way to solve the problem is to take advantage of the block structure of the Kronecker product $$P = M_1otimes M_2$$
      Simply extract the $(1,1)$-block of the product and divide it by the $(1,1)$ element of the first matrix. Assuming the dimensions of the matrices are
      $$M_1in{mathbb R}^{mtimes n},quad M_2in{mathbb R}^{ptimes q}$$
      one can isolate the second matrix as follows
      $$eqalign{
      M_2 &= frac{pmatrix{I_{p} & 0}Big(M_1otimes M_2Big)pmatrix{I_{q}cr 0}}{pmatrix{1&0}M_1pmatrix{1cr 0}} cr
      }$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Another way to solve the problem is to take advantage of the block structure of the Kronecker product $$P = M_1otimes M_2$$
        Simply extract the $(1,1)$-block of the product and divide it by the $(1,1)$ element of the first matrix. Assuming the dimensions of the matrices are
        $$M_1in{mathbb R}^{mtimes n},quad M_2in{mathbb R}^{ptimes q}$$
        one can isolate the second matrix as follows
        $$eqalign{
        M_2 &= frac{pmatrix{I_{p} & 0}Big(M_1otimes M_2Big)pmatrix{I_{q}cr 0}}{pmatrix{1&0}M_1pmatrix{1cr 0}} cr
        }$$






        share|cite|improve this answer









        $endgroup$



        Another way to solve the problem is to take advantage of the block structure of the Kronecker product $$P = M_1otimes M_2$$
        Simply extract the $(1,1)$-block of the product and divide it by the $(1,1)$ element of the first matrix. Assuming the dimensions of the matrices are
        $$M_1in{mathbb R}^{mtimes n},quad M_2in{mathbb R}^{ptimes q}$$
        one can isolate the second matrix as follows
        $$eqalign{
        M_2 &= frac{pmatrix{I_{p} & 0}Big(M_1otimes M_2Big)pmatrix{I_{q}cr 0}}{pmatrix{1&0}M_1pmatrix{1cr 0}} cr
        }$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 26 '18 at 15:16









        lynnlynn

        1,766177




        1,766177






























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