Solving the matrix equation
$begingroup$
How can I solve the matrix equation of the form
$$
mathbf{SXK} + mathbf{X} = mathbf{Y}
$$
Here $mathbf{S}$ and $mathbf{K}$ are symmetric matrices, in addition $mathbf{K}$ is a sparse symmetric matrix. $X$ is the variable. Though $mathbf{S}$ and $mathbf{K}$ are symmetric, it is not invertible in general and $mathbf{X}$ is not symmetric. Is it possible to find a closed-form solution for $mathbf{X}$ ?. Is there any relevant literature study about solving such equations ?
linear-algebra matrices matrix-equations
asked Dec 23 '18 at 19:33
ShewShew
596413
$endgroup$
add a comment |
$begingroup$
How can I solve the matrix equation of the form
$$
mathbf{SXK} + mathbf{X} = mathbf{Y}
$$
Here $mathbf{S}$ and $mathbf{K}$ are symmetric matrices, in addition $mathbf{K}$ is a sparse symmetric matrix. $X$ is the variable. Though $mathbf{S}$ and $mathbf{K}$ are symmetric, it is not invertible in general and $mathbf{X}$ is not symmetric. Is it possible to find a closed-form solution for $mathbf{X}$ ?. Is there any relevant literature study about solving such equations ?
linear-algebra matrices matrix-equations
asked Dec 23 '18 at 19:33
ShewShew
596413
$endgroup$
1
$begingroup$
Are all square matrices? $S,K$ are invertible?
$endgroup$
– Exodd
Dec 23 '18 at 19:45
$begingroup$
@Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
$endgroup$
– Shew
Dec 23 '18 at 20:05
add a comment |
$begingroup$
How can I solve the matrix equation of the form
$$
mathbf{SXK} + mathbf{X} = mathbf{Y}
$$
Here $mathbf{S}$ and $mathbf{K}$ are symmetric matrices, in addition $mathbf{K}$ is a sparse symmetric matrix. $X$ is the variable. Though $mathbf{S}$ and $mathbf{K}$ are symmetric, it is not invertible in general and $mathbf{X}$ is not symmetric. Is it possible to find a closed-form solution for $mathbf{X}$ ?. Is there any relevant literature study about solving such equations ?
linear-algebra matrices matrix-equations
asked Dec 23 '18 at 19:33
ShewShew
596413
$endgroup$
How can I solve the matrix equation of the form
$$
mathbf{SXK} + mathbf{X} = mathbf{Y}
$$
Here $mathbf{S}$ and $mathbf{K}$ are symmetric matrices, in addition $mathbf{K}$ is a sparse symmetric matrix. $X$ is the variable. Though $mathbf{S}$ and $mathbf{K}$ are symmetric, it is not invertible in general and $mathbf{X}$ is not symmetric. Is it possible to find a closed-form solution for $mathbf{X}$ ?. Is there any relevant literature study about solving such equations ?
linear-algebra matrices matrix-equations
linear-algebra matrices matrix-equations
asked Dec 23 '18 at 19:33
ShewShew
596413
asked Dec 23 '18 at 19:33
ShewShew
596413
edited Dec 23 '18 at 20:08
Shew
asked Dec 23 '18 at 19:33
ShewShew
596413
asked Dec 23 '18 at 19:33
ShewShew
596413
asked Dec 23 '18 at 19:33
ShewShew
596413
596413
1
$begingroup$
Are all square matrices? $S,K$ are invertible?
$endgroup$
– Exodd
Dec 23 '18 at 19:45
$begingroup$
@Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
$endgroup$
– Shew
Dec 23 '18 at 20:05
add a comment |
1
$begingroup$
Are all square matrices? $S,K$ are invertible?
$endgroup$
– Exodd
Dec 23 '18 at 19:45
$begingroup$
@Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
$endgroup$
– Shew
Dec 23 '18 at 20:05
1
1
$begingroup$
Are all square matrices? $S,K$ are invertible?
$endgroup$
– Exodd
Dec 23 '18 at 19:45
$begingroup$
Are all square matrices? $S,K$ are invertible?
$endgroup$
– Exodd
Dec 23 '18 at 19:45
$begingroup$
@Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
$endgroup$
– Shew
Dec 23 '18 at 20:05
$begingroup$
@Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
$endgroup$
– Shew
Dec 23 '18 at 20:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The symmetry of $S$ and $K$ allow for you to write them as diagonalized matrices. Suppose we have
$$S=PDP^{-1},;;;;K=QCQ^{-1}$$
for diagonal matrices $C$ and $D$. We also may write $X$ as
$$X=PP^{-1}XQQ^{-1}$$
which gives us
$$PDP^{-1}XQCQ^{-1}+PP^{-1}XQQ^{-1}=Y$$
Letting $Z=P^{-1}XQ$, we have
$$PDZCQ^{-1}+PQ^{-1}Z=Y$$
or
$$DZC+Z=P^{-1}YQ$$
The matrix $DZC$ is the matrix $Z$ with rows scaled by elements of $D$ and columns scaled by elements of $C$. This leads to a system of equations that look like
$$[D]_i[Z]_{ij}[C]_j+[Z]_{ij}=[P^{-1}YQ]_{ij}$$
The existence and uniqueness of a solution can now be seen, as it is dependent upon the value of $[D]_i[C]_j$. Provided that this is not $-1$, then
$$[Z]_{ij}=frac{[P^{-1}YQ]_{ij}}{[D]_i[C]_j+1}$$
The matrix $X$ may the be recovered by $X=PZQ^{-1}$.
answered Dec 23 '18 at 20:33
Josh B.Josh B.
1,0298
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add a comment |
$begingroup$
This equation is similar to the discrete Lyapunov equation and can be solved in a similar way. Using the equality
$$ operatorname{vec}(ABC)=(C^{T} otimes A)operatorname{vec}(B) $$
one obtains the system of linear equations
$$
left( K^T otimes S+I_{n^2} right)operatorname{vec}(X)=operatorname{vec}(Y).
$$
answered Dec 23 '18 at 20:42
AVKAVK
2,1561518
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The symmetry of $S$ and $K$ allow for you to write them as diagonalized matrices. Suppose we have
$$S=PDP^{-1},;;;;K=QCQ^{-1}$$
for diagonal matrices $C$ and $D$. We also may write $X$ as
$$X=PP^{-1}XQQ^{-1}$$
which gives us
$$PDP^{-1}XQCQ^{-1}+PP^{-1}XQQ^{-1}=Y$$
Letting $Z=P^{-1}XQ$, we have
$$PDZCQ^{-1}+PQ^{-1}Z=Y$$
or
$$DZC+Z=P^{-1}YQ$$
The matrix $DZC$ is the matrix $Z$ with rows scaled by elements of $D$ and columns scaled by elements of $C$. This leads to a system of equations that look like
$$[D]_i[Z]_{ij}[C]_j+[Z]_{ij}=[P^{-1}YQ]_{ij}$$
The existence and uniqueness of a solution can now be seen, as it is dependent upon the value of $[D]_i[C]_j$. Provided that this is not $-1$, then
$$[Z]_{ij}=frac{[P^{-1}YQ]_{ij}}{[D]_i[C]_j+1}$$
The matrix $X$ may the be recovered by $X=PZQ^{-1}$.
answered Dec 23 '18 at 20:33
Josh B.Josh B.
1,0298
$endgroup$
add a comment |
$begingroup$
The symmetry of $S$ and $K$ allow for you to write them as diagonalized matrices. Suppose we have
$$S=PDP^{-1},;;;;K=QCQ^{-1}$$
for diagonal matrices $C$ and $D$. We also may write $X$ as
$$X=PP^{-1}XQQ^{-1}$$
which gives us
$$PDP^{-1}XQCQ^{-1}+PP^{-1}XQQ^{-1}=Y$$
Letting $Z=P^{-1}XQ$, we have
$$PDZCQ^{-1}+PQ^{-1}Z=Y$$
or
$$DZC+Z=P^{-1}YQ$$
The matrix $DZC$ is the matrix $Z$ with rows scaled by elements of $D$ and columns scaled by elements of $C$. This leads to a system of equations that look like
$$[D]_i[Z]_{ij}[C]_j+[Z]_{ij}=[P^{-1}YQ]_{ij}$$
The existence and uniqueness of a solution can now be seen, as it is dependent upon the value of $[D]_i[C]_j$. Provided that this is not $-1$, then
$$[Z]_{ij}=frac{[P^{-1}YQ]_{ij}}{[D]_i[C]_j+1}$$
The matrix $X$ may the be recovered by $X=PZQ^{-1}$.
answered Dec 23 '18 at 20:33
Josh B.Josh B.
1,0298
$endgroup$
add a comment |
$begingroup$
The symmetry of $S$ and $K$ allow for you to write them as diagonalized matrices. Suppose we have
$$S=PDP^{-1},;;;;K=QCQ^{-1}$$
for diagonal matrices $C$ and $D$. We also may write $X$ as
$$X=PP^{-1}XQQ^{-1}$$
which gives us
$$PDP^{-1}XQCQ^{-1}+PP^{-1}XQQ^{-1}=Y$$
Letting $Z=P^{-1}XQ$, we have
$$PDZCQ^{-1}+PQ^{-1}Z=Y$$
or
$$DZC+Z=P^{-1}YQ$$
The matrix $DZC$ is the matrix $Z$ with rows scaled by elements of $D$ and columns scaled by elements of $C$. This leads to a system of equations that look like
$$[D]_i[Z]_{ij}[C]_j+[Z]_{ij}=[P^{-1}YQ]_{ij}$$
The existence and uniqueness of a solution can now be seen, as it is dependent upon the value of $[D]_i[C]_j$. Provided that this is not $-1$, then
$$[Z]_{ij}=frac{[P^{-1}YQ]_{ij}}{[D]_i[C]_j+1}$$
The matrix $X$ may the be recovered by $X=PZQ^{-1}$.
answered Dec 23 '18 at 20:33
Josh B.Josh B.
1,0298
$endgroup$
The symmetry of $S$ and $K$ allow for you to write them as diagonalized matrices. Suppose we have
$$S=PDP^{-1},;;;;K=QCQ^{-1}$$
for diagonal matrices $C$ and $D$. We also may write $X$ as
$$X=PP^{-1}XQQ^{-1}$$
which gives us
$$PDP^{-1}XQCQ^{-1}+PP^{-1}XQQ^{-1}=Y$$
Letting $Z=P^{-1}XQ$, we have
$$PDZCQ^{-1}+PQ^{-1}Z=Y$$
or
$$DZC+Z=P^{-1}YQ$$
The matrix $DZC$ is the matrix $Z$ with rows scaled by elements of $D$ and columns scaled by elements of $C$. This leads to a system of equations that look like
$$[D]_i[Z]_{ij}[C]_j+[Z]_{ij}=[P^{-1}YQ]_{ij}$$
The existence and uniqueness of a solution can now be seen, as it is dependent upon the value of $[D]_i[C]_j$. Provided that this is not $-1$, then
$$[Z]_{ij}=frac{[P^{-1}YQ]_{ij}}{[D]_i[C]_j+1}$$
The matrix $X$ may the be recovered by $X=PZQ^{-1}$.
answered Dec 23 '18 at 20:33
Josh B.Josh B.
1,0298
answered Dec 23 '18 at 20:33
Josh B.Josh B.
1,0298
answered Dec 23 '18 at 20:33
Josh B.Josh B.
1,0298
answered Dec 23 '18 at 20:33
Josh B.Josh B.
1,0298
1,0298
add a comment |
add a comment |
$begingroup$
This equation is similar to the discrete Lyapunov equation and can be solved in a similar way. Using the equality
$$ operatorname{vec}(ABC)=(C^{T} otimes A)operatorname{vec}(B) $$
one obtains the system of linear equations
$$
left( K^T otimes S+I_{n^2} right)operatorname{vec}(X)=operatorname{vec}(Y).
$$
answered Dec 23 '18 at 20:42
AVKAVK
2,1561518
$endgroup$
add a comment |
$begingroup$
This equation is similar to the discrete Lyapunov equation and can be solved in a similar way. Using the equality
$$ operatorname{vec}(ABC)=(C^{T} otimes A)operatorname{vec}(B) $$
one obtains the system of linear equations
$$
left( K^T otimes S+I_{n^2} right)operatorname{vec}(X)=operatorname{vec}(Y).
$$
answered Dec 23 '18 at 20:42
AVKAVK
2,1561518
$endgroup$
add a comment |
$begingroup$
This equation is similar to the discrete Lyapunov equation and can be solved in a similar way. Using the equality
$$ operatorname{vec}(ABC)=(C^{T} otimes A)operatorname{vec}(B) $$
one obtains the system of linear equations
$$
left( K^T otimes S+I_{n^2} right)operatorname{vec}(X)=operatorname{vec}(Y).
$$
answered Dec 23 '18 at 20:42
AVKAVK
2,1561518
$endgroup$
This equation is similar to the discrete Lyapunov equation and can be solved in a similar way. Using the equality
$$ operatorname{vec}(ABC)=(C^{T} otimes A)operatorname{vec}(B) $$
one obtains the system of linear equations
$$
left( K^T otimes S+I_{n^2} right)operatorname{vec}(X)=operatorname{vec}(Y).
$$
answered Dec 23 '18 at 20:42
AVKAVK
2,1561518
answered Dec 23 '18 at 20:42
AVKAVK
2,1561518
answered Dec 23 '18 at 20:42
AVKAVK
2,1561518
answered Dec 23 '18 at 20:42
AVKAVK
2,1561518
2,1561518
add a comment |
add a comment |
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1
$begingroup$
Are all square matrices? $S,K$ are invertible?
$endgroup$
– Exodd
Dec 23 '18 at 19:45
$begingroup$
@Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
$endgroup$
– Shew
Dec 23 '18 at 20:05