How to solve $A^{frac 12} B A^{frac 12} = C$ for $A$?Is the following product of matrices...
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How to solve $A^{frac 12} B A^{frac 12} = C$ for $A$?
Is the following product of matrices symmetric?Non-Symmetric Positive Definite MatricesAre only symmetric matrices definite?Help with showing that the following is positive definitePositive definite and block matrixSingly indexed basis for symmetric matricesMatrix equation with symmetric and positive definite matricesFor what kind of matrix $A$, there is a (symmetric) positive definite matrix $B$ such that $BA$ is symmetricPositive definite squareshow to prove the following linear matrix inequality
$begingroup$
Suppose that matrices $A,B,C$ are symmetric and positive definite. Then, $A$ has a unique, positive square root, which we call $A^{frac 12}$. If $$A^{frac 12} B A^{frac 12} = C$$ then can we write an expression for $A$ in terms of $B, C$?
matrices matrix-equations positive-definite symmetric-matrices
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
Greg Ver SteegGreg Ver Steeg
1625
$endgroup$
add a comment |
$begingroup$
Suppose that matrices $A,B,C$ are symmetric and positive definite. Then, $A$ has a unique, positive square root, which we call $A^{frac 12}$. If $$A^{frac 12} B A^{frac 12} = C$$ then can we write an expression for $A$ in terms of $B, C$?
matrices matrix-equations positive-definite symmetric-matrices
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
Greg Ver SteegGreg Ver Steeg
1625
$endgroup$
2
$begingroup$
Note that this equation is nonlinear in $A$ (or any function of $A$), so existence of a closed expression for the general solution seems unlikely.
$endgroup$
– Blazej
2 days ago
$begingroup$
Let me mention that it is easy to give a solution when $B$ commutes with $C$ (just work in the common eigenbasis), but it's not clear to me whether this is the only solution.
$endgroup$
– Blazej
2 days ago
$begingroup$
Good point... in a way, the square root is superfluous. We could ask if $A B A = C$ has a solution for $A$. This is a quadratic equation in $A$... for scalars we have closed form solutions for quadratic equations, but do we have them for matrices?
$endgroup$
– Greg Ver Steeg
2 days ago
$begingroup$
Thanks for getting me to look at it this way. This gives a searchable keyword with papers like this. I believe this paper has the solution, which differs from the proposed ones so far. I will add this solution when I understand it (or I'll happily accept the solution if someone else explains it).
$endgroup$
– Greg Ver Steeg
2 days ago
$begingroup$
Another keyword comment for future searches, this is related to the "matrix Riccati equation".
$endgroup$
– Greg Ver Steeg
2 days ago
add a comment |
$begingroup$
Suppose that matrices $A,B,C$ are symmetric and positive definite. Then, $A$ has a unique, positive square root, which we call $A^{frac 12}$. If $$A^{frac 12} B A^{frac 12} = C$$ then can we write an expression for $A$ in terms of $B, C$?
matrices matrix-equations positive-definite symmetric-matrices
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
Greg Ver SteegGreg Ver Steeg
1625
$endgroup$
Suppose that matrices $A,B,C$ are symmetric and positive definite. Then, $A$ has a unique, positive square root, which we call $A^{frac 12}$. If $$A^{frac 12} B A^{frac 12} = C$$ then can we write an expression for $A$ in terms of $B, C$?
matrices matrix-equations positive-definite symmetric-matrices
matrices matrix-equations positive-definite symmetric-matrices
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
Greg Ver SteegGreg Ver Steeg
1625
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
Greg Ver SteegGreg Ver Steeg
1625
edited 2 days ago
Rodrigo de Azevedo
13k41960
edited 2 days ago
Rodrigo de Azevedo
13k41960
edited 2 days ago
Rodrigo de Azevedo
13k41960
13k41960
asked 2 days ago
Greg Ver SteegGreg Ver Steeg
1625
asked 2 days ago
Greg Ver SteegGreg Ver Steeg
1625
asked 2 days ago
Greg Ver SteegGreg Ver Steeg
1625
1625
2
$begingroup$
Note that this equation is nonlinear in $A$ (or any function of $A$), so existence of a closed expression for the general solution seems unlikely.
$endgroup$
– Blazej
2 days ago
$begingroup$
Let me mention that it is easy to give a solution when $B$ commutes with $C$ (just work in the common eigenbasis), but it's not clear to me whether this is the only solution.
$endgroup$
– Blazej
2 days ago
$begingroup$
Good point... in a way, the square root is superfluous. We could ask if $A B A = C$ has a solution for $A$. This is a quadratic equation in $A$... for scalars we have closed form solutions for quadratic equations, but do we have them for matrices?
$endgroup$
– Greg Ver Steeg
2 days ago
$begingroup$
Thanks for getting me to look at it this way. This gives a searchable keyword with papers like this. I believe this paper has the solution, which differs from the proposed ones so far. I will add this solution when I understand it (or I'll happily accept the solution if someone else explains it).
$endgroup$
– Greg Ver Steeg
2 days ago
$begingroup$
Another keyword comment for future searches, this is related to the "matrix Riccati equation".
$endgroup$
– Greg Ver Steeg
2 days ago
add a comment |
2
$begingroup$
Note that this equation is nonlinear in $A$ (or any function of $A$), so existence of a closed expression for the general solution seems unlikely.
$endgroup$
– Blazej
2 days ago
$begingroup$
Let me mention that it is easy to give a solution when $B$ commutes with $C$ (just work in the common eigenbasis), but it's not clear to me whether this is the only solution.
$endgroup$
– Blazej
2 days ago
$begingroup$
Good point... in a way, the square root is superfluous. We could ask if $A B A = C$ has a solution for $A$. This is a quadratic equation in $A$... for scalars we have closed form solutions for quadratic equations, but do we have them for matrices?
$endgroup$
– Greg Ver Steeg
2 days ago
$begingroup$
Thanks for getting me to look at it this way. This gives a searchable keyword with papers like this. I believe this paper has the solution, which differs from the proposed ones so far. I will add this solution when I understand it (or I'll happily accept the solution if someone else explains it).
$endgroup$
– Greg Ver Steeg
2 days ago
$begingroup$
Another keyword comment for future searches, this is related to the "matrix Riccati equation".
$endgroup$
– Greg Ver Steeg
2 days ago
2
2
$begingroup$
Note that this equation is nonlinear in $A$ (or any function of $A$), so existence of a closed expression for the general solution seems unlikely.
$endgroup$
– Blazej
2 days ago
$begingroup$
Note that this equation is nonlinear in $A$ (or any function of $A$), so existence of a closed expression for the general solution seems unlikely.
$endgroup$
– Blazej
2 days ago
$begingroup$
Let me mention that it is easy to give a solution when $B$ commutes with $C$ (just work in the common eigenbasis), but it's not clear to me whether this is the only solution.
$endgroup$
– Blazej
2 days ago
$begingroup$
Let me mention that it is easy to give a solution when $B$ commutes with $C$ (just work in the common eigenbasis), but it's not clear to me whether this is the only solution.
$endgroup$
– Blazej
2 days ago
$begingroup$
Good point... in a way, the square root is superfluous. We could ask if $A B A = C$ has a solution for $A$. This is a quadratic equation in $A$... for scalars we have closed form solutions for quadratic equations, but do we have them for matrices?
$endgroup$
– Greg Ver Steeg
2 days ago
$begingroup$
Good point... in a way, the square root is superfluous. We could ask if $A B A = C$ has a solution for $A$. This is a quadratic equation in $A$... for scalars we have closed form solutions for quadratic equations, but do we have them for matrices?
$endgroup$
– Greg Ver Steeg
2 days ago
$begingroup$
Thanks for getting me to look at it this way. This gives a searchable keyword with papers like this. I believe this paper has the solution, which differs from the proposed ones so far. I will add this solution when I understand it (or I'll happily accept the solution if someone else explains it).
$endgroup$
– Greg Ver Steeg
2 days ago
$begingroup$
Thanks for getting me to look at it this way. This gives a searchable keyword with papers like this. I believe this paper has the solution, which differs from the proposed ones so far. I will add this solution when I understand it (or I'll happily accept the solution if someone else explains it).
$endgroup$
– Greg Ver Steeg
2 days ago
$begingroup$
Another keyword comment for future searches, this is related to the "matrix Riccati equation".
$endgroup$
– Greg Ver Steeg
2 days ago
$begingroup$
Another keyword comment for future searches, this is related to the "matrix Riccati equation".
$endgroup$
– Greg Ver Steeg
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The problem is: find the solutions of the equation
(*) $XBX=C$ where the unknown $X$ and the given matrices $B,C$ are $ntimes n$ symmetric $>0$.
$textbf{Proposition.}$ (*) has the unique solution
$X=B^{-1/2}S^{1/2}B^{-1/2}$, where $S=B^{1/2}CB^{1/2}$.
$textbf{Proof}.$ (*) is equivalent to $(XB)^2=CB$, that is,
$(XB)^2=B^{-1/2}SB^{1/2}$, where $S$ is symmetric $>0$.
We put $XB=B^{-1/2}ZB^{1/2}$ where $Z^2=S$.
Then $Z=B^{1/2}XB^{1/2}$ is symmetric $>0$ and $Z=S^{1/2}$.
Finally $X=B^{-1/2}S^{1/2}B^{-1/2}$ as required. $square$
answered 2 days ago
loup blancloup blanc
23.6k21851
$endgroup$
add a comment |
$begingroup$
Since the matrices are symmetric and positive definite, you shall be able and express
$B$ and $C$ as
$$
B = U;Sigma ;overline U quad C = V;Lambda ;overline V = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} ;Sigma ;overline {left( {Lambda Sigma ^{, - 1} } right)^{,1/2} } ;overline V
$$
where the overbar indicates the transpose, and $Sigma, ;Lambda$ are diagonal.
Thereupon you get
$$
eqalign{
& A^{,1/2} ,U;Sigma ;overline U ,A^{,1/2} = A^{,1/2} ,U;Sigma ;overline U ,overline {A^{,1/2} } = cr
& = left( {A^{,1/2} ,U} right);Sigma ;overline {left( {A^{,1/2} ,U} right)} = V;Lambda ;overline V = cr
& = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} ;Sigma ;overline {left( {Lambda Sigma ^{, - 1} } right)^{,1/2} } cr}
$$
i.e.:
$$
A^{,1/2} = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} U^{, - 1}
$$
But from here I don't see a way to express $A^{1/2}$ directly in function of $B$ and $C$.
answered 2 days ago
G CabG Cab
19.9k31340
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem is: find the solutions of the equation
(*) $XBX=C$ where the unknown $X$ and the given matrices $B,C$ are $ntimes n$ symmetric $>0$.
$textbf{Proposition.}$ (*) has the unique solution
$X=B^{-1/2}S^{1/2}B^{-1/2}$, where $S=B^{1/2}CB^{1/2}$.
$textbf{Proof}.$ (*) is equivalent to $(XB)^2=CB$, that is,
$(XB)^2=B^{-1/2}SB^{1/2}$, where $S$ is symmetric $>0$.
We put $XB=B^{-1/2}ZB^{1/2}$ where $Z^2=S$.
Then $Z=B^{1/2}XB^{1/2}$ is symmetric $>0$ and $Z=S^{1/2}$.
Finally $X=B^{-1/2}S^{1/2}B^{-1/2}$ as required. $square$
answered 2 days ago
loup blancloup blanc
23.6k21851
$endgroup$
add a comment |
$begingroup$
The problem is: find the solutions of the equation
(*) $XBX=C$ where the unknown $X$ and the given matrices $B,C$ are $ntimes n$ symmetric $>0$.
$textbf{Proposition.}$ (*) has the unique solution
$X=B^{-1/2}S^{1/2}B^{-1/2}$, where $S=B^{1/2}CB^{1/2}$.
$textbf{Proof}.$ (*) is equivalent to $(XB)^2=CB$, that is,
$(XB)^2=B^{-1/2}SB^{1/2}$, where $S$ is symmetric $>0$.
We put $XB=B^{-1/2}ZB^{1/2}$ where $Z^2=S$.
Then $Z=B^{1/2}XB^{1/2}$ is symmetric $>0$ and $Z=S^{1/2}$.
Finally $X=B^{-1/2}S^{1/2}B^{-1/2}$ as required. $square$
answered 2 days ago
loup blancloup blanc
23.6k21851
$endgroup$
add a comment |
$begingroup$
The problem is: find the solutions of the equation
(*) $XBX=C$ where the unknown $X$ and the given matrices $B,C$ are $ntimes n$ symmetric $>0$.
$textbf{Proposition.}$ (*) has the unique solution
$X=B^{-1/2}S^{1/2}B^{-1/2}$, where $S=B^{1/2}CB^{1/2}$.
$textbf{Proof}.$ (*) is equivalent to $(XB)^2=CB$, that is,
$(XB)^2=B^{-1/2}SB^{1/2}$, where $S$ is symmetric $>0$.
We put $XB=B^{-1/2}ZB^{1/2}$ where $Z^2=S$.
Then $Z=B^{1/2}XB^{1/2}$ is symmetric $>0$ and $Z=S^{1/2}$.
Finally $X=B^{-1/2}S^{1/2}B^{-1/2}$ as required. $square$
answered 2 days ago
loup blancloup blanc
23.6k21851
$endgroup$
The problem is: find the solutions of the equation
(*) $XBX=C$ where the unknown $X$ and the given matrices $B,C$ are $ntimes n$ symmetric $>0$.
$textbf{Proposition.}$ (*) has the unique solution
$X=B^{-1/2}S^{1/2}B^{-1/2}$, where $S=B^{1/2}CB^{1/2}$.
$textbf{Proof}.$ (*) is equivalent to $(XB)^2=CB$, that is,
$(XB)^2=B^{-1/2}SB^{1/2}$, where $S$ is symmetric $>0$.
We put $XB=B^{-1/2}ZB^{1/2}$ where $Z^2=S$.
Then $Z=B^{1/2}XB^{1/2}$ is symmetric $>0$ and $Z=S^{1/2}$.
Finally $X=B^{-1/2}S^{1/2}B^{-1/2}$ as required. $square$
answered 2 days ago
loup blancloup blanc
23.6k21851
answered 2 days ago
loup blancloup blanc
23.6k21851
answered 2 days ago
loup blancloup blanc
23.6k21851
answered 2 days ago
loup blancloup blanc
23.6k21851
23.6k21851
add a comment |
add a comment |
$begingroup$
Since the matrices are symmetric and positive definite, you shall be able and express
$B$ and $C$ as
$$
B = U;Sigma ;overline U quad C = V;Lambda ;overline V = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} ;Sigma ;overline {left( {Lambda Sigma ^{, - 1} } right)^{,1/2} } ;overline V
$$
where the overbar indicates the transpose, and $Sigma, ;Lambda$ are diagonal.
Thereupon you get
$$
eqalign{
& A^{,1/2} ,U;Sigma ;overline U ,A^{,1/2} = A^{,1/2} ,U;Sigma ;overline U ,overline {A^{,1/2} } = cr
& = left( {A^{,1/2} ,U} right);Sigma ;overline {left( {A^{,1/2} ,U} right)} = V;Lambda ;overline V = cr
& = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} ;Sigma ;overline {left( {Lambda Sigma ^{, - 1} } right)^{,1/2} } cr}
$$
i.e.:
$$
A^{,1/2} = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} U^{, - 1}
$$
But from here I don't see a way to express $A^{1/2}$ directly in function of $B$ and $C$.
answered 2 days ago
G CabG Cab
19.9k31340
$endgroup$
add a comment |
$begingroup$
Since the matrices are symmetric and positive definite, you shall be able and express
$B$ and $C$ as
$$
B = U;Sigma ;overline U quad C = V;Lambda ;overline V = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} ;Sigma ;overline {left( {Lambda Sigma ^{, - 1} } right)^{,1/2} } ;overline V
$$
where the overbar indicates the transpose, and $Sigma, ;Lambda$ are diagonal.
Thereupon you get
$$
eqalign{
& A^{,1/2} ,U;Sigma ;overline U ,A^{,1/2} = A^{,1/2} ,U;Sigma ;overline U ,overline {A^{,1/2} } = cr
& = left( {A^{,1/2} ,U} right);Sigma ;overline {left( {A^{,1/2} ,U} right)} = V;Lambda ;overline V = cr
& = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} ;Sigma ;overline {left( {Lambda Sigma ^{, - 1} } right)^{,1/2} } cr}
$$
i.e.:
$$
A^{,1/2} = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} U^{, - 1}
$$
But from here I don't see a way to express $A^{1/2}$ directly in function of $B$ and $C$.
answered 2 days ago
G CabG Cab
19.9k31340
$endgroup$
add a comment |
$begingroup$
Since the matrices are symmetric and positive definite, you shall be able and express
$B$ and $C$ as
$$
B = U;Sigma ;overline U quad C = V;Lambda ;overline V = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} ;Sigma ;overline {left( {Lambda Sigma ^{, - 1} } right)^{,1/2} } ;overline V
$$
where the overbar indicates the transpose, and $Sigma, ;Lambda$ are diagonal.
Thereupon you get
$$
eqalign{
& A^{,1/2} ,U;Sigma ;overline U ,A^{,1/2} = A^{,1/2} ,U;Sigma ;overline U ,overline {A^{,1/2} } = cr
& = left( {A^{,1/2} ,U} right);Sigma ;overline {left( {A^{,1/2} ,U} right)} = V;Lambda ;overline V = cr
& = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} ;Sigma ;overline {left( {Lambda Sigma ^{, - 1} } right)^{,1/2} } cr}
$$
i.e.:
$$
A^{,1/2} = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} U^{, - 1}
$$
But from here I don't see a way to express $A^{1/2}$ directly in function of $B$ and $C$.
answered 2 days ago
G CabG Cab
19.9k31340
$endgroup$
Since the matrices are symmetric and positive definite, you shall be able and express
$B$ and $C$ as
$$
B = U;Sigma ;overline U quad C = V;Lambda ;overline V = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} ;Sigma ;overline {left( {Lambda Sigma ^{, - 1} } right)^{,1/2} } ;overline V
$$
where the overbar indicates the transpose, and $Sigma, ;Lambda$ are diagonal.
Thereupon you get
$$
eqalign{
& A^{,1/2} ,U;Sigma ;overline U ,A^{,1/2} = A^{,1/2} ,U;Sigma ;overline U ,overline {A^{,1/2} } = cr
& = left( {A^{,1/2} ,U} right);Sigma ;overline {left( {A^{,1/2} ,U} right)} = V;Lambda ;overline V = cr
& = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} ;Sigma ;overline {left( {Lambda Sigma ^{, - 1} } right)^{,1/2} } cr}
$$
i.e.:
$$
A^{,1/2} = V;left( {Lambda Sigma ^{, - 1} } right)^{,1/2} U^{, - 1}
$$
But from here I don't see a way to express $A^{1/2}$ directly in function of $B$ and $C$.
answered 2 days ago
G CabG Cab
19.9k31340
answered 2 days ago
G CabG Cab
19.9k31340
answered 2 days ago
G CabG Cab
19.9k31340
answered 2 days ago
G CabG Cab
19.9k31340
19.9k31340
add a comment |
add a comment |
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2
$begingroup$
Note that this equation is nonlinear in $A$ (or any function of $A$), so existence of a closed expression for the general solution seems unlikely.
$endgroup$
– Blazej
2 days ago
$begingroup$
Let me mention that it is easy to give a solution when $B$ commutes with $C$ (just work in the common eigenbasis), but it's not clear to me whether this is the only solution.
$endgroup$
– Blazej
2 days ago
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Good point... in a way, the square root is superfluous. We could ask if $A B A = C$ has a solution for $A$. This is a quadratic equation in $A$... for scalars we have closed form solutions for quadratic equations, but do we have them for matrices?
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– Greg Ver Steeg
2 days ago
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Thanks for getting me to look at it this way. This gives a searchable keyword with papers like this. I believe this paper has the solution, which differs from the proposed ones so far. I will add this solution when I understand it (or I'll happily accept the solution if someone else explains it).
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– Greg Ver Steeg
2 days ago
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Another keyword comment for future searches, this is related to the "matrix Riccati equation".
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– Greg Ver Steeg
2 days ago