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How to solve this quadratic equation in matrix $X$?
The Next CEO of Stack OverflowIs it possible to solve for scalar in this multiplication of two quadratic forms involving inverse matrix?How to solve the optimization problem $mathbfF=argmin_mathbfF tr((mathbfG^HmathbfF^HmathbfFmathbfG+alphamathbfI)^-1)$How to solve this quadratic matrix equation with maximum eigenvalue within it?How to solve for the matrices in this equation?Upperbound with Frobenius and Euclidean NormsMinimizing quadratic objective subject to a quadratic equality constraintMatrix Notation Form of Roots of a Quadratic EquationInversion of matrix with the sum of three Kronecker productsWeighted PSD matrix minimizationConvergence analysis for Solving quadratic matrix equation $mathbfXBX=A+lambda mathbfX$
$begingroup$
Given $mathbfA$ and $mathbfB$ two $m times n$ real matrices, is there a closed form for the matrix equation
beginequation
|mathbfX|^2_F - 2 cdot mboxtrace (mathbfX^TmathbfA) +|mathbfB|^2_F = 0,
endequation
where $mathbfX in mathbbR^mtimes n$ and $| cdot |_F$ is the Frobenius norm? Notice that this equation can be viewed as a generalization of a simple quadratic equation.
matrices quadratics matrix-equations
edited Mar 19 at 20:58
Rodrigo de Azevedo
13k41960
asked Sep 22 '14 at 16:30
Alex SilvaAlex Silva
2,76031432
$endgroup$
add a comment |
$begingroup$
Given $mathbfA$ and $mathbfB$ two $m times n$ real matrices, is there a closed form for the matrix equation
beginequation
|mathbfX|^2_F - 2 cdot mboxtrace (mathbfX^TmathbfA) +|mathbfB|^2_F = 0,
endequation
where $mathbfX in mathbbR^mtimes n$ and $| cdot |_F$ is the Frobenius norm? Notice that this equation can be viewed as a generalization of a simple quadratic equation.
matrices quadratics matrix-equations
edited Mar 19 at 20:58
Rodrigo de Azevedo
13k41960
asked Sep 22 '14 at 16:30
Alex SilvaAlex Silva
2,76031432
$endgroup$
add a comment |
$begingroup$
Given $mathbfA$ and $mathbfB$ two $m times n$ real matrices, is there a closed form for the matrix equation
beginequation
|mathbfX|^2_F - 2 cdot mboxtrace (mathbfX^TmathbfA) +|mathbfB|^2_F = 0,
endequation
where $mathbfX in mathbbR^mtimes n$ and $| cdot |_F$ is the Frobenius norm? Notice that this equation can be viewed as a generalization of a simple quadratic equation.
matrices quadratics matrix-equations
edited Mar 19 at 20:58
Rodrigo de Azevedo
13k41960
asked Sep 22 '14 at 16:30
Alex SilvaAlex Silva
2,76031432
$endgroup$
Given $mathbfA$ and $mathbfB$ two $m times n$ real matrices, is there a closed form for the matrix equation
beginequation
|mathbfX|^2_F - 2 cdot mboxtrace (mathbfX^TmathbfA) +|mathbfB|^2_F = 0,
endequation
where $mathbfX in mathbbR^mtimes n$ and $| cdot |_F$ is the Frobenius norm? Notice that this equation can be viewed as a generalization of a simple quadratic equation.
matrices quadratics matrix-equations
matrices quadratics matrix-equations
edited Mar 19 at 20:58
Rodrigo de Azevedo
13k41960
asked Sep 22 '14 at 16:30
Alex SilvaAlex Silva
2,76031432
edited Mar 19 at 20:58
Rodrigo de Azevedo
13k41960
asked Sep 22 '14 at 16:30
Alex SilvaAlex Silva
2,76031432
edited Mar 19 at 20:58
Rodrigo de Azevedo
13k41960
edited Mar 19 at 20:58
Rodrigo de Azevedo
13k41960
edited Mar 19 at 20:58
Rodrigo de Azevedo
13k41960
13k41960
asked Sep 22 '14 at 16:30
Alex SilvaAlex Silva
2,76031432
asked Sep 22 '14 at 16:30
Alex SilvaAlex Silva
2,76031432
asked Sep 22 '14 at 16:30
Alex SilvaAlex Silva
2,76031432
2,76031432
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$||X||_F^2=Tr(X^TX)$. So you can rewrite your equation in the following form:
$$
Tr(X^TX-2X^TA+B^TB)=Tr(X^TX-X^TA-A^TX+A^TA+B^TB-A^TA)=Tr((X-A)^T(X-A)+B^TB-A^TA)=0
$$
So
$$
Tr(X-A)^T(X-A))=Tr(A^TA-B^TB);
$$
At least we can understand now that solution may exists only if $a=Tr(A^TA-B^TB)ge 0$.
Now all matrices such that $||Y||_F^2=a$ will generate solution as $X=Y+A$.
answered Sep 24 '14 at 19:31
Alexander VigodnerAlexander Vigodner
2,0491713
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$||X||_F^2=Tr(X^TX)$. So you can rewrite your equation in the following form:
$$
Tr(X^TX-2X^TA+B^TB)=Tr(X^TX-X^TA-A^TX+A^TA+B^TB-A^TA)=Tr((X-A)^T(X-A)+B^TB-A^TA)=0
$$
So
$$
Tr(X-A)^T(X-A))=Tr(A^TA-B^TB);
$$
At least we can understand now that solution may exists only if $a=Tr(A^TA-B^TB)ge 0$.
Now all matrices such that $||Y||_F^2=a$ will generate solution as $X=Y+A$.
answered Sep 24 '14 at 19:31
Alexander VigodnerAlexander Vigodner
2,0491713
$endgroup$
add a comment |
$begingroup$
$||X||_F^2=Tr(X^TX)$. So you can rewrite your equation in the following form:
$$
Tr(X^TX-2X^TA+B^TB)=Tr(X^TX-X^TA-A^TX+A^TA+B^TB-A^TA)=Tr((X-A)^T(X-A)+B^TB-A^TA)=0
$$
So
$$
Tr(X-A)^T(X-A))=Tr(A^TA-B^TB);
$$
At least we can understand now that solution may exists only if $a=Tr(A^TA-B^TB)ge 0$.
Now all matrices such that $||Y||_F^2=a$ will generate solution as $X=Y+A$.
answered Sep 24 '14 at 19:31
Alexander VigodnerAlexander Vigodner
2,0491713
$endgroup$
add a comment |
$begingroup$
$||X||_F^2=Tr(X^TX)$. So you can rewrite your equation in the following form:
$$
Tr(X^TX-2X^TA+B^TB)=Tr(X^TX-X^TA-A^TX+A^TA+B^TB-A^TA)=Tr((X-A)^T(X-A)+B^TB-A^TA)=0
$$
So
$$
Tr(X-A)^T(X-A))=Tr(A^TA-B^TB);
$$
At least we can understand now that solution may exists only if $a=Tr(A^TA-B^TB)ge 0$.
Now all matrices such that $||Y||_F^2=a$ will generate solution as $X=Y+A$.
answered Sep 24 '14 at 19:31
Alexander VigodnerAlexander Vigodner
2,0491713
$endgroup$
$||X||_F^2=Tr(X^TX)$. So you can rewrite your equation in the following form:
$$
Tr(X^TX-2X^TA+B^TB)=Tr(X^TX-X^TA-A^TX+A^TA+B^TB-A^TA)=Tr((X-A)^T(X-A)+B^TB-A^TA)=0
$$
So
$$
Tr(X-A)^T(X-A))=Tr(A^TA-B^TB);
$$
At least we can understand now that solution may exists only if $a=Tr(A^TA-B^TB)ge 0$.
Now all matrices such that $||Y||_F^2=a$ will generate solution as $X=Y+A$.
answered Sep 24 '14 at 19:31
Alexander VigodnerAlexander Vigodner
2,0491713
answered Sep 24 '14 at 19:31
Alexander VigodnerAlexander Vigodner
2,0491713
answered Sep 24 '14 at 19:31
Alexander VigodnerAlexander Vigodner
2,0491713
answered Sep 24 '14 at 19:31
Alexander VigodnerAlexander Vigodner
2,0491713
2,0491713
add a comment |
add a comment |
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