Solve $A = C B C^t$ for $B$Find $C$, if $A=CBC$, where $A$,$B$,$C$ are symmetric matrices.Is symmetry a necessary condition for positive (or negative) definiteness?Variant of Cholesky Decomposition: solve $B^TB=A$ for general square matrix $A$Finding Rank of the following MatrixProve that a square matrix $A$ is positive definite if and only if $A$+ $A^T$ is positive definiteFinding an orthogonal matrix for a 3x2How to solve for square matrix $y$ given $x = A'yA$ for $A'A = I$Find Condition on the matrix $X in mathbbR^n times m$ where $mleq n$Necessary condition for a matrix equation to be zeroPrincipal-Minors-Test for Positivity of Complex Hermitian MatricesHow to solve for unknown matrix?
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Solve $A = C B C^t$ for $B$
Find $C$, if $A=CBC$, where $A$,$B$,$C$ are symmetric matrices.Is symmetry a necessary condition for positive (or negative) definiteness?Variant of Cholesky Decomposition: solve $B^TB=A$ for general square matrix $A$Finding Rank of the following MatrixProve that a square matrix $A$ is positive definite if and only if $A$+ $A^T$ is positive definiteFinding an orthogonal matrix for a 3x2How to solve for square matrix $y$ given $x = A'yA$ for $A'A = I$Find Condition on the matrix $X in mathbbR^n times m$ where $mleq n$Necessary condition for a matrix equation to be zeroPrincipal-Minors-Test for Positivity of Complex Hermitian MatricesHow to solve for unknown matrix?
$begingroup$
I know this question, but I would like to know the middle square matrix $B$.
Given positive definite matrix $A in mathbb R^2 times 2$ and non-zero matrix $C in mathbb R^2 times 3$, find $3 times3$ matrix $B$
$$A = C B C^t$$
matrices systems-of-equations matrix-equations
edited Mar 21 at 7:28
Rodrigo de Azevedo
13.1k41960
asked Mar 21 at 6:33
JimSDJimSD
114
$endgroup$
|
show 5 more comments
$begingroup$
I know this question, but I would like to know the middle square matrix $B$.
Given positive definite matrix $A in mathbb R^2 times 2$ and non-zero matrix $C in mathbb R^2 times 3$, find $3 times3$ matrix $B$
$$A = C B C^t$$
matrices systems-of-equations matrix-equations
edited Mar 21 at 7:28
Rodrigo de Azevedo
13.1k41960
asked Mar 21 at 6:33
JimSDJimSD
114
$endgroup$
$begingroup$
Are A, B and C real or complex matrices?
$endgroup$
– PerelMan
Mar 21 at 6:37
$begingroup$
yes, real only, just simplest case...
$endgroup$
– JimSD
Mar 21 at 6:37
$begingroup$
$CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
$endgroup$
– Saucy O'Path
Mar 21 at 6:38
1
$begingroup$
This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 6:54
1
$begingroup$
@JimSD Then un-vectorize!! In MATLAB, usereshapeto do both vectorization and un-vectorization.
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 7:25
|
show 5 more comments
$begingroup$
I know this question, but I would like to know the middle square matrix $B$.
Given positive definite matrix $A in mathbb R^2 times 2$ and non-zero matrix $C in mathbb R^2 times 3$, find $3 times3$ matrix $B$
$$A = C B C^t$$
matrices systems-of-equations matrix-equations
edited Mar 21 at 7:28
Rodrigo de Azevedo
13.1k41960
asked Mar 21 at 6:33
JimSDJimSD
114
$endgroup$
I know this question, but I would like to know the middle square matrix $B$.
Given positive definite matrix $A in mathbb R^2 times 2$ and non-zero matrix $C in mathbb R^2 times 3$, find $3 times3$ matrix $B$
$$A = C B C^t$$
matrices systems-of-equations matrix-equations
matrices systems-of-equations matrix-equations
edited Mar 21 at 7:28
Rodrigo de Azevedo
13.1k41960
asked Mar 21 at 6:33
JimSDJimSD
114
edited Mar 21 at 7:28
Rodrigo de Azevedo
13.1k41960
asked Mar 21 at 6:33
JimSDJimSD
114
edited Mar 21 at 7:28
Rodrigo de Azevedo
13.1k41960
edited Mar 21 at 7:28
Rodrigo de Azevedo
13.1k41960
edited Mar 21 at 7:28
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Mar 21 at 6:33
JimSDJimSD
114
asked Mar 21 at 6:33
JimSDJimSD
114
asked Mar 21 at 6:33
JimSDJimSD
114
114
$begingroup$
Are A, B and C real or complex matrices?
$endgroup$
– PerelMan
Mar 21 at 6:37
$begingroup$
yes, real only, just simplest case...
$endgroup$
– JimSD
Mar 21 at 6:37
$begingroup$
$CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
$endgroup$
– Saucy O'Path
Mar 21 at 6:38
1
$begingroup$
This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 6:54
1
$begingroup$
@JimSD Then un-vectorize!! In MATLAB, usereshapeto do both vectorization and un-vectorization.
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 7:25
|
show 5 more comments
$begingroup$
Are A, B and C real or complex matrices?
$endgroup$
– PerelMan
Mar 21 at 6:37
$begingroup$
yes, real only, just simplest case...
$endgroup$
– JimSD
Mar 21 at 6:37
$begingroup$
$CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
$endgroup$
– Saucy O'Path
Mar 21 at 6:38
1
$begingroup$
This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 6:54
1
$begingroup$
@JimSD Then un-vectorize!! In MATLAB, usereshapeto do both vectorization and un-vectorization.
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 7:25
$begingroup$
Are A, B and C real or complex matrices?
$endgroup$
– PerelMan
Mar 21 at 6:37
$begingroup$
Are A, B and C real or complex matrices?
$endgroup$
– PerelMan
Mar 21 at 6:37
$begingroup$
yes, real only, just simplest case...
$endgroup$
– JimSD
Mar 21 at 6:37
$begingroup$
yes, real only, just simplest case...
$endgroup$
– JimSD
Mar 21 at 6:37
$begingroup$
$CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
$endgroup$
– Saucy O'Path
Mar 21 at 6:38
$begingroup$
$CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
$endgroup$
– Saucy O'Path
Mar 21 at 6:38
1
1
$begingroup$
This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 6:54
$begingroup$
This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 6:54
1
1
$begingroup$
@JimSD Then un-vectorize!! In MATLAB, use
reshape to do both vectorization and un-vectorization.$endgroup$
– Rodrigo de Azevedo
Mar 21 at 7:25
$begingroup$
@JimSD Then un-vectorize!! In MATLAB, use
reshape to do both vectorization and un-vectorization.$endgroup$
– Rodrigo de Azevedo
Mar 21 at 7:25
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let (*) $A=CBC^T$. Since $A$ is sym. $>0$, necessarily $rank(C)=2$ ( otherwise, there are no solutions in $B$).
Then $C^+=C^T(CC^T)^-1$ ( cf. the Moore Penrose inverse of $C$) and $CC^+=I_2$. Moreover, $S=C^+C$ is sym. and similar to $diag(1,1,0)$. Up to an orthonormal change of basis, we may assume that $S=C^+C=diag(1,1,0),C=[D_2,2,0_2,1],C^+=beginpmatrixD^-1\0_1,2endpmatrix$.
(*) is equivalent to $C^+A(C^+)^T=SBS$ or
$beginpmatrixD^-1A(D^-1)^T&0\0&0endpmatrix=beginpmatrixa&b&0\d&e&0\0&0&0endpmatrix$ where the unknown $B=beginpmatrixa&b&c\d&e&f\g&h&pendpmatrix$.
Note that $U=[u_i,j]=D^-1A(D^-1)^T$ is a known sym. $>0$ matrix. Then $B$ is a solution of (*) iff the following conditions are satisfied
$a=u_1,1,e=u_2,2,b=d=u_1,2$; $g,h,p,c,f$ are $5=9-4$ arbitrary parameters.
answered Mar 22 at 11:58
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
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$begingroup$
Let (*) $A=CBC^T$. Since $A$ is sym. $>0$, necessarily $rank(C)=2$ ( otherwise, there are no solutions in $B$).
Then $C^+=C^T(CC^T)^-1$ ( cf. the Moore Penrose inverse of $C$) and $CC^+=I_2$. Moreover, $S=C^+C$ is sym. and similar to $diag(1,1,0)$. Up to an orthonormal change of basis, we may assume that $S=C^+C=diag(1,1,0),C=[D_2,2,0_2,1],C^+=beginpmatrixD^-1\0_1,2endpmatrix$.
(*) is equivalent to $C^+A(C^+)^T=SBS$ or
$beginpmatrixD^-1A(D^-1)^T&0\0&0endpmatrix=beginpmatrixa&b&0\d&e&0\0&0&0endpmatrix$ where the unknown $B=beginpmatrixa&b&c\d&e&f\g&h&pendpmatrix$.
Note that $U=[u_i,j]=D^-1A(D^-1)^T$ is a known sym. $>0$ matrix. Then $B$ is a solution of (*) iff the following conditions are satisfied
$a=u_1,1,e=u_2,2,b=d=u_1,2$; $g,h,p,c,f$ are $5=9-4$ arbitrary parameters.
answered Mar 22 at 11:58
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
$begingroup$
Let (*) $A=CBC^T$. Since $A$ is sym. $>0$, necessarily $rank(C)=2$ ( otherwise, there are no solutions in $B$).
Then $C^+=C^T(CC^T)^-1$ ( cf. the Moore Penrose inverse of $C$) and $CC^+=I_2$. Moreover, $S=C^+C$ is sym. and similar to $diag(1,1,0)$. Up to an orthonormal change of basis, we may assume that $S=C^+C=diag(1,1,0),C=[D_2,2,0_2,1],C^+=beginpmatrixD^-1\0_1,2endpmatrix$.
(*) is equivalent to $C^+A(C^+)^T=SBS$ or
$beginpmatrixD^-1A(D^-1)^T&0\0&0endpmatrix=beginpmatrixa&b&0\d&e&0\0&0&0endpmatrix$ where the unknown $B=beginpmatrixa&b&c\d&e&f\g&h&pendpmatrix$.
Note that $U=[u_i,j]=D^-1A(D^-1)^T$ is a known sym. $>0$ matrix. Then $B$ is a solution of (*) iff the following conditions are satisfied
$a=u_1,1,e=u_2,2,b=d=u_1,2$; $g,h,p,c,f$ are $5=9-4$ arbitrary parameters.
answered Mar 22 at 11:58
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
$begingroup$
Let (*) $A=CBC^T$. Since $A$ is sym. $>0$, necessarily $rank(C)=2$ ( otherwise, there are no solutions in $B$).
Then $C^+=C^T(CC^T)^-1$ ( cf. the Moore Penrose inverse of $C$) and $CC^+=I_2$. Moreover, $S=C^+C$ is sym. and similar to $diag(1,1,0)$. Up to an orthonormal change of basis, we may assume that $S=C^+C=diag(1,1,0),C=[D_2,2,0_2,1],C^+=beginpmatrixD^-1\0_1,2endpmatrix$.
(*) is equivalent to $C^+A(C^+)^T=SBS$ or
$beginpmatrixD^-1A(D^-1)^T&0\0&0endpmatrix=beginpmatrixa&b&0\d&e&0\0&0&0endpmatrix$ where the unknown $B=beginpmatrixa&b&c\d&e&f\g&h&pendpmatrix$.
Note that $U=[u_i,j]=D^-1A(D^-1)^T$ is a known sym. $>0$ matrix. Then $B$ is a solution of (*) iff the following conditions are satisfied
$a=u_1,1,e=u_2,2,b=d=u_1,2$; $g,h,p,c,f$ are $5=9-4$ arbitrary parameters.
answered Mar 22 at 11:58
loup blancloup blanc
24.1k21851
$endgroup$
Let (*) $A=CBC^T$. Since $A$ is sym. $>0$, necessarily $rank(C)=2$ ( otherwise, there are no solutions in $B$).
Then $C^+=C^T(CC^T)^-1$ ( cf. the Moore Penrose inverse of $C$) and $CC^+=I_2$. Moreover, $S=C^+C$ is sym. and similar to $diag(1,1,0)$. Up to an orthonormal change of basis, we may assume that $S=C^+C=diag(1,1,0),C=[D_2,2,0_2,1],C^+=beginpmatrixD^-1\0_1,2endpmatrix$.
(*) is equivalent to $C^+A(C^+)^T=SBS$ or
$beginpmatrixD^-1A(D^-1)^T&0\0&0endpmatrix=beginpmatrixa&b&0\d&e&0\0&0&0endpmatrix$ where the unknown $B=beginpmatrixa&b&c\d&e&f\g&h&pendpmatrix$.
Note that $U=[u_i,j]=D^-1A(D^-1)^T$ is a known sym. $>0$ matrix. Then $B$ is a solution of (*) iff the following conditions are satisfied
$a=u_1,1,e=u_2,2,b=d=u_1,2$; $g,h,p,c,f$ are $5=9-4$ arbitrary parameters.
answered Mar 22 at 11:58
loup blancloup blanc
24.1k21851
answered Mar 22 at 11:58
loup blancloup blanc
24.1k21851
answered Mar 22 at 11:58
loup blancloup blanc
24.1k21851
answered Mar 22 at 11:58
loup blancloup blanc
24.1k21851
24.1k21851
add a comment |
add a comment |
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$begingroup$
Are A, B and C real or complex matrices?
$endgroup$
– PerelMan
Mar 21 at 6:37
$begingroup$
yes, real only, just simplest case...
$endgroup$
– JimSD
Mar 21 at 6:37
$begingroup$
$CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
$endgroup$
– Saucy O'Path
Mar 21 at 6:38
1
$begingroup$
This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 6:54
1
$begingroup$
@JimSD Then un-vectorize!! In MATLAB, use
reshapeto do both vectorization and un-vectorization.$endgroup$
– Rodrigo de Azevedo
Mar 21 at 7:25