Mixing
Solving matrix equation $C=A B A^mathrm{T}$
$begingroup$
I am writing a code on Matlab to calculate the matrix $B$, given $C$ and $A$, following the equation
$$C=A B A^mathrm{T}$$
$A^mathrm{T}$ is non-invertible, so I can't just multiply $C$ by the inverse of the matrices.
I can't solve this by hand either because $C$ is a $3144×3144$ matrix...any help will be greatly appreciated!
linear-algebra matlab
edited Jan 29 at 12:29
YuiTo Cheng
2,1862937
asked Jan 29 at 11:08
Lisa GohLisa Goh
62
$endgroup$
add a comment |
$begingroup$
I am writing a code on Matlab to calculate the matrix $B$, given $C$ and $A$, following the equation
$$C=A B A^mathrm{T}$$
$A^mathrm{T}$ is non-invertible, so I can't just multiply $C$ by the inverse of the matrices.
I can't solve this by hand either because $C$ is a $3144×3144$ matrix...any help will be greatly appreciated!
linear-algebra matlab
edited Jan 29 at 12:29
YuiTo Cheng
2,1862937
asked Jan 29 at 11:08
Lisa GohLisa Goh
62
$endgroup$
3
$begingroup$
If $A$ is not invertible, in general, there may be infinite $B$ which yield the same $C$. Give us more context, please.
$endgroup$
– Harnak
Jan 29 at 11:45
$begingroup$
Do you know the dimension of A? The rank of A?
$endgroup$
– Bertrand
Jan 29 at 12:34
$begingroup$
A is a 740*3144 matrix of the form A= 1 1 1 0 0 0 0 0 0...; 0 0 0 1 1 1 0 0 0...; 0 0 0 0 0 0 1 1 1...; ... The constraint is that C is a symmetric matrix, and B has to be a symmetric matrix too.
$endgroup$
– Lisa Goh
Jan 30 at 3:55
add a comment |
$begingroup$
I am writing a code on Matlab to calculate the matrix $B$, given $C$ and $A$, following the equation
$$C=A B A^mathrm{T}$$
$A^mathrm{T}$ is non-invertible, so I can't just multiply $C$ by the inverse of the matrices.
I can't solve this by hand either because $C$ is a $3144×3144$ matrix...any help will be greatly appreciated!
linear-algebra matlab
edited Jan 29 at 12:29
YuiTo Cheng
2,1862937
asked Jan 29 at 11:08
Lisa GohLisa Goh
62
$endgroup$
I am writing a code on Matlab to calculate the matrix $B$, given $C$ and $A$, following the equation
$$C=A B A^mathrm{T}$$
$A^mathrm{T}$ is non-invertible, so I can't just multiply $C$ by the inverse of the matrices.
I can't solve this by hand either because $C$ is a $3144×3144$ matrix...any help will be greatly appreciated!
linear-algebra matlab
linear-algebra matlab
edited Jan 29 at 12:29
YuiTo Cheng
2,1862937
asked Jan 29 at 11:08
Lisa GohLisa Goh
62
edited Jan 29 at 12:29
YuiTo Cheng
2,1862937
asked Jan 29 at 11:08
Lisa GohLisa Goh
62
edited Jan 29 at 12:29
YuiTo Cheng
2,1862937
edited Jan 29 at 12:29
YuiTo Cheng
2,1862937
edited Jan 29 at 12:29
YuiTo Cheng
2,1862937
2,1862937
asked Jan 29 at 11:08
Lisa GohLisa Goh
62
asked Jan 29 at 11:08
Lisa GohLisa Goh
62
asked Jan 29 at 11:08
Lisa GohLisa Goh
62
62
3
$begingroup$
If $A$ is not invertible, in general, there may be infinite $B$ which yield the same $C$. Give us more context, please.
$endgroup$
– Harnak
Jan 29 at 11:45
$begingroup$
Do you know the dimension of A? The rank of A?
$endgroup$
– Bertrand
Jan 29 at 12:34
$begingroup$
A is a 740*3144 matrix of the form A= 1 1 1 0 0 0 0 0 0...; 0 0 0 1 1 1 0 0 0...; 0 0 0 0 0 0 1 1 1...; ... The constraint is that C is a symmetric matrix, and B has to be a symmetric matrix too.
$endgroup$
– Lisa Goh
Jan 30 at 3:55
add a comment |
3
$begingroup$
If $A$ is not invertible, in general, there may be infinite $B$ which yield the same $C$. Give us more context, please.
$endgroup$
– Harnak
Jan 29 at 11:45
$begingroup$
Do you know the dimension of A? The rank of A?
$endgroup$
– Bertrand
Jan 29 at 12:34
$begingroup$
A is a 740*3144 matrix of the form A= 1 1 1 0 0 0 0 0 0...; 0 0 0 1 1 1 0 0 0...; 0 0 0 0 0 0 1 1 1...; ... The constraint is that C is a symmetric matrix, and B has to be a symmetric matrix too.
$endgroup$
– Lisa Goh
Jan 30 at 3:55
3
3
$begingroup$
If $A$ is not invertible, in general, there may be infinite $B$ which yield the same $C$. Give us more context, please.
$endgroup$
– Harnak
Jan 29 at 11:45
$begingroup$
If $A$ is not invertible, in general, there may be infinite $B$ which yield the same $C$. Give us more context, please.
$endgroup$
– Harnak
Jan 29 at 11:45
$begingroup$
Do you know the dimension of A? The rank of A?
$endgroup$
– Bertrand
Jan 29 at 12:34
$begingroup$
Do you know the dimension of A? The rank of A?
$endgroup$
– Bertrand
Jan 29 at 12:34
$begingroup$
A is a 740*3144 matrix of the form A= 1 1 1 0 0 0 0 0 0...; 0 0 0 1 1 1 0 0 0...; 0 0 0 0 0 0 1 1 1...; ... The constraint is that C is a symmetric matrix, and B has to be a symmetric matrix too.
$endgroup$
– Lisa Goh
Jan 30 at 3:55
$begingroup$
A is a 740*3144 matrix of the form A= 1 1 1 0 0 0 0 0 0...; 0 0 0 1 1 1 0 0 0...; 0 0 0 0 0 0 1 1 1...; ... The constraint is that C is a symmetric matrix, and B has to be a symmetric matrix too.
$endgroup$
– Lisa Goh
Jan 30 at 3:55
add a comment |
1 Answer
1
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$begingroup$
For $A(Ntimes K)$ and $rank(A)=K leq N$ we have $$B = (A^TA)^{-1}A^TCA(A^TA)^{-1}. $$
answered Jan 29 at 12:56
BertrandBertrand
45815
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1 Answer
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$begingroup$
For $A(Ntimes K)$ and $rank(A)=K leq N$ we have $$B = (A^TA)^{-1}A^TCA(A^TA)^{-1}. $$
answered Jan 29 at 12:56
BertrandBertrand
45815
$endgroup$
add a comment |
$begingroup$
For $A(Ntimes K)$ and $rank(A)=K leq N$ we have $$B = (A^TA)^{-1}A^TCA(A^TA)^{-1}. $$
answered Jan 29 at 12:56
BertrandBertrand
45815
$endgroup$
add a comment |
$begingroup$
For $A(Ntimes K)$ and $rank(A)=K leq N$ we have $$B = (A^TA)^{-1}A^TCA(A^TA)^{-1}. $$
answered Jan 29 at 12:56
BertrandBertrand
45815
$endgroup$
For $A(Ntimes K)$ and $rank(A)=K leq N$ we have $$B = (A^TA)^{-1}A^TCA(A^TA)^{-1}. $$
answered Jan 29 at 12:56
BertrandBertrand
45815
answered Jan 29 at 12:56
BertrandBertrand
45815
answered Jan 29 at 12:56
BertrandBertrand
45815
answered Jan 29 at 12:56
BertrandBertrand
45815
45815
add a comment |
add a comment |
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3
$begingroup$
If $A$ is not invertible, in general, there may be infinite $B$ which yield the same $C$. Give us more context, please.
$endgroup$
– Harnak
Jan 29 at 11:45
$begingroup$
Do you know the dimension of A? The rank of A?
$endgroup$
– Bertrand
Jan 29 at 12:34
$begingroup$
A is a 740*3144 matrix of the form A= 1 1 1 0 0 0 0 0 0...; 0 0 0 1 1 1 0 0 0...; 0 0 0 0 0 0 1 1 1...; ... The constraint is that C is a symmetric matrix, and B has to be a symmetric matrix too.
$endgroup$
– Lisa Goh
Jan 30 at 3:55