solving the matrix equation A*B=A*D*B for D
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I have a matrix equation like this: A * B=A * D * B
- A is 3 by 200.
- B is a 200 by 1 column vector.
- All the elements of A and B are positive (non-negative and non-zero)
- D is a 200 by 200 diagonal matrix.
- D must be a linear combination of these two matrices:
D1=diag([1,2,3,...,200]) , D2=diag([1,4,9,...,40000])
(in other words, D=a * D1 +b * D2, a and b are real numbers but can not be zero both at the same time.)
For any given A and B, I want to prove that there is no solution for the diagonal matrix D satisfying all the above conditions.
Do you have any idea about it? Any suggestion or a counter example?
If we multiply both sides with A' (A transpose) and B'( B transpose) we will have:
A' * A*B * B'=A' * A * D *B * B'
which can be written as :
P*Q=P * D * Q , P=A' * A, Q=B * B'.
P and Q are symmetric and singular matrices. Does this help?
matrix-equations
asked Nov 17 at 11:16
Maria
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up vote
-1
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I have a matrix equation like this: A * B=A * D * B
- A is 3 by 200.
- B is a 200 by 1 column vector.
- All the elements of A and B are positive (non-negative and non-zero)
- D is a 200 by 200 diagonal matrix.
- D must be a linear combination of these two matrices:
D1=diag([1,2,3,...,200]) , D2=diag([1,4,9,...,40000])
(in other words, D=a * D1 +b * D2, a and b are real numbers but can not be zero both at the same time.)
For any given A and B, I want to prove that there is no solution for the diagonal matrix D satisfying all the above conditions.
Do you have any idea about it? Any suggestion or a counter example?
If we multiply both sides with A' (A transpose) and B'( B transpose) we will have:
A' * A*B * B'=A' * A * D *B * B'
which can be written as :
P*Q=P * D * Q , P=A' * A, Q=B * B'.
P and Q are symmetric and singular matrices. Does this help?
matrix-equations
asked Nov 17 at 11:16
Maria
11
New contributor
Maria is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If you allow $0$s in the linear combination, just choose $D_1+0D_2$ and it works.
– Matt Samuel
Nov 17 at 11:28
Thank you so much @MattSamuel. Zero is not allowed. Maybe I was not clear enough so I modified the question.
– Maria
Nov 17 at 20:47
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have a matrix equation like this: A * B=A * D * B
- A is 3 by 200.
- B is a 200 by 1 column vector.
- All the elements of A and B are positive (non-negative and non-zero)
- D is a 200 by 200 diagonal matrix.
- D must be a linear combination of these two matrices:
D1=diag([1,2,3,...,200]) , D2=diag([1,4,9,...,40000])
(in other words, D=a * D1 +b * D2, a and b are real numbers but can not be zero both at the same time.)
For any given A and B, I want to prove that there is no solution for the diagonal matrix D satisfying all the above conditions.
Do you have any idea about it? Any suggestion or a counter example?
If we multiply both sides with A' (A transpose) and B'( B transpose) we will have:
A' * A*B * B'=A' * A * D *B * B'
which can be written as :
P*Q=P * D * Q , P=A' * A, Q=B * B'.
P and Q are symmetric and singular matrices. Does this help?
matrix-equations
asked Nov 17 at 11:16
Maria
11
New contributor
Maria is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a matrix equation like this: A * B=A * D * B
- A is 3 by 200.
- B is a 200 by 1 column vector.
- All the elements of A and B are positive (non-negative and non-zero)
- D is a 200 by 200 diagonal matrix.
- D must be a linear combination of these two matrices:
D1=diag([1,2,3,...,200]) , D2=diag([1,4,9,...,40000])
(in other words, D=a * D1 +b * D2, a and b are real numbers but can not be zero both at the same time.)
For any given A and B, I want to prove that there is no solution for the diagonal matrix D satisfying all the above conditions.
Do you have any idea about it? Any suggestion or a counter example?
If we multiply both sides with A' (A transpose) and B'( B transpose) we will have:
A' * A*B * B'=A' * A * D *B * B'
which can be written as :
P*Q=P * D * Q , P=A' * A, Q=B * B'.
P and Q are symmetric and singular matrices. Does this help?
matrix-equations
matrix-equations
asked Nov 17 at 11:16
Maria
11
New contributor
Maria is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 17 at 11:16
Maria
11
New contributor
Maria is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 17 at 20:51
asked Nov 17 at 11:16
Maria
11
New contributor
Maria is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 17 at 11:16
Maria
11
asked Nov 17 at 11:16
Maria
11
11
New contributor
Maria is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Maria is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Maria is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If you allow $0$s in the linear combination, just choose $D_1+0D_2$ and it works.
– Matt Samuel
Nov 17 at 11:28
Thank you so much @MattSamuel. Zero is not allowed. Maybe I was not clear enough so I modified the question.
– Maria
Nov 17 at 20:47
add a comment |
If you allow $0$s in the linear combination, just choose $D_1+0D_2$ and it works.
– Matt Samuel
Nov 17 at 11:28
Thank you so much @MattSamuel. Zero is not allowed. Maybe I was not clear enough so I modified the question.
– Maria
Nov 17 at 20:47
If you allow $0$s in the linear combination, just choose $D_1+0D_2$ and it works.
– Matt Samuel
Nov 17 at 11:28
If you allow $0$s in the linear combination, just choose $D_1+0D_2$ and it works.
– Matt Samuel
Nov 17 at 11:28
Thank you so much @MattSamuel. Zero is not allowed. Maybe I was not clear enough so I modified the question.
– Maria
Nov 17 at 20:47
Thank you so much @MattSamuel. Zero is not allowed. Maybe I was not clear enough so I modified the question.
– Maria
Nov 17 at 20:47
add a comment |
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Maria is a new contributor. Be nice, and check out our Code of Conduct.
Maria is a new contributor. Be nice, and check out our Code of Conduct.
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If you allow $0$s in the linear combination, just choose $D_1+0D_2$ and it works.
– Matt Samuel
Nov 17 at 11:28
Thank you so much @MattSamuel. Zero is not allowed. Maybe I was not clear enough so I modified the question.
– Maria
Nov 17 at 20:47