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




  1. A is 3 by 200.

  2. B is a 200 by 1 column vector.

  3. All the elements of A and B are positive (non-negative and non-zero)

  4. D is a 200 by 200 diagonal matrix.

  5. 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?










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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















up vote
-1
down vote

favorite












I have a matrix equation like this: A * B=A * D * B




  1. A is 3 by 200.

  2. B is a 200 by 1 column vector.

  3. All the elements of A and B are positive (non-negative and non-zero)

  4. D is a 200 by 200 diagonal matrix.

  5. 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?










share|cite|improve this question
























  • 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













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











I have a matrix equation like this: A * B=A * D * B




  1. A is 3 by 200.

  2. B is a 200 by 1 column vector.

  3. All the elements of A and B are positive (non-negative and non-zero)

  4. D is a 200 by 200 diagonal matrix.

  5. 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?










share|cite|improve this question















I have a matrix equation like this: A * B=A * D * B




  1. A is 3 by 200.

  2. B is a 200 by 1 column vector.

  3. All the elements of A and B are positive (non-negative and non-zero)

  4. D is a 200 by 200 diagonal matrix.

  5. 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






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edited Nov 17 at 20:51

























asked Nov 17 at 11:16









Maria

11




11












  • 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










  • 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















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