How to efficiently solve a system with an infinity number of solutions?
Consider A to be a square matrix of size n and S to be a symmetric square matrix of size n. I know the first row of A and I know each element in S. I need to find the remaining elements (from row 2 to row n) of A such that A’A = S. This system allows an infinite number of solutions and I am not interested in a particular solution (any of them is fine). Does it exist an efficient procedure to find one among these solutions? In other words, I do not want to use an "fsolve" procedure because it is computationally too demanding. I would like to find a solution in the same spirit of chol(S)’*chol(S) = S. Nevertheless, chol(S) does not work because does not guarantee that the first row of A is the one I need. For another similar problem, I remember that null(A(1,:)) was useful. The problem is that [A(1,:); null(A(1,:))]’*[A(1,:); null(A(1,:))] = eye(n) and I need the right-hand side to be equal to S (instead of eye(n)).
I hope the problem is clear. Many thanks for the help.Consider A to be a square matrix of size n and S to be a symmetric square matrix of size n. I know the first row of A and I know each element in S. I need to find the remaining elements (from row 2 to row n) of A such that A’A = S. This system allows an infinite number of solutions and I am not interested in a particular solution (any of them is fine). Does it exist an efficient procedure to find one among these solutions? In other words, I do not want to use an "fsolve" procedure because it is computationally too demanding. I would like to find a solution in the same spirit of chol(S)’*chol(S) = S. Nevertheless, chol(S) does not work because does not guarantee that the first row of A is the one I need. For another similar problem, I remember that null(A(1,:)) was useful. The problem is that [A(1,:); null(A(1,:))]’*[A(1,:); null(A(1,:))] = eye(n) and I need the right-hand side to be equal to S (instead of eye(n)).
I hope the problem is clear. Many thanks for the help. Consider A to be a square matrix of size n and S to be a symmetric square matrix of size n. I know the first row of A and I know each element in S. I need to find the remaining elements (from row 2 to row n) of A such that A’A = S. This system allows an infinite number of solutions and I am not interested in a particular solution (any of them is fine). Does it exist an efficient procedure to find one among these solutions? In other words, I do not want to use an "fsolve" procedure because it is computationally too demanding. I would like to find a solution in the same spirit of chol(S)’*chol(S) = S. Nevertheless, chol(S) does not work because does not guarantee that the first row of A is the one I need. For another similar problem, I remember that null(A(1,:)) was useful. The problem is that [A(1,:); null(A(1,:))]’*[A(1,:); null(A(1,:))] = eye(n) and I need the right-hand side to be equal to S (instead of eye(n)).
I hope the problem is clear. Many thanks for the help. matrix algebra, cholesky, null MATLAB Answers — New Questions