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| Syntax |
linlsq(A, b, bound) |
| See Also | nlsq , linlsqb |
x, that solves the problem
2
minimize | A x - b | with respect to x
The matrix A is real, double-precision, or complex,
b is a column vector
with the same type and row dimension as A,
and bound is a scalar with the same type as A
such that absolute singular values of the matrix A
that are less than bound are replaced by 0 before solving the
minimization problem.
If A is not complex,
bound must have the same type as A.
If A is complex,
bound must be double-precision.
The returned vector, x, has the same type as A.
If more than one solution exists,
the solution of minimum norm is returned
(i.e., the value of x that minimizes |x|
over the set of solutions).
The operation "\" uses a
QR factorization to solve similar problems.
If the rank of A is less than both its row and column dimension,
the "\" operator will generate an error
if you compute A \ b.
The linlsq function can solve such problems.
A = {1., 1.}
b = {1., 2.}
eps = 1e-7
linlsq(A, b, eps)
returns
1.5