What is Fredholm alternative theorem?
In mathematics, the Fredholm alternative, named after Ivar Fredholm, is one of Fredholm’s theorems and is a result in Fredholm theory. Part of the result states that a non-zero complex number in the spectrum of a compact operator is an eigenvalue.
What is solvability condition?
Abstract. Solvability conditions for linear differential equations are usually formulated in terms of orthogonality of the right-hand side to solutions of the homogeneous adjoint equation. These functionals are solutions of the homogeneous adjoint equation L∗φ = 0.
What are the conditions for solvability of linear equations?
To understand the condition for solvability of linear simultaneous equations in two variables, if linear simultaneous equations in two variables have no solution, they are called inconsistent whereas if they have solution, they are called consistent.
What is a solvable equation?
Solvable equation, a polynomial equation whose Galois group is solvable, or equivalently, one whose solutions may be expressed by nested radicals. Solvable Lie algebra, a Lie algebra whose derived series reaches the zero algebra in finitely many steps.
Is Banach space compact?
(In particular, in this case the unit sphere is compact.) Equivalently, any bounded sequence in E has a convergent subsequence. By the Heine-Borel theorem, finite-dimensional Banach spaces are locally compact. The converse is true.
What are inconsistent pair of linear equations?
A system of two linear equations can have one solution, an infinite number of solutions, or no solution. If a system has no solution, it is said to be inconsistent . The graphs of the lines do not intersect, so the graphs are parallel and there is no solution.
What are the conditions for consistency?
A linear system is consistent if and only if its coefficient matrix has the same rank as does its augmented matrix (the coefficient matrix with an extra column added, that column being the column vector of constants).
Is integral operator compact?
Hilbert–Schmidt integral operators are both continuous (and hence bounded) and compact (as with all Hilbert–Schmidt operators). then K is also self-adjoint and so the spectral theorem applies.
