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--- fredholm_alternative [2024/07/22 15:46] – created spencer
+++ fredholm_alternative [2024/07/22 16:10] (current) – spencer
@@ Line 14: / Line 14: @@
 It follows that either both the kernel and cokernel of $aI - K$ are zero, or both the kernel and cokernel of $aI - K$ are nonzero.
 Said differently, either $(aI-K)u = 0$ has a nontrivial solution (that is, Ku = au is an eigenvector) or else the cokernel of $I-K$ is non-zero; that is, it is surjective, so $(I-aK)u = v$ has a unique solution for all v.
+There is another theorem called the Fredholm alternative: if $L$ is a bounded operator on a Hilbert space, then $Ly = f$ has a solution if and only if $\langle f, z \rangle = 0$ for every $z$ with $L^* z = 0$.
+Indeed, the perp to the image of $L$ is the kernel of the adjoint.
+Thus if the adjoint is injective, $L$ is surjective. If $L$ is not surjective, the adjoint has some kernel.