---
title: "Fredholm alternative"
date: 2021-05-29
categories:
- Mathematics
layout: "concept"
---

The **Fredholm alternative** is a theorem regarding equations involving
a linear operator $\hat{L}$ on a [Hilbert space](/know/concept/hilbert-space/),
and is useful in the context of multiple-scale perturbation theory.
It is an *alternative* because it gives two mutually exclusive options,
given here in [Dirac notation](/know/concept/dirac-notation/):

1.  $\hat{L} \Ket{u} = \Ket{f}$ has a unique solution $\Ket{u}$ for every $\Ket{f}$.
2.  $\hat{L}^\dagger \Ket{w} = 0$ has non-zero solutions.
    Then regarding $\hat{L} \Ket{u} = \Ket{f}$:
    1.  If $\Inprod{w}{f} = 0$ for all $\Ket{w}$, then it has infinitely many solutions $\Ket{u}$.
    2.  If $\Inprod{w}{f} \neq 0$ for any $\Ket{w}$, then it has no solutions $\Ket{u}$.

Where $\hat{L}^\dagger$ is the adjoint of $\hat{L}$.
In other words, $\hat{L} \Ket{u} = \Ket{f}$ has non-trivial solutions if
and only if for all $\Ket{w}$ (including the trivial case $\Ket{w} = 0$)
it holds that $\Inprod{w}{f} = 0$.

As a specific example,
if $\hat{L}$ is a matrix and the kets are vectors,
this theorem can alternatively be stated as follows using the determinant:

1.  If $\mathrm{det}(\hat{L}) \neq 0$, then $\hat{L} \vec{u} = \vec{f}$
    has a unique solution $\vec{u}$ for every $\vec{f}$.
2.  If $\mathrm{det}(\hat{L}) = 0$,
    then $\hat{L}^\dagger \vec{w} = \vec{0}$ has non-zero solutions.
    Then regarding $\hat{L} \vec{u} = \vec{f}$:
    1.  If $\vec{w} \cdot \vec{f} = 0$ for all $\vec{w}$, then it has
        infinitely many solutions $\vec{u}$.
    2.  If $\vec{w} \cdot \vec{f} \neq 0$ for any $\vec{w}$, then it has
        no solutions $\vec{u}$.

Consequently, the Fredholm alternative is also brought up
in the context of eigenvalue problems.
Define $\hat{M} = (\hat{L} - \lambda \hat{I})$,
where $\lambda$ is an eigenvalue of $\hat{L}$
if and only if $\mathrm{det}(\hat{M}) = 0$.
Then for the equation $\hat{M} \Ket{u} = \Ket{f}$, we can say that:

1.  If $\lambda$ is *not* an eigenvalue,
    then there is a unique solution $\Ket{u}$ for each $\Ket{f}$.
2.  If $\lambda$ is an eigenvalue, then $\hat{M}^\dagger \Ket{w} = 0$
    has non-zero solutions. Then:
    1.  If $\Inprod{w}{f} = 0$ for all $\Ket{w}$, then there are
        infinitely many solutions $\Ket{u}$.
    2.  If $\Inprod{w}{f} \neq 0$ for any $\Ket{w}$, then there are no
        solutions $\Ket{u}$.



## References
1.  O. Bang,
    *Nonlinear mathematical physics: lecture notes*, 2020,
    unpublished.