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---
title: "Fredholm alternative"
sort_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.
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