Categories: Mathematics.

# Fredholm alternative

The **Fredholm alternative** is a theorem regarding equations involving
a linear operator $\hat{L}$ on a 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:

- $\hat{L} \Ket{u} = \Ket{f}$ has a unique solution $\Ket{u}$ for every $\Ket{f}$.
- $\hat{L}^\dagger \Ket{w} = 0$ has nonzero solutions.
Then regarding $\hat{L} \Ket{u} = \Ket{f}$:
- If $\Inprod{w}{f} = 0$ for all $\Ket{w}$, then it has infinitely many solutions $\Ket{u}$.
- 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:

- If $\mathrm{det}(\hat{L}) \neq 0$, then $\hat{L} \vec{u} = \vec{f}$ has a unique solution $\vec{u}$ for every $\vec{f}$.
- If $\mathrm{det}(\hat{L}) = 0$,
then $\hat{L}^\dagger \vec{w} = \vec{0}$ has nonzero solutions.
Then regarding $\hat{L} \vec{u} = \vec{f}$:
- If $\vec{w} \cdot \vec{f} = 0$ for all $\vec{w}$, then it has infinitely many solutions $\vec{u}$.
- 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:

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

## References

- O. Bang,
*Nonlinear mathematical physics: lecture notes*, 2020, unpublished.