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+---
+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.