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diff --git a/source/know/concept/fredholm-alternative/index.md b/source/know/concept/fredholm-alternative/index.md index 5d53e79..c954272 100644 --- a/source/know/concept/fredholm-alternative/index.md +++ b/source/know/concept/fredholm-alternative/index.md @@ -8,51 +8,51 @@ layout: "concept" --- The **Fredholm alternative** is a theorem regarding equations involving -a linear operator $\hat{L}$ on a [Hilbert space](/know/concept/hilbert-space/), +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 $\Ket{u}$. +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 $\Inprod{w}{f} = 0$. +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, +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 $\vec{u}$. +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 $\hat{M}^\dagger \Ket{w} = 0$ +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 $\Ket{u}$. + 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 $$\Ket{u}$$. |