From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

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 source/know/concept/fredholm-alternative/index.md | 64 +++++++++++------------
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diff --git a/source/know/concept/fredholm-alternative/index.md b/source/know/concept/fredholm-alternative/index.md
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--- 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 $$\Ket{u}$$.
 
 
 
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