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Diffstat (limited to 'source/know/concept/fredholm-alternative')
| -rw-r--r-- | source/know/concept/fredholm-alternative/index.md | 6 |
1 files changed, 3 insertions, 3 deletions
diff --git a/source/know/concept/fredholm-alternative/index.md b/source/know/concept/fredholm-alternative/index.md index c954272..fdc90be 100644 --- a/source/know/concept/fredholm-alternative/index.md +++ b/source/know/concept/fredholm-alternative/index.md @@ -14,7 +14,7 @@ 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. +2. $$\hat{L}^\dagger \Ket{w} = 0$$ has nonzero 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 $$\Ket{u}$$. @@ -31,7 +31,7 @@ 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 $$\hat{L}^\dagger \vec{w} = \vec{0}$$ has nonzero 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 $$\vec{u}$$. @@ -48,7 +48,7 @@ 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: + has nonzero 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 |