01 - Fredholm Alternative (Vector Space Version)

Fredholm Alternative Theorem ( ${\mathbb{R}}^n$ version)

Consider homogeneous and inhomogeneous problems

$$\begin{array}{rlrl}A\mathbf{x}& =0& & (\mathcal{H})\\ A\mathbf{x}& =\mathbf{b}& & (\mathcal{N})\end{array}$$

Exactly one of the following occur

1. Homogeneous Equation $(\mathcal{H})$ $A\mathbf{x}=0$ only has zero solution
   Hence solution of $(\mathcal{N})$ $A\mathbf{x}=\mathbf{b}$ is unique

2. Homogeneous Equation $(\mathcal{H})$ $A\mathbf{x}=0$ has non-trivial solutions and so does $({\mathcal{H}}^{\ast })$ $A^{\ast }\mathbf{w}=0$
   Hence there are two sub-possibilities:

   2a) If $⟨\mathbf{b},\mathbf{w}⟩=0$ for all $\mathbf{w}$ satisfying $({\mathcal{H}}^{\ast })$ then $(\mathcal{N})$ has a non-unique solution

   2b) Otherwise $(\mathcal{N})$ has no solution

Motivation for Fredholm Alternative

Consider homogeneous and inhomogeneous problems

$$\begin{array}{rlrl}A\mathbf{x}& =0& & (\mathcal{H})\\ A\mathbf{x}& =\mathbf{b}& & (\mathcal{N})\end{array}$$

Linear Algebra Version $A\in {\mathbb{R}}^{n\times n}$ Let $\mathbf{x},\mathbf{b}\in {\mathbb{R}}^n$

Let

If $A$ is invertible (non-zero determinant) then
$(\mathcal{H})$ has only trivial solution $\mathbf{x}=0$
So $(\mathcal{N})$ has unique solution $\mathbf{x}=A^{-1}\mathbf{b}$

However if $\mathcal{H}$ has solution $\mathbf{x}={\mathbf{x}}_1\ne 0$ then
$A$ is singular and $\mathbf{b}$ doesn’t have a general solution for $(\mathcal{N})$

If for a specific $\mathbf{b}$ has solution $\mathbf{x}$ for $(\mathcal{N})$ then there are infinite solutions in form

$$\mathbf{x}+\alpha {\mathbf{x}}_1$$

Linear Transformation Version

Let $A$ be a linear transformation on ${\mathbb{R}}^n$ then
Let $A^{\ast }$ be the corresponding adjoint transformation such that

$$⟨A\mathbf{x},\mathbf{w}⟩\equiv ⟨\mathbf{x},A^{\ast }\mathbf{w}⟩$$

where $⟨\mathbf{x},\mathbf{w}⟩\equiv \mathbf{x}\cdot \mathbf{w}$ denotes the usual Cartesian Inner Product

If $(\mathcal{H})$ has non-trivial solutions for $\mathbf{x}$ then corresponding adjoint problem

$$A^{\ast }\mathbf{w}=0$$

also has non-trivial solutions for $\mathbf{w}$

Taking the inner product of $\mathcal{N}$ with $\mathbf{w}$ and using adjoint inner product relation then

Necessary Condition for $\mathcal{N}$ to be solvable is

$$⟨\mathbf{b},\mathbf{w}⟩=0\text{ for all }\mathbf{w}\text{ satisfying }({\mathcal{H}}^{\ast })$$