Fredholm alternative

Remarks

The first part follows from the rank–nullity theorem: the map $x \mapsto Ax$ on $F^n$ is injective if and only if it is surjective. For the solvability condition, the column space of $A$ and the solution space of $A^{T} y = 0$ have dimensions $r$ and $n - r$, where $r$ is the rank of $A$. Every column $a$ of $A$ satisfies $y^{T} a = 0$ for such $y$. So the column space lies in the space of all $b$ with $y^{T} b = 0$ for all such $y$. Both spaces have dimension $r$, so they are equal. Example: for $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, case 2 holds with $x = (1, -1)$. The solutions of $A^{T} y = 0$ are the multiples of $y = (1, -1)$, so $Ax = b$ is solvable exactly when $b_1 = b_2$. The name comes from Fredholm's theory of integral equations, where the same alternative holds for compact perturbations of the identity on Banach spaces.

Going deeper

For a student

The theorem says that for square systems, uniqueness and existence go together. To show that $Ax = b$ is always solvable, it is enough to show that $Ax = 0$ has only the zero solution. The solvability condition in case 2 is the statement that the column space of $A$ is the annihilator of the null space of $A^{T}$. It is the finite-dimensional model for the Fredholm alternative for integral equations and for elliptic boundary value problems.