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  "body": "\n## Statement\n\nLet $F$ be a [[field|04C5ZBG1WT]], let $n \\ge 1$ and let $A$ be an $n \\times n$ matrix over $F$, with [[transpose|09DXWYRD43]] $A^{T}$.\nThen exactly one of the following two cases holds.\n\n1. For every $b \\in F^n$, the [[system|0GT4WWGWH2]] $Ax = b$ has exactly one solution $x \\in F^n$.\n2. The [[homogeneous system|04KGXJG1A2]] $Ax = 0$ has a solution $x \\ne 0$.\n\nIn case 2, for $b \\in F^n$, the system $Ax = b$ has a solution if and only if $y^{T} b = 0$ for every $y \\in F^n$ with $A^{T} y = 0$.\n\n## Remarks\n\nThe first part follows from the [[rank–nullity theorem|0W4FSW1TVB]]: the map $x \\mapsto Ax$ on $F^n$ is injective if and only if it is surjective.\nFor the solvability condition, the [[column space|03AJKXH1AV]] of $A$ and the solution space of $A^{T} y = 0$ have dimensions $r$ and $n - r$, where $r$ is the [[rank|0ZC9YTXZRY]] of $A$.\nEvery column $a$ of $A$ satisfies $y^{T} a = 0$ for such $y$.\nSo the column space lies in the space of all $b$ with $y^{T} b = 0$ for all such $y$.\nBoth spaces have dimension $r$, so they are equal.\nExample: for $A = \\begin{pmatrix} 1 \u0026 1 \\\\ 1 \u0026 1 \\end{pmatrix}$, case 2 holds with $x = (1, -1)$.\nThe solutions of $A^{T} y = 0$ are the multiples of $y = (1, -1)$, so $Ax = b$ is solvable exactly when $b_1 = b_2$.\nThe name comes from Fredholm's theory of integral equations, where the same alternative holds for compact perturbations of the identity on Banach spaces.\n\n## Explanations\n\n### L1 human\n\nA square linear system behaves in one of two ways.\nEither it is perfect: every right-hand side has one and only one solution.\nOr the homogeneous system has a nonzero solution, and then some right-hand sides have no solution at all.\nIn the second case, a simple test tells which right-hand sides work: $y^{T} b$ must be $0$ for every solution $y$ of the transposed homogeneous system $A^{T} y = 0$.\n\n### L2 human\n\nThe theorem says that for square systems, uniqueness and existence go together.\nTo show that $Ax = b$ is always solvable, it is enough to show that $Ax = 0$ has only the zero solution.\nThe solvability condition in case 2 is the statement that the column space of $A$ is the annihilator of the null space of $A^{T}$.\nIt is the finite-dimensional model for the Fredholm alternative for integral equations and for elliptic boundary value problems.\n",
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