Generalized inverse

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Suppose LaTeX: A is any LaTeX: m \times n matrix. LaTeX: A^+ is a generalized inverse of LaTeX: A if LaTeX: A^+ is LaTeX: n \times m and LaTeX: AA^+A = A. Then, a fundamental theorem for linear equations is:

The equation, LaTeX: Ax = b, has a solution if and only if LaTeX: AA^+b = b for any generalized inverse, LaTeX: A^+, in which case the solutions are of the form:
LaTeX: x = A^+b + (I - A^+A)y   for any   LaTeX: y \in \mathbb{R}^n.

Here is an Example.

The Moore-Penrose class additionally requires that LaTeX: AA^+ and LaTeX: A^+A be symmetric (hermitian, if LaTeX: A is complex). In particular, if LaTeX: \mbox{rank}(A) = m, LaTeX: A^+ = A^T (AA^T)^{-1} is a Moore-Penrose inverse. Note that LaTeX: AA^+ = I, and LaTeX: A^+A is a projection matrix for the subspace LaTeX: \{x: x = A^T y \mbox{ for some } y \in \mathbb{R}^m\}, since LaTeX: x = A^T y implies LaTeX: A^+Ax = A^T (AA^T)^{-1} AA^T y = A^T y = x. Further, LaTeX: I-A^+A projects into its orthogonal complement, which is the null space of LaTeX: A, i.e. LaTeX: A(I-A^+A)x = (A-A)x = 0 for any LaTeX: x \in \mathbb{R}^n.


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