Interesting bit about Least-Squares

Does this make sense?

Let’s talk about the least-square solution to linear systems of equations. Let’s say we have the problem

$latex Ax=b$

where A is an m-by-n matrix (m rows, n columns), x is an n-element vector, and b is an m-element vector. This is just a standard form of a linear algebra problem. Let’s also assume A is full-rank.

Now’s let’s say $latex m>n$, or A is a tall matrix. We call A an overdetermined matrix. Simply put, there are way too many unique linear equations to solve for a unique x vector answer. So what do we do? We can use least-squares formulation of the problem to get some sort of an acceptable answer. If we multiply by sides of the equation by the Hermitian transpose of A, we get

$latex A^{*}Ax = A^{*}b \\ x = (A^{*}A)^{-1}A^{*}b$

So we have a least-squares solution for x. For people well-versed in linear…

View original post 187 more words

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