If A^T b lies in the column space of A^ then should b not also lie in the column space of A? With application to least squares.

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I am trying to get a somewhat intuitive understanding of the Least Mean Square estimator and ran into the following problem.

I start with a matrix A, a target point/vector b and a vector x that takes linear combinations of the columns of A.

Then comes the argument that you have your vector Ax closest to b when the vector $ b - Ax $ is orthogonal to the column space of A (if that is the right entity that the orthogonal vector is orthogonal to).

I.e. $$ A^\intercal (b - Ax) = 0 $$

Leading straightforwardly to $$ A^\intercal b = A^\intercal Ax $$

Now, $ A^\intercal A $ is the square symmetric matrix which has across each row the inner product of a column vector of A with all column vectors in A, with the self inner products along the diagonal, so: $$ A^\intercal Ax = \begin{bmatrix} a_1 \cdot a_1 & a_1\cdot a_2 & ...\\ a_2 \cdot a_1 & a_2 \cdot a_2 & ...\\ ... & ... & ... \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ ... \end{bmatrix} $$

Each corresponding row element in $ A^\intercal b $ must be equal to the corresponding element in $ A^\intercal A x $, or: $$ a_1 \cdot a_1 x_1 + a_1 \cdot a_2 x_2+ ... + a_1 \cdot a_n x_n = a_1 \cdot (a_1 x_1 + a_2 x_2+ ... + a_n x_n) = a_1 \cdot b $$ Therefore $$ b = (a_1 x_1 + a_2 x_2+ ... + a_n x_n) $$

Which means that b lies in the column space of A, i.e. it is a linear combination of the columns of A. But I believe LMS works when b does not lie in the column space of A. Was it not allowed to factor out the dot product? What's wrong here?

Thanks.

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Your conclusion 'Therefore' simply doesn't follow.

The equations only say that $a_i\cdot b=a_i\cdot (Ax)$ for each column $a_i$ of $A$, and this is not enough to uniquely determine $b$, unless the vectors $a_i$ form a basis.