Context
An over-determined system of equations can not be generally solved by elimination operated on $Ax = b$.
Instead, it normally involves solving $$(A^t A) x = A^t b$$
which is obtained in the OLS method., this is, assuming we choose the sum of squared errors as the cost function.
Simplifying the $A$s product inverse to just $B$ then we could to LU decomposition $ B = LU $ and if $U$ has some zeroes in the diagonal then:
- some columns are linearly dependent
- it means $r<n$,
- and $Det(B)=0$,
- and $B^{-1}$ does not exist.
But does this means we can't find a solution ? Could you hint me why it is so conceptually, and how it relates to the obvious properties listed above ?