inverting non-square matrices

Inverting non-square matrices PhL 3.3.16 [See v2 of this doc for a more sophisticated analysis. ] So now I am off to Linear Algebra for Non Square Matrices world. I have some notes somewhere on this. I can find no PL notes, but I do find this in wiki on invertible matrix "Non-square matrices (m-by-n matrices for which m ≠ n) do not have an inverse. However, in some cases such a matrix may have a left inverse or right inverse. If A is m-by-n and the rank of A is equal to n, then A has a left inverse: an n-by-m matrix B such that BA = I. If A has rank m, then it has a right inverse: an n-by-m matrix B such that AB = I." So assuming this is correct, I will translate what it says my the current wedge doc Chapter 10 context. In this context, R is a matrix with m rows and n columns with m > n, so R is m x n. So R has m rows and n columns and so is an m x n matrix. The max rank that this matrix could have I know is the smaller of the two, which for me means rank ≤ n. If R has this full rank n, then (R-1l) exists and then of course we have (R-1l) R = 1n . Since m > n matrix R can never have rank m, so the second claim made above does not apply, and I guess I conclude that (R-1r) does not exist. Theorem: If R is a "tall" matrix of full rank, then (R-1l) exists and (R-1r) does not exist. Consider again the graphic above where S is on the right, Question: Does the matrix S = (R-1l) exist? Answer: yes, as shown below If you write out all the equations, you get n2 equations (one for each entry on the right). If you regard the elements of S as unknowns, there are then n*m unknowns. If you want to solve for the elements of S you have n2 linear equations in n*m unknowns. Since n < m, there are then fewer equations than there are unknowns, so there are then many solutions and then you can say that you can always a candidate for S = (R-1l), in agreement with the claim made above. If you instead regard the elements of S as unknowns, you still have n2 equations in n*m unknowns, so you always have a candidate for (Sr-1). On the other hand, consider Now there are m2 equations in n*m unknowns (one for each entry on the right), so in general there are too many constraints on those unknowns, so there will be no solution. Then (S-1l) and (R-1r) do not exist. So what does rank have to do with it? I need to add this section later. I think a matrix has to be full rank for the inverse to really exist, but maybe not. ****** Related topic. Consider this equation Rvt = vx where as drawn, m > n. Can we solve this for vt given vx? Two methods to answer this question: (1) There are clearly m equations. If vx is known and we seek vt which has n unknowns, then we have m equations in n unknowns. Problem is overspecified so there is no R-1 . (2) We showed above that (R-1l) exists. So we apply that to both sides of Rvt = vx to get Rvt = vx (R-1l) Rvt = (R-1l) vx vt = (R-1l) vx So yes, you can invert the equation Rvt = vx Question: Why do my two methods give conflicting results? Here is the answer. In the first method, we treat the vector vx as if it were completely arbitrary. But in fact, vx must be a vector that is obtained by application of R to some vector vt. So it is NOT completely arbitrary. Consider R: Rn → Rm. The vectors vx must be in the range of operator R and this range is some dimension-n subspace of Rm. If we consider arbitrary vectors vx in Rm , you will NOT be able to solve for vt just as (1) says. But for vectors vx in the range, you CAN find a solution, that given by (2).