tensor doc for n NE m REVIEWED

Tensor Doc Changes for n ≠ m 5.17.16: OK, here is where I first deal head-on with the non-square R matrix stuff. Tensor doc has only square matrices. I thought maybe I could trivially "update" tensor doc to apply to non-square R, but then I decided that would be a huge undertaking and would complicate tensor doc beyond readability. It is really only needed in wedge doc, and I now have this all written up in Section 10.6. Suppose we have this situation, Rn n ≤ m Rm instead of the case n = m. How does this change the presentation in tensor doc? I will just page through and make changes as needed. 2.1 Linear Local Transformations dx'i = Σk=1m(∂x'i/∂xk) dxk = Σk=1m Rik dxk where Rik ≡ (∂x'i/∂xk) . (2.1.4) Doing the same operation in the other direction gives dxi = Σk=1n( ∂xi/∂x'k) dx'k = Σk=1n Sik dx'k where Sik ≡ (∂xi/∂x'k) . (2.1.5) The above are completely valid now as stated, just the usual notion of a differential! The idea that S = R-1 is now a little more complicated. So: R is the right inverse of S relative to 1m R = (S-1r) S (S-1r) = 1m S is the left inverse of R relative to 1m S = (R-1l) (R-1l) R = 1m S is the right inverse of R relative to 1n S = (R-1r) R (R-1r) = 1n R is the left inverse of S relative to 1n R = (S-1l) (S-1l) S = 1n In the nxn case, we could compute S if we were given R according to S = R-1 where we just compute the inverse of a square matrix using standard methods. But now how can we compute S from R? In other words, how do we compute the inverse of a non-square matrix, for example in evaluating S = (R-1r) I don't prove but I show in Appendix A below that Fact: If rank(S) = n (the max it can be), then (S-1l) and (Rr-1) exist (and I think in fact are not unique.) Therefore, if we assume that our S matrix has full rank, then (Rr-1) exists. Suppose then we have the matrix equation v' = Rv Since (Rr-1) exists but (Rl-1) does not exist, we cannot invert the equation! Here is a picture There are n equations but there are m unknowns (components of v) and m > n so not solvable. But suppose you have the equation v = S v' I know that (S-1l) exists if S has full rank n, so the above can be inverted to get v' = (S-1l) v = R v. –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– Appendix A : Inverting non -square matrices. 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 to the current context: S 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 S has this full rank n, then (S-1l) exists and then of course we have (S-1l) S = 1n . Since m > n matrix S can never have rank m, so the second claim made above does not apply, and I guess I conclude that (S-1r) does not exist. Consider again the graphic above where S is on the right, If you write out all the equations, you get n2 equations. If you regard the elements of R as unknowns, there are then n*m unknowns. If you want to solve for the elements of R you have n2 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 R = (S-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 (Rr-1). On the other hand, consider Now there are m2 equations in n*m unknowns, so in general there are too many constraints on those unknowns, so there will be no solution. Then (R-1l) and (S-1r) do not exist. So what does rank have to do with it? A square matrix has to be full rank to be invertible, meaning to have a non-zero determinant. Maybe you can present this argument with minors in the non-square case. How would you actually go about solving the equations? What are the equations you need to solve? Go back to the solvable problem Here are the equations you need to solve R11S11 + R12S21 + ..... R1mSm1 = 1 R21S11 + R22S21 + ..... R2mSm1 = 0 ... Rn1S11 + Rn2S21 + ..... RnmSm1 = 0 R11S12 + R12S22 + ..... R1mSm2 = 0 R21S12 + R22S22 + ..... R2mSm2 = 1 ... Rn1S12 + Rn2S22 + ..... RnmSm2 = 0 .... R11S1n + R12S2n + ..... R1mSmn = 0 R21S1n + R22S2n + ..... R2mSmn = 0 ... Rn1S1n + Rn2S2n+ ..... RnmSmn = 1 There are n groups of equations, and each group has n equations, so n2 equations as noted above. Imagine that Rij are known and you are trying to compute the Sij. The first n equations each have n unknowns S11, S21....Sm1 . Suppose we just pick arbitrary values for the last few unknowns and then regard S11, S21....Sn1 as a set of n unknowns. Then the first group of equations above becomes n equations in n unknowns. In order for that set of equations to have a solution, we must have the det ≠ 0 for this matrix of "coefficients" R11 R12 ... R1n R21 R22 ... R2n ... Rn1 Rn2 ... Rnn n x n Although we have only partial rows in this matrix, the columns are complete columns. Suppose we knew that matrix R had rank less than n. Then any set of n columns is linearly dependent. STOP. I have given this an hour or two and I am finding no solution and it is distracting me from the main stream today, so put this proof on indefinite hold. I believe the wiki claim.