WilliamC

Often the authors of research articles will have PDF files freely available for download on their personal or laboratory web sites, even if they are published. Typically these are the unedited articles and not the final edited proof-read version that was published but the information is still there. Also using Pubmed or such you can find some recent articles and email the authors directly for reprints or, again, they might send you a private PDF file for free. This is good for a few critical articles but not for large numbers.

Well I got a few intro linear algebra texts and their solution manuals, and I suppose I have to get past more of this before I can get deeper. But it's so frustrating not being able to solve a problem, as if I am defeated somehow. And as I'm sure you all are aware the number of people in the real world willing to even talk about math is quite small, and of those finding someone able to help me outside a University is impossible. I can't get as good an understanding of everything else I read without being able to deal with more advanced math than calculus and basic differential equations, and I was hoping I could learn tensors and differential geometry and sort of pick up the linear algebra I needed as I went. yea, right.

I have an embarrassing abundance of resources and books, and I've gotten some ways into linear algebra, but I really don't have much choice as I am not in any position to take classes anymore. I just have to cram in as much as possible while I can. At least the Schaum's outline series provides me with problems, and only by solving problems do I learn.

Learning as I go. Thanks. Let me try again. With matrix B defined as the matrix of coefficients of vectors u1, u2, u3 (what I mistakenly called matrix A originally) , then would C = [BT]-1, that is the inverse of the transpose of B? When I work out the result I get something frustratingly close to the correct answer but not quite, and sometimes the Schaum's outline series has incorrect answers so I don't know if my calculations are off or I'm completely on the wrong idea. I don't know how to write matrices, but I'm taking the inverse of BT by calculating the 9 minors, constructing a 3 x 3 matrix of these values, and multiplying by det BT. Am I on the right track? I've lost more than a week on this problem and I have to get past it.

OK, see where I confused you and myself. With the vector a = 2u1 - u3 the matrix A I described should relate to the vector a, but it doesn't. Question, is it possible to solve the problem without knowing anything about vector a, in which case matrix A I previously described would be the coefficients of vectors u1, u2, u3 and matrix B the coefficients of vectors v1, v2, v3? Or do I have to consider matrix A to be the row vector a and then a new matrix, C which are the coefficients of v1, v2, v3 when expressed as u1 = c11v1 + c12v2 + c12v3, u2 = c21v1 + c22v2 + c23v3, u3 = c31v1 + c32v2 + c33v3? As for the tensor notation I am starting to learn it but I am not proficient with it yet. Perhaps if I can work through this problem in matrix notation then I can redo it in tensor notation, and I know for anything more than 3D I will need to be able to understand the tensor notation, but I'd like to understand it in terms of matrices first. But it just doesn't leap out at me and I can't find anyone who knows anything about it to talk to and as difficult as it is to have to type all this (I don't even know where you get the summation symbols and such, I can write things I can't express with a keyboard) it seriously irritates me but I can't figure it out on my own. So much so I finally broke down and sought out this resource. Thanks for helping.

I am trying to work through multiple Schaums outline books on my own including Schaum's Outline of Differential Geometry. I cannot seem to get how to change basis in 3D vectors. From chapter 1, problem 1.47, Given vectors u1, u2, and u3 form a basis in E3 and v1 = -u1 + u2 -u3, v2 = u1 +2u2 - u3, v3 = 2u1 + u3, show that v1, v2, v3 are linearly independent and find the components of a = 2u1 - u3 in terms of v1, v2, and v3. END OF PROBLEM. Now that v1, v2, v3 are linearly independent is obvious since there is a zero coefficient for u2 in the equation for v3, and I can show that the determinate of the matrix formed by the row vectors (-1, 1, -1), (1, 2, -1), (2, 0, 1) is not equal to zero. I get that the matrix A = [aij] i,j, = 1,2,3 are the coefficients a11 = -1, a12 = 2, a 13 = -1, a21 = 1, a22 = 2, a23 = -1, a31 = 2, a32 = 0, a33 = 1 and that there is a matrix B = [bij] which will have the coefficients allowing the expressions of the equations u1 = b11v1 + b12v2 +b13v3, u2 = b21v1 + b22v2 + b23v3, u3 = b31v1 + b32v2 + b33v3. What I can't get is how to determine the values of the [bij]'s. I've tried cross multiplying matrix A by the 3 x 3 identity matrix to get 3 sets of equations and setting them all equal to zero, which seems to make sense, but i can't get the same answer in the book. Any suggestions? Please, this is driving me insane and I can't get past it to chapter two and I need to also get back to making progress in Schaum's Outline of Tensor Calculus, Outline of Linear Algebra, and Outline of Group Theory too. The answer is supposed to be u1 = -2v1 + v2 -v3, u2 = 3v1 -v2 + 2v2, u3 = 4v1 -2v2 +3v3. P.S. typing all the formatting takes too much time, wish I could more easily communicate with someone more easily.