Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: Notesfiles $Revision: 1.6.2.17 $; site uiucdcsb.UUCP Path: utzoo!watmath!clyde!burl!ulysses!mhuxj!ihnp4!inuxc!pur-ee!uiucdcsb!robison From: robison@uiucdcsb.UUCP Newsgroups: net.math Subject: Re: Some Math problems Message-ID: <9700030@uiucdcsb.UUCP> Date: Sun, 28-Oct-84 15:39:00 EST Article-I.D.: uiucdcsb.9700030 Posted: Sun Oct 28 15:39:00 1984 Date-Received: Tue, 30-Oct-84 01:38:20 EST References: <689@ihuxt.UUCP> Lines: 57 Nf-ID: #R:ihuxt:-68900:uiucdcsb:9700030:000:2041 Nf-From: uiucdcsb!robison Oct 28 14:39:00 1984 I originally sent the answer to question 2 directly to Barry. Since no-one else posted an answer, I'll "reprint" mine below: ------------------------------------------------------------------------- - 2. Given a matrix A where - - - .68 -.6928 .24 0 - .6928 .5 -.52 0 - A = .24 .52 .82 0 - 0 0 0 1 - - find a transformation B such that BB=A. This is the problem of finding the square-root of matrix A. (Notation: "*" means multiplication, "**" means exponentiation. I wish I could have superscripts/subscripts or a least an APL character set!) First find the matrix Z, such that L = (1/Z)*A*Z, where L is a diagonal matrix: l1 0 0 0 L = 0 l2 0 0 0 0 l3 0 0 0 0 l4 Then A = Z*L*(1/Z) and B = Z*(L**.5)*(1/Z). (Why the answer for B? Try multiplying out BB.) Since L is a diagonal matrix, its square-root is trivially: l1**.5 0 0 0 0 l2**.5 0 0 0 0 l3**.5 0 0 0 0 l4**.5 The values l1,l2,l3, and l4 are the eigenvalues of A. The columns of Z are the eigenvectors of A. Methods for finding eigenvectors and eigenvalues may be found in a linear algebra textbook. The eigenstuff has many other uses. You can, for instance, take the "tangent" of a matrix by finding L and Z, then taking the tangent of each element of L, and then transforming back in the same manner as for finding the square-root. (This works for any function with a McLauren expansion). One neat application involves taking e raised to a matrix power. One can take the formula for solving a single linear differential equation and use it for solving a system of linear differential equations by substituting matrix exponentiation for scalar exponentiation. ----------------------------------------------------------------------------- Added note: the Newton-Ralphson method probably only works for matrices with positive real determinants, but I can't prove this. - Arch @ uiucdcs