Orthogonal MatricesDate: 01/07/98 at 23:54:34 From: Thomas Dehler Subject: orthogonal matrices If A is an orthogonal matrix, then the determinant of A is either 1 or -1. How do I prove this? Date: 01/09/98 at 10:03:03 From: Doctor Joe Subject: Re: orthogonal matrices Hi Thomas, The problem you have asked belongs to a set of just about the nicest set of problems in linear algebra. I hope you already know the following theorems and I can use them without proofs: Theorem A Let A and B be any two square matrices with the same dimension. Denoting the determinant of a matrix A by det A, (likewise for B), we have: det AB = det A * det B. In other words, the determinant preserves the product of matrices. (Remark: This gives an insight to why the determinant gives rise to many homomorphisms between certain subgroups of square matrices and the non-zero real number line.) Theorem B Denote the transpose of A as A^T for a given square matrix A. The determinant of A^T is the same as the determinant of A itself; i.e. det A^T = det A. Now, let's prove your statement: "If A is orthogonal, then det A = either 1 or -1." Proof: By definition, A is orthogonal means that A^T = A^(-1) (read as the transpose of A is equal to its inverse). Of course, from the definition of inverse of a matrix, A * A^(-1) = I, where I denotes the usual identity matrix. It follows that for this orthogonal matrix A, A * A^T = I. Now taking determinant on both sides, det (A * A^T) = det I Then, invoking Thm A, det A * det A^T = 1 (since det I = 1. Recall that the determinant of a diagonal matrix is the product of its diagonal terms.) Next, invoking Thm B, det A * det A = 1 (since det A^T = det A). It follows that (det A)^2 = 1. Hence, det A = 1 or -1. (proven) Remarks. When restricted to the case of A being dimension 2, an orthogonal matrix is some rotation or reflection. (As an exercise, try to find out exactly which rotation and which reflection.) Thus, det of such matrices are one (they are area-preserving). -Doctor Joe, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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