Skew-SymmetryDate: 02/17/97 at 22:12:15 From: ANDY Subject: linear algebra A matrix A is symmetric if A(T) = a and a matrix b is skew-symmetric if B(T) = -B. Let A be any symmetric matrix and B be any skew-symmetric matrix. Prove the following statements: a) A is a square matrix. b) B is a square matrix. c) The main diagonal entries in B are all zeros and thus tr(B) = 0. d) If C is any n*n matrix, then the matrix C - C(T) is skew-symmetric. Date: 02/18/97 at 14:44:45 From: Doctor Keith Subject: Re: linear algebra Hi, This is the real fun part of linear algebra, in my opinion. We can handle parts (a) and (b) by a simple dimensional analysis. I will go through it for (a) and you can do it for (b). In general, we can say that a matrix A is mxn where m is the number of rows and n is the number of columns. Now, since A = A^T and we know A^ T is a nxm matrix, we know that m and n must be equal (otherwise we would contradict the definition of equality of matrices, which requires both sides to have the same dimension, thus the mxn and nxm matrices must be of the same dimension and so m=n). But since A is a nxn matix, it is square by the definition of square matrix and we are done. In part (c), let b(i,j) be the element in the i_th_ row and j_th_ column of B. We know that B = -B^T, so we have b(i,j)=-b(j,i). On the diagonal i = j so the formula specializes to b(i,i) = -b(i,i), which is only true when b(i,i)=0. Since i was arbitrary, this holds for all diagonal elements, and the first part is done. Since the trace of a matrix is the sum of the diagonal elements, which are all zero, the second part follows directly (i.e., the sum of a bunch of zeros is zero). In part (d), you get to apply the definition directly. You are given a square matrix, C, and you need to show that C-C^T is skew-symmetric. Take the transpose of C-C^T and see if it is C^T-C: (C-C^T)^T = C^T - C^T^T = C^T - C = -(C-C^T) And you are done. Remember, most problems like this are testing knowledge of a definition or theorem, so always see if you can test the problem directly in light of some definition or theorem. Linear algebra is lots of fun, so keep at it and you will do great. Let us know if we can help you with anything else. -Doctor Keith, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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