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Unformatted text preview: D.S.G. POLLOCK: TOPICS IN ECONOMETRICS DIAGONALISATION OF A SYMMETRIC MATRIX Characteristic Roots and Characteristic Vectors. Let A be an n n symmetric matrix such that A = A , and imagine that the scalar and the vector x satisfy the equation Ax = x . Then is a characteristic root of A and x is a corresponding characteristic vector. We also refer to characteristic roots as latent roots or eigenvalues. The characteristic vectors are also called eigenvectors. (1) The characteristic vectors corresponding to two distinct characteristic roots are orthogonal. Thus, if Ax 1 = 1 x 1 and Ax 2 = 2 x 2 with 1 6 = 2 , then x 1 x 2 = 0. Proof. Premultiplying the defining equations by x 2 and x 1 respectively, gives x 2 Ax 1 = 1 x 2 x 1 and x 1 Ax 2 = 2 x 1 x 2 . But A = A implies that x 2 Ax 1 = x 1 Ax 2 , whence 1 x 2 x 1 = 2 x 1 x 2 . Since 1 6 = 2 , it must be that x 1 x 2 = 0. The characteristic vector corresponding to a particular root is defined only up to a factor of proportionality. For let x be a characteristic vector of A such that Ax = x . Then multiplying the equation by a scalar gives A ( x ) = ( x ) or Ay = y ; so y = x is another characteristic vector corresponding to . (2) If P = P = P 2 is a symmetric idempotent matrix, then its characteristic roots can take only the values of 0 and 1. Proof. Since P = P 2 , it follows that, if Px = x , then P 2 x = x or P ( Px ) = P ( x ) = 2 x = x , which implies that = 2 . This is possible only when = 0 , 1. Diagonalisation of a Symmetric Matrix. Let A be an n n symmetric matrix, and let x 1 ,...,x n be a set of n linearly independent characteristic vectors corresponding to its roots 1 ,..., n . Then we can form a set of normalised vectors c 1 = x 1 p x 1 x 1 , ... , c n = x n p x n x n , which have the property that c i c j = , if i 6 = j ; 1 , if i = j ....
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