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Unformatted text preview: ECON 4261: Introduction to Econometrics Fall 2009 Problem Set 1 Answer Key Exercise 1 Let A be an ( m n ) matrix, with m > n . Show that the matrix A A is symmetric, and that if A has full rank, A A is positive definite. Let A = a 11 a 12 ... a 1 n a 21 a 22 ... a 2 n . . . . . . . . . . . . a m 1 a m 2 ... a mn . Then A A = a 11 a 21 ... a m 1 a 12 a 22 ... a mn . . . . . . . . . . . . a 1 n a 2 n ... a mn a 11 a 12 ... a 1 n a 21 a 22 ... a 2 n . . . . . . . . . . . . a m 1 a m 2 ... a mn = m j =1 a 2 j 1 m j =1 a j 1 a j 2 ... m j =1 a j 1 a jn m j =1 a j 2 a j 1 m j =1 a 2 j 2 ... m j =1 a j 2 a jn . . . . . . . . . . . . m j =1 a jn a j 1 m j =1 a jn a j 2 ... m j =1 a 2 jn and note that the elements off the diagonal of this matrix are the same. Hence A A is symmetric. To prove A A is positive definite, we must prove that q = x ( A A ) x > 0 for all x 6 = 0 . Suppose this does not happen, i.e. we have x ( A A ) x 0 for some x 6 = 0 . Note that q = x ( A A ) x = ( x A )( Ax ) = ( Ax ) ( Ax ) and Ax is an ( m 1) vector. So ( Ax ) ( Ax ) is the inner product of the vector Ax with itself. Hence, its a sum of squares. So it cannot be the case that q < 0. Hence the only possibility is to have q = 0 . Since q is a sum of squares, the only possibility is that all the components of the sum are zero. This would imply that Ax = 0. But since we are assuming that A has full rank (and m > n ), we must have x = 0 . But this contradicts our initial hypothesis that x 6 = 0. Hence we must have q = x ( A A ) x > 0 for all x 6 = 0 . So A A is positive definite. 1 Exercise 2 Characterize each of the following statements as true or false. If a statement is true, show the proof. If it is false, give a counterexample. Assume that A and B are square matrices....
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This note was uploaded on 02/23/2011 for the course ECON 4261 taught by Professor Staff during the Fall '08 term at Minnesota.
 Fall '08
 Staff
 Econometrics

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