STAT_151A_HW2 solution

# STAT_151A_HW2 solution - STAT 151A Linear Modeling Homework...

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STAT 151A Linear Modeling Homework 2 Solution March 13, 2016 Problem 1 Suppose G is positive definite. Show that (a) G is invertible and G - 1 is positive definite. Solution: We will prove by contradiction that G is invertible. Suppose G is not invertible, then A ˆ x = 0 must have non-trivial solution (e.g. ˆ x = ˆ 0 ). Then, A ˆ x = ˆ 0 = 0 ˆ x , so by definition 0 is an eigenvalue which contradicts with the assumption that G has all positive eigenvalues. Alternatively, we can show that G is invertible by showing that the nullspace of G is { ˆ 0 } . Let ˆ x null ( G ) . Thus, M ˆ x = ˆ 0 and ˆ x T G ˆ x = ˆ x T ˆ 0 = 0 . Since G is positive definite, which means ˆ x T G ˆ x > 0 for all ˆ x = 0 . Therefore, ˆ x = 0 and null ( G ) = { ˆ 0 } . Again, there are many ways to show that G - 1 is also positive definite. The easiest way, in my opinion, is to argue that the eigenvalues of G - 1 are simply the inverse of the eigenvalues of G . To see why, let λ be an eigenvalue of G , then G ˆ x = λ ˆ x = G - 1 A ˆ x = λG - 1 ˆ x = G - 1 ˆ x = 1 λ ˆ x And since the inverse of a positive number is still positive, the eigenvalues of G - 1 are all positive. So G - 1 is positive definite. Alternatively, to show that G - 1 is positive definite, define y = Gx , then y T G - 1 y = ( Gx ) T G - 1 ( Gx ) = x T G T G - 1 Gx = x T Gx > 0 (b) G has a positive definite square root G 1 / 2 . Solution: Since G = PDP T where D is a diagonal matrix whose entries are eigenvalues of G , we can construct G 1 / 2 = PD 1 / 2 P T where D 1 / 2 can be computed elementwise. 1

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Problem 2 Suppose G is n × n non-negative definite, and α is n × 1 . Let V be n × 1 vector of indepedent N (0 , 1) variables.
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