# HW3_Sol - y-ˆ y 1 k 2 2 y-ˆ y t(ˆ y-ˆ y 1 = k y-ˆ y k...

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STAT424 Spring 2008 Solution to Homework 3 1) Let X = [ X 1 X 2 ] where X 1 is n × p and X 2 is n × q , so X is n × ( p + q ). Let M be the projection matrix for C ( X ) and let M 1 be the projection matrix for C ( X 1 ). X 1 and X 2 are not necessarily of full rank, and the columns of X 1 and X 2 are not necessarily orthogonal to each other. (a) Show that MM 1 = M 1 . ( Hint : C ( X 1 ) C ( X )) Sol) Since C ( I - M ) = C ( X ) and C ( X ) C ( X 1 ) , we have ( I - M ) M 1 = 0. Therefore, MM 1 = M 1 . (b) Let y be a vector in R n . Let ˆ y = M y and ˆ y 1 = M 1 y . Show that k y - ˆ y 1 k 2 = k y - ˆ y k 2 + k ˆ y - ˆ y 1 k 2 . Sol) Since y - ˆ y C ( X ) and ˆ y - ˆ y 1 C ( X ) C ( X 1 ) , y - ˆ y and ˆ y - ˆ y 1 are orthogonal to each other. Thus, k y - ˆ y 1 k 2 = k y - ˆ y + ˆ y - ˆ y 1 k 2 = k y - ˆ y k 2 + k ˆ
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Unformatted text preview: y-ˆ y 1 k 2 + 2( y-ˆ y ) t (ˆ y-ˆ y 1 ) = k y-ˆ y k 2 + k ˆ y-ˆ y 1 k 2 . 2) Show that any symmetric and idempotent matrix M is a projection matrix. Sol) Suppose M has the two properties. Note that ( I-M ) M = M-M 2 = M-M = 0 . So, for any y in R k , its residual vector y-M y is orthogonal to C ( M ) because ( y-M y ) t M = y t ( I-M ) t M = y t ( I-M ) M = 0 . Therefore, M is a projection matrix....
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## This note was uploaded on 04/29/2010 for the course STAT stat 426 taught by Professor Xe during the Spring '10 term at University of Illinois at Urbana–Champaign.

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