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Unformatted text preview: Math 54  Homework 8 Solutions 5.4.2 ker ( A T ) = ker 1 1 1 1 2 3 1 1 1 1 2 3 1 1 1 1 2 1 1 1 2 1 1 1 2 1 2 1 = 0 Basis for ker ( A T ) = 1 2 1 For the sketch illustrating ( Im ( A )) = ker ( A T ), Im ( A ) is the plane spanned by 1 1 1 , 1 2 3 . Notice that 1 2 1 is orthogonal to 1 1 1 and 1 2 3 5.4.7 A is a n n symmetric matrix. This implies that A = A T ( Im ( A )) = ker ( A T ) = ker ( A ) Or alternatively, Im ( A ) = ( ker ( A )) 1 5.4.8 A has dimension m n y has dimension m 1 ker ( A ) = 0 A T A is invertible Least squares solution of L ( x ) = y is ( A T A ) 1 A T y (a) L + ( y ) = ( A T A ) 1 A T y A + = ( A T A ) 1 A T It is easy to verify that L + is linear L + ( y + z ) = L + ( y ) + L + ( z ) L + ( k y ) = kL + ( y ) (b) If A is invertible, then ( A T A ) 1 = A 1 ( A T ) 1 then, A + = ( A T A ) 1 A T = A 1 ( A T ) 1 A T = A 1 If L is invertible, L + = L 1 (c) L + ( L ( x ) L + ( A x ) ( A T A ) 1 A T A x = x (d) L ( L + ( y ) = L (( A T A ) 1 A T y ) = A ( A T A ) 1 A T y 2 5.4.10 (a)5....
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This note was uploaded on 09/14/2011 for the course MAT 211 taught by Professor Movshev during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 MOVSHEV
 Math

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