# hw4sol - ECE 130C Homework 4 Solution May 3 2007 2.3.14 The...

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ECE 130C Homework 4 Solution May 3, 2007 2 . 3 . 14 The dimension of S is (a) 0 when x = 0 (b) 1 when x = (1 , 1 , 1 , 1) (c) 3 when x = (1 , 1 , - 1 , - 1) because all arrangements of this x are perpendicular to (1 , 1 , 1 , 1) (d) 4 when the x ’s are not equal and don’t add to zero. No x gives dim S =2. 2 . 3 . 26 (a) True. (b) False because the basis vectors may not be in S . 2 . 3 . 34 If v 1 , v 2 , v 3 is a basis for V , and w 1 , w 2 , w 3 is a basis for W , then these six vectors cannot be independent and some combination is zero: c i v i + d i w i = 0. This puts c i v i = - d i w i in both subspaces. 2 . 3 . 36 n - r = 17 - 11 = 6 is nullspace dimension; 64 - 11 = 53 is dim N ( A T ) 2 . 3 . 40 (a) y ( x ) = e 2 x (b) y = x (one basis vector in each case). 2 . 4 . 28 1 1 0 2 1 0 " 1 0 1 1 2 0 # = 2 2 1 2 4 0 1 0 1 ; r + ( n - r ) = n = 3 but 2 + 2 is 4. 2 . 4 . 32 Row space = yz plane; column space = xy plane; nullspace =

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## This note was uploaded on 08/08/2010 for the course ECE PROF. VOLK taught by Professor Volkanrodoplu during the Summer '10 term at UCSB.

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hw4sol - ECE 130C Homework 4 Solution May 3 2007 2.3.14 The...

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