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MATH 330 Q4142sol

# MATH 330 Q4142sol - Math 330 Chapter 4 Quizzes Q41...

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Math 330 - Chapter 4 Quizzes Q41 Take-home due March 13,2009 All but one of the following is not a vector space. Explain why the fake ones are not vector spaces and indicate which one is a vector space (You don’t have to prove that it is): 2 points each The subset of R 2 consisting of vectors bracketleftBigg m n bracketrightBigg where both m and n are integers. SOLUTION: This is not a vector space because scalar multiplication fails. For example, although bracketleftBigg 1 1 bracketrightBigg meets the criterion, 1 2 bracketleftBigg 1 1 bracketrightBigg = bracketleftBigg 1 2 1 2 bracketrightBigg does not. The subset of R 2 consisting of vectors bracketleftBigg x y bracketrightBigg such that xy = 0 . SOLUTION: This is not a vector space because vector addition fails. Although bracketleftBigg 1 0 bracketrightBigg and bracketleftBigg 0 1 bracketrightBigg meet the criterion, their sum bracketleftBigg 1 0 bracketrightBigg + bracketleftBigg 0 1 bracketrightBigg = bracketleftBigg 1 1 bracketrightBigg does not. The set of 2 × 2 matrices A such that XA = AX where X = bracketleftBigg 1 - 1 1 1 bracketrightBigg . SOLUTION: This is a vector space. While you weren’t expected to show this, the reasons are not difficult: if AX = XA and BX = XB then ( A + B ) X = AX + BX = XA + XB = X ( A + B ) if AX = XA and λ is any scalar, then ( λA ) X = λ ( AX ) = λ ( XA ) = ( λX ) A = ( ) A = X ( λA ) The set of continuous functions f : R R which are non-decreasing (i.e. x y implies that f ( x ) f ( y ) .) SOLUTION: The is not a vector space because scalar multiplication fails. For an example, suppose that f ( x ) = x +1. This is surely non decreasing since x y implies that f ( x ) = x +1 y +1 = f ( y ). However, ( - 1) f ( x ) = - x - 1 is spectacularly decreasing.

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MATH 330 Q4142sol - Math 330 Chapter 4 Quizzes Q41...

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