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Unformatted text preview: Math 330  Chapter 4 Quizzes Q41 Takehome 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 dont 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 bracketrightBigg and bracketleftBigg 1 bracketrightBigg meet the criterion, their sum bracketleftBigg 1 bracketrightBigg + bracketleftBigg 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 werent 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 is any scalar, then ( A ) X = ( AX ) = ( XA ) = ( X ) A = ( X ) A = X ( A ) The set of continuous functions f : R R which are nondecreasing (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 )....
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This note was uploaded on 09/19/2009 for the course MATH 330 taught by Professor Johnson,j during the Spring '08 term at Nevada.
 Spring '08
 Johnson,J
 Math, Linear Algebra, Algebra, Vector Space

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