Solution01_121

# Solution01_121 - Problem Set 1 Solutions J Scholtz October...

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Problem Set 1 - Solutions J. Scholtz October 1, 2006 Question 1 (1.1/2a) Let ~a = (3 , - 2 , 4) and ~ b = ( - 5 , 7 , 1). Then ~ b - ~a = ( - 8 , 9 , - 3) and so the equation of the line is: l : ~v = ( ~ b - ~a ) t + ~a = ( - 8 , 9 , - 3) t + (3 , - 2 , 4). Question 2 (1.1/3a) This is almost the same as previous question only in one more dimension: The can be parametrized as P : ~v = ( ~ b - ~a ) t + ( ~ c - ~a ) s + ~a . Plugging in the numbers gives us: ~v = ( - 2 , 9 , 7) t + ( - 5 , 12 , 2) s + (2 , - 5 , - 1). Question 3 (1.2/8) Given that x,y V and a,b F then we know we can use various Vectorspace and Field axioms to manipulate this expression. Since a + b F then we can use VS7 to transform ( a + b )( x + y ) = ( a + b ) x + ( a + b ) y . Then by applying VS8 twice we get ax + ay + bx + by . Finally by VS1 we can rearange to get ax + ay + bx + by . Question 4 (1.2/16) If V is a set of all m × n matrices with real coeﬃcients is V a vectorspace over Q ? We need to check whether all the axioms are satisﬁed. However, from the example 2 of the book we know that real matrices form a vectorspace over real numbers so we need to check a little less: VS1, VS2, VS3, VS4 form an abelian group VS5: 1 Q VS6, VS7, VS8: Since Q R as a subﬁeld then these axioms are satisﬁed. Question 5 (1.2/19) This construction does not form a vectorspace because it violates VS8. Consider: ( a + b )( x,y ) = (( a + b ) x, y a + b ). However, by VS7 and VS8, ( a + b )( x,y ) = a ( x,y ) + b ( x,y ) = ( ax,y/a ) + ( bx,y/b ) = (( a + b ) x, y a + y b ) 6 = (( a + b ) x, y a + b ). 1

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Question 6 (1.3/8 c,e) In order to check that a set is a subspace we need to check whether it is closed under vector addition, scalar multiplication, existence of zero and additive in- verses. c) W = { ( a 1 ,a 2 ,a 3 ) , 2 a 1 - 7 a 2 + a 3 = 0 } W = { ( a 1 ,a 2 ,a 3 ) , 2 a 1 - 7 a 2 + a 3 = 0 } is a subspace. To make all the notation shorter we can note that if ~ z = (2 , - 7 , 1), then ~v V iﬀ ~ z.~v = 0 (meaning standard dot product). Then it follows that if both ~a.~ z = ~ b.~ z = 0 then also ( ~a + ~ b ) .~ z = 0. Similarly if ~a.~ z = 0 then ( c~a ) .~ z = c ( ~a.~ z ) = c. 0 = 0. Hence W is closed under vector addition and scalar multiplication. Note that
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Solution01_121 - Problem Set 1 Solutions J Scholtz October...

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