HW3_sol - CHAPTER 3 SECTION 1 3. To show that C is a vector...

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CHAPTER 3 SECTION 1 3. To show that C is a vector space we must show that all eight axioms are satisfed. A1. ( a + bi )+( c + di )=( a + c b + d ) i =( c + a d + b ) i c + di a + bi ) A2. ( x + y )+ z =[( x 1 + x 2 i y 1 + y 2 i )] + ( z 1 + z 2 i ) x 1 + y 1 + z 1 x 2 + y 2 + z 2 ) i x 1 + x 2 i )+[( y 1 + y 2 i z 1 + z 2 i )] = x +( y + z ) A3. ( a + bi )+(0+0 i a + bi ) A4. IF z = a + bi then defne - z = - a - bi . It Follows that z - z a + bi - a - bi )=0+0 i = 0 A5. α [( a + bi c + di )] = ( αa + αc αb + αd ) i = α ( a + bi α ( c + di ) A6. ( α + β )( a + bi α + β ) a α + β ) bi = α ( a + bi β ( a + bi ) A7. ( αβ )( a + bi αβ ) a αβ ) bi = α ( βa + βbi ) 37

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38 CHAPTER 3 A8. 1 · ( a + bi )=1 · a +1 · bi = a + bi 4. Let A =( a ij ), B b ij ) and C c ij ) be arbitrary elements of R m × n . A1. Since a ij + b ij = b ij + a ij for each i and j it follows that A + B = B + A . A2. Since ( a ij + b ij )+ c ij = a ij +( b ij + c ij ) for each i and j it follows that ( A + B C = A B + C ) A3. Let O be the m × n matrix whose entries are all 0. If M = A + O then m ij = a ij +0= a ij Therefore A + O = A . A4. DeFne - A to be the matrix whose ij th entry is - a ij . Since a ij - a ij )=0 for each i and j it follows that A - A )= O A5. Since α ( a ij + b ij αa ij + αb ij for each i and j it follows that α ( A + B αA + αB A6. Since ( α + β ) a ij = αa ij + βa ij for each i and j it follows that ( α + β ) A = αA + βA A7. Since ( αβ ) a ij = α ( ij ) for each i and j it follows that ( αβ ) A = α ( ) A8. Since 1 · a ij = a ij for each i and j it follows that 1 A = A 5. Let f , g and h be arbitrary elements of C [ a,b ]. A1. ±or all x in [ ] ( f + g )( x f ( x g ( x g ( x f ( x )=( g + f )( x ) . Therefore f + g = g + f
Section 1 39 A2. For all x in [ a,b ], [( f + g )+ h ]( x )=( f + g )( x h ( x ) = f ( x g ( x h ( x ) = f ( x )+( g + h )( x ) =[ f +( g + h )]( x ) Therefore [( f + g h ]=[ f g + h )] A3. If z ( x ) is identically 0 on [ ], then for all x in [ ] ( f + z )( x )= f ( x z ( x f ( x )+0= f ( x ) Thus f + z = f A4. De±ne - f by ( - f )( x - f ( x ) for all x in [ ] Since ( f - f ))( x f ( x ) - f ( x )=0 for all x in [ ] it follows that f - f z A5. For each x in [ ] [ α ( f + g )]( x αf ( x αg ( x ) =( αf )( x αg )( x ) Thus α ( f + g αf + αg A6. For each x in [ ] [( α + β ) f ]( x α + β ) f ( x ) = αf ( x βf ( x ) αf )( x )( x ) Therefore ( α + β ) f = αf + A7. For each x in [ ], [( αβ ) f ]( x αβf ( x α [ ( x )] = [ α ( )]( x ) Therefore ( αβ ) f = α ( ) A8. For each x in [ ] 1 f ( x f ( x ) Therefore 1 f = f

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40 CHAPTER 3 6. The proof is exactly the same as in Exercise 5. 9. (a) If y = β 0 then y + y = β 0 + β 0 = β ( 0 + 0 )= β 0 = y and it follows that ( y + y )+( - y y +( - y ) y +[ y - y )] = 0 y + 0 = 0 y = 0 (b) If α x = 0 and α ± = 0 then it follows from part (a), A7 and A8 that 0 = 1 α 0 = 1 α ( α x ± 1 α α ² x =1 x = x 10. Axiom 6 fails to hold. ( α + β ) x =( ( α + β ) x 1 , ( α + β ) x 2 ) α x + β x ( α + β ) x 1 , 0) 12. A1. x y = x · y = y · x = y x A2. ( x y ) z = x · y · z = x ( y z ) A3. Since x 1= x · x for all x, it follows that 1 is the zero vector. A4. Let - x = - 1 x = x - 1 = 1 x It follows that x ( - x x · 1 x = 1 (the zero vector) . Therefore 1 x is the additive inverse of x for the operation . A5. α ( x y )=( x y ) α x · y ) α = x α · y α α x α y = x α y α = x α · y α A6. ( α + β ) x = x ( α + β ) = x α · x β α x β x = x α x β = x α · x β A7. ( αβ ) x = x αβ α ( β x α x β x β ) α = x αβ A8. 1 x = x 1 = x Since all eight axioms hold, R + is a vector space under the operations of and .
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This note was uploaded on 04/18/2010 for the course FINANCE 1231854365 taught by Professor Wuyiling during the Spring '10 term at Nashville State Community College.

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HW3_sol - CHAPTER 3 SECTION 1 3. To show that C is a vector...

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