# sol7 - MATH 235 – Algebra I Solutions to Assignment 7...

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Unformatted text preview: MATH 235 – Algebra I Solutions to Assignment 7 November 5, 2007 Solution of 1. (1) To see that f is bijective, note that a matrix in the set z 1 z 2- z 2 z 1 : z 1 ,z 2 ∈ C is uniquely determined by its first row ( z 1 ,z 2 ). Also note that each first row ( a + bi,c + di ) comes, via f , from exactly one expression of the form a + bi + cj + dk , so every matrix in the set comes from exactly one element of H and therefore f is a bijection. Now let x = a x + b x i + c x j + d x k , and y = a y + b y i + c y j + d y k , so f ( x + y ) = f (( a x + a y ) + ( b x + b y ) i + ( c x + c y ) j + ( d x + d y ) k ) = ( a x + a y ) + ( b x + b y ) i ( c x + c y ) + ( d x + d y ) i- ( c x + c y ) + ( d x + d y ) i ( a x + a y )- ( b x + b y ) i = a x + b x i c x + d x i- c x + d x i a x- b x i + a y + b y i c y + d y i- c y + d y i a y- b y i = f ( x ) + f ( y ) . Also, once we notice that f ( i ) = i- i , f ( j ) = 1- 1 0 , f ( k ) = i i , simple computations (which we omit here) show that f ( i ) 2 = f ( j ) 2 = f ( k ) 2 =- I 2 and f ( i ) f ( j ) =- f ( j ) f ( i ) = f ( k ). (2) To show that H satisfies the ring axioms, we use the fact that the ring axioms hold in M 2 ( C ), and the fact that the multiplication and addition in H are “the same” as that in M 2 ( C ). We already proved in (1) that f ( x + y ) = f ( x )+ f ( y ), which shows that the addition is “the same” in both structures. To see that the multiplication is “the same”, we must prove that f ( xy ) = f ( x ) f ( y ). First observe that if r is a real number, we have f ( rx ) = ra x + rb x i rc x + rd x i- rc x + rd x i ra x- rb x i = r a x + b x i c x + d x i- c x + d x i a x- b x i = rf ( x ) . 1 If x = a x + b x i + c x j + d x k and y = a y + b y i + c y j + d y k , then by the fact that f ( x + y ) = f ( x ) + f ( y ) and f ( rx ) = rf ( x ), we have f ( x ) = a x I 2 + b x f ( i ) + c x f ( j ) + d x f ( k ), and f ( y ) = a y I 2 + b y f ( i ) + c y f ( j ) + d y f ( k ). Then xy = a x a y + a x b y i + a x c y j + a x d y k + b x a y i + b x b y i 2 + b x c y ij + b x d y ik + c x a y j + c x b y ji + c x c y j 2 + c x d y jk + d x a y k + d x b y ki + d x c y kj + d x d y k 2 = a x a y- b x b y- c x c y- d x d y + ( a x b y + b x a y + c x d y- d x c y ) i +( a x c y + c x a y + d x b y- b x d y ) j + ( a x d y + d x a y + b x c y- c x b y ) k where the last step uses multiplication in H . For example ik = i ( ij ) = i 2 j =- j . So, using f ( x + y ) = f ( x ) + f ( y ) and f ( rx ) = rf ( x ), f ( xy ) = ( a x a y- b x b y- c x c y- d x d y ) I 2 + ( a x b y + b x a y + c x d y- d x c y ) f ( i ) +( a x c y + c x a y + d x b y- b x d y ) f ( j ) + ( a x d y + d x a y + b x c y- c x b y ) f ( k ) ....
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sol7 - MATH 235 – Algebra I Solutions to Assignment 7...

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