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Unformatted text preview: Let A i M n n ( R ), prove that n i =1 A i T = n i=1 A T i . (We will give two similar proofs illustrating the differences between strong and weak induction. In strong induction, we assume the induction hypothesis holds for k n and establish the result for n + 1. In weak induction, we assume the induction hypothesis holds for k = n and establish the result for n + 1. Explanatory notes will be placed in parentheses.) Proof 1 (strong induction). We begin by considering the case n = 1, and write 1 X i =1 A i T = A 1 T = A T 1 = 1 X i=1 A T i so that the result is trivially true in this case. Next assume that the result k i =1 A i T = k i=1 A T i holds for k n (here is the strong induc tion) and consider the case k = n + 1: n +1 X i =1 A i T = n X i=1 A i + A n + 1 T (Note, by regrouping the first n matrices and considering this to be a single matrix, we have reduced the problem to the case of just two matrices. Here is where we needed strong induction since the result is required to also be true for two matrices, not just for n matrices. Thus we are using the induction hypothesis twice to obtain the following sequence of equalities.) By the induction hypothesis we may therefore write n +1 X i =1 A i T = n X i=1 A i T + A T n + 1 = n X i=1 A T i + A T n + 1 = n+1 X i=1 A T i . Thus the result is established for k = n + 1 and therefore by induction, it is true for all n . Proof 2 (weak induction). We begin by considering the case n = 2, and write 2 X i =1 A i T ij = A 1 + A 2 T ij = A 1 + A 2 ji = A 1 ji + A 2 ji = A T 1 ij + A T 2 ij = A T 1 + A T 2 ij = 2 X i=1 A T i ij . 1 Since the sums of the matrices are component wise equal, they are equal and the result is established. Next assume that the result k i =1 A i T = k i=1 A T i holds for k = n (here is the weak induc tion) and consider the case k = n + 1: n +1 X i =1 A i T = n X i=1 A i + A n + 1 T (Note, by regrouping the first n matrices and considering this to be a single matrix, we have reduced the problem to the case of just two matrices. The case for two matrices was our base case so that the strong induction is not needed here. Thus we are using the induction hypothesis just once to obtain the following sequence of equalities.) By the induction hypothesis we may therefore write n +1 X i =1 A i T = n X i=1 A i T + A T n + 1 = n X i=1 A T i + A T n + 1 = n+1 X i=1 A T i . Thus the result is established for k = n + 1 and therefore by induction, it is true for all n . 2 Section 1.6, Problem 26. For a fixed a R , determine the dimension of the subspace of P n ( R ) defined by W = { f(x) P n ( R ) : f(a) = 0 } . Proof. Consider the set = { ( x a ) , ( x a ) 2 , . . . , ( x a ) n } P n ( R ). By the definition of W , W . By the lemma, the vectors in are linearly independent so the Replacement Theorem gives n dim( W ). By Theorem 1.11, dim( W ) dim( P n ( R )) = n + 1. Next consider the vector 1 P n ( R ); this vector is not an element of...
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This note was uploaded on 02/14/2012 for the course MAS 4105 taught by Professor Rudyak during the Spring '09 term at University of Florida.
 Spring '09
 RUDYAK

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