Unformatted text preview: SOLUTION To see what is going on here, we square the matrix 1 2a . Cubing it we next ﬁnd 01 1 0 1a . The result is 01 3a . The pattern is clear. We claim that 1
n 1a 01 = 1 na 01 . (∗) We have checked this for n = 2 and n = 3, and it is a tautology for n = 1. To handle the general case, we proceed by induction. Assuming that (∗) holds with n replaced by n − 1, we have n n−1 1a 1a 1a 1 (n − 1)a 1a = = . 01 01 01 0 1 01 Multiplying out the last two matrices we see that (∗) is true provided it is true with n replaced by n − 1. This give us the inductive step, and establishes (∗) for all n. Now it is easy to solve (a): With a = 1/n, we see from (∗) that 1 0 1/n 1
n = 1 0 1 1 for all n. It is also easy to solve (b): With a = 1/n3 , we see from (∗) that 1 0 for all n. Therefore,
n→∞ 1/n3 1 n = 1 0 1/n2 1 lim 1 1/n3 0 1 n 1 = 0 0 1 n . 3.27 For any n × n matrix A, deﬁne the trace of A to be the number tr(A) given by
N tr(A) =
i=1 Ai,i . Show that for any two n × n matrices B and C , tr(BC ) = tr(CB ) even if it is the case that BC = CB . SOLUTION We have that (BC )i,i =
j =1 Bi,j Cj,i (∗) 21/september/2005; 22:09 26 ...
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.
 Fall '08
 Gladue
 Calculus

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