Adv Alegbra HW Solutions 140

Adv Alegbra HW Solutions 140 - C A n be an ( n , M , d...

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140 Solution. If A v = c v , then A m v = c m v for all m 1. Hence, if A m = 0, then c m = 0 (because eigenvectors are nonzero). Thus, c = 0. Conversely, if all the eigenvalues of an n × n matrix A are 0, then the characteristic polynomial is h A ( x ) = ( x 0 ) ··· ( x 0 ) = x n . By the Cayley-Hamilton theorem, A n = 0, and so A is nilpotent. 4.65 If N is a nilpotent matrix, prove that I + N is nonsingular. Solution. De f ne N 1 = I N + N 2 N 3 +···+ N m 1 , where N m = 0. 4.66 Let A be an alphabet with | A |= q 2, let T : A n A n be a transmission function, and let the probability of error in each letter of a transmitted word be p , where 0 < p < 1. (i) Prove that the probability P of the occurrence of exactly ` erro- neous letters in a transmitted word of length n is P = µ p q 1 ` ( 1 p ) n ` . Solution. Absent. (ii) Prove that P = ³ n ` ´ p ` ( 1 p ) n ` , and conclude that the probabil- ity P is independent of q . Solution. Absent. 4.67 Prove that d 3, where d is the minimum distance of the two-dimensional parity code in Example 4.105(iii). Solution. Absent. 4.68 Let A be an alphabet with | A |= q , and let
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Unformatted text preview: C A n be an ( n , M , d )-code. (i) Prove that the projection : C A n d + 1 , de f ned by ( c 1 , . . . , c n ) = ( c d , . . . , c n ), is an injection. Solution. Let c = ( c 1 , . . . , c n ) and c = ( c 1 , . . . , c n ) lie in C . If ( c ) = ( c ) , then ( c d , . . . , c n ) = ( c d , . . . , c n ) ; that is, c and c agree in at least n d + 1 positions, and so ( c , c ) d 1 &lt; d . Since d is the minimum distance, c = c and is an injection. (ii) ( Singleton bound . ) Prove that M q n d + 1 . Solution. M = | C | , and part (i) gives | C | q n d + 1 . 4.69 (i) If A is an alphabet with | A | = q and is the Hamming distance, prove that w A n : ( u , w) = i = n i ( q 1 ) i ....
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