101st_tut2sol

101st_tut2sol - MATH1111/2010-11/Tutorial I Solution 1...

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MATH1111/2010-11/Tutorial I Solution 1 Tutorial I Suggested Solution 1. Let A and B be two matrices satisfying AB = ± BA . (a) Must A and B be square matrices of the same order? Explain. (b) Show that if the order is odd, then A or B is singular. (c) Must A or B be singular if the order is even? Justify your answer. Ans . (a) Yes. Let A be of size m ² r and B be n ² s . As AB and BA are well-de±ned matrix products, r = n and m = s . Then AB is of dimension m ² s and BA is n ² r . From AB = ± BA , we conclude m = n and s = r . So m = r = n = s . A and B are square matrices of the same order. (b) Let the order of A and B be n . Taking determinant, then det( ± BA ) = ( ± 1) n det( BA ) = ( ± 1) n (det B )(det A ) : Given AB = ± BA and n is odd (so ( ± 1) n = ± 1). We deduce that (det A )(det B ) = det( AB ) = det( ± BA ) = ± (det B )(det A ) : Thus 2(det A )(det B ) = 0. At least one of det A and det B equal 0. i.e. A or B (maybe both) is singular. (c) No. Here is a counterexample: A = ± 0 1 1 0 ² and B = ± 1 ± 2 2 ± 1 ² . 2. Let V be a vector space and x 2 V . A student says that it is ambiguous to write ± 2 x , because one may interpret ± 2 x in di²erent ways, such as ± (2 x ) ; ( ± 1)(2 x ) ; ( ± 2) x ; 2( ± x ) ; (2 ³ ( ± 1)) x ; ³³³ What is your comment? Give your justi±cation. Ans . No ambiguity. It is because all di²erent interpretations are equal. For instance, ± (2 x ) is the element satisfying 2 x + ³ ± (2 x ) ´ = 0 (the zero vector in V ). So ± (2 x ) = ( ± 1)(2 x ) by Theorem 3.1.1 (iii). For your own good, prove that it is equal to all the others.
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MATH1111/2010-11/Tutorial I Solution 2 3. Are the following sets vector spaces with the indicated operations? If not, why not?
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This note was uploaded on 05/04/2011 for the course MATH 1111 taught by Professor Dr,li during the Spring '10 term at HKU.

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101st_tut2sol - MATH1111/2010-11/Tutorial I Solution 1...

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