10aAss1sol

10aAss1sol - AS1sol/MATH1111/YKL/10-11 THE UNIVERSITY OF...

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AS1sol/MATH1111/YKL/10-11 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1111: Linear Algebra Assignment 1 Suggested Solution Remark : You may ±nd that the solutions (especially the proofs) are rather lengthy. It is because we hope to explain every step detailedly. After reading each answer, you should dig out the central ideas and work out the answer again by your words. 1. (a) Let x = ( x y ) T 2 R 2 . To show that x is a linear combination of x 1 and x 2 , we need to prove that x = c 1 x 1 + c 2 x 2 is solvable for some scalars c 1 ;c 2 , i.e. ± x = c 1 c + c 2 a y = c 1 d + c 2 b has a solution (for c 1 ;c 2 ). The augmented matrix is ² c a x d b y ³ : As x 2 6 = 0 , at least one of c or d is nonzero. Case 1. c 6 = 0. Then the row operations 1 c R 1 and ± dR 1 + R 2 gives ² 1 a c x c 0 b ± da c y ± dx c ³ : Note that b ± da c 6 = 0, for otherwise, x 2 = ( a b ) T = a c ( c d ) T = a c x 1 (contradicting to the condition x 1 6 = ±x 2 for any ± 2 R ). Thus the operation ( b ± da c ) ± 1 R 2 further transforms the augmented matrix into ² 1 a c x c 0 1 ( b ± da c ) ± 1 ( y ± dx c ) ³ : From it we solve for c 1 and c 2 . Case 2. d 6 = 0. The proof is similar and the details are left to you. (b) In case x 1 is a scalar multiple of x 2 , there are (plenty of) vectors that cannot be written as a linear combination of x 1 and x 2 . Below is an example. Choose u;v 2 R such that av 6 = bu , which can be done because at least one of a;b is not equal to 0. Take x = ( a + u b + v ) T . We prove the following. Claim: x is not a linear combination of x 1 and x 2 . 1
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Proof. Suppose not. i.e. x = ±x 1 + ²x 2 where ±;² 2 R . As x 1 = cx 2 for some constant c , we have x = ( c + 1) x 2 . This implies a + u = ( c + 1) a and b + v = ( c + 1) b: Or equivalently ca = u and cb = v . This implies cav = cbu ; as av 6 = bu , we conclude c = 0 and thus x 1 = cx 2 = 0 . Contradiction arises. 2. As A is nonsingular, we get by multiplying A ± 1 to both sides of A 3 = A . We have A 2 = I . [If then, you answer A = ± I . Well this question is signi±cant to you because you are WRONG . Life is not so simple in matrices.] To see what A can satisfy the equation, we carry out a brute-force calculation. Let A = ± a b c d ² : Then A 2 = ± a 2 + bc b ( a + d ) c ( a + d ) d 2 + bc ² . From A 2 = I , we conclude a 2 + bc = 1 ; b ( a + d ) = 0 = c ( a + d ) ; d 2 + bc = 1 : Case 1. b = 0. Then a 2 = 1 = d 2 , so a;d can be ± 1. We have b a d a + d c A 0 1 1 2 0 ± 1 0 0 1 ² 0 1 ² 1 0 any ± 1 0 c ² 1 ² 0 ² 1 1 0 any ± ² 1 0 c 1 ² 0 ² 1 ² 1 ² 2 0 ± ² 1 0 0 ² 1 ² Case 2. c = 0. We consider A T . Note that A 2 = I implies ( A T ) 2 = I . Applying Case 1, we get A = ± ± 1 0 0 1 ² ; or A = ± ± 1 c 0 ² 1 ² where c can be any real number.
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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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10aAss1sol - AS1sol/MATH1111/YKL/10-11 THE UNIVERSITY OF...

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