10aAss2sol

10aAss2sol - AS2sol/MATH1111/YKL/10-11 THE UNIVERSITY OF...

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AS2sol/MATH1111/YKL/10-11 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1111: Linear Algebra Assignment 2 Suggested Solution 1. (a) True. adj A T = ± ( A T ) ij ² T = ± A ji ² T = ± adj A ² T . (b) True. "Only if" part: r ( A ) = n ) A = E k ±±± E 1 where E i ’s are elementary matrices ) det A 6 = 0 ) det(adj A ) 6 = 0 ) adj A is invertible ) adj A = F k ±±± F 1 I where F i ’s are elementary matrices ) r (adj A ) = n: "If" part: Check that the last three implications can obviously go backward. r (adj A ) = n ) adj A is invertible : Claim : adj A is invertible ) A is invertible. Proof . Assume A is not invertible. Then det A = 0, so A (adj A ) = (det A ) I = 0 I = 0 (a zero matrix). So adj A is invertible ) A = 0(adj A )) ± 1 = 0. When A = 0, it is found from de±nition that adj A = 0. So adj A is not invertible. Contradiction arises. Hence, r (adj A ) = n ) A is invertible ) A is row equivalent to I . So r ( A ) = n . (c) False. Counterexample: Let A = 0 @ 1 0 0 0 0 0 0 0 0 1 A . Then r ( A ) = 1 and adj A = 0 so r (adj A ) = 0. Remark. "Only if" part is true: r ( A ) = 0 ) A = E k ±±± E 1 0 where E i ’s are elementary matrices ) A = 0 ) adj A = 0 ) r (adj A ) = 0 : 1

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2. Only (A3), (A4) and (A5) hold. Veri±cation: (A3): Set 0 = (0 ±±± 0) T 2 R n , then x ± 0 = x for all x 2 R n (A4): For each x 2 R n , set ² x = x , then x ± ( ² x ) = 0 ; the zero element in (A3). (Note: here ² x is not the usual ² x , you should view ² x as a symbol (or an element in R n ).) (A5): For each x ; y 2 R n , and any scalar ± , ± ² ( x ± y ) = ± ² z where z = x ² y = ² ± z = ² ± ( x ² y ) = ² ± x + ± y ; ( ± ² x ) ± ( ± ² y ) = ( ² ± x ) ± ( ² ± y ) = ² ± x + ± y : ) ± ² ( x ± y ) = ( ± ² x ) ± ( ± ² y ). Qn. Why do other axioms not hold? 3. (a) Yes. (A3) Take 0 = 1 2 V , then for any x 2 V , we have x ± 0 = x ± 1 = x : (A4) Let x 2 V . Set ² x = x ± 1 (the usual reciprocal, i.e. 1 = x), then x ± ( ² x) = x ± x ± 1 = 1 = 0 : (b) Yes.
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