1. Find a basis for the row space of each of the following matrices. Your basis should consist of rows of the original matrix.


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1 Mh 7 Exm  Prcice Prolem Solions. Find sis for he row spce of ech of he following mrices. Yor sis shold consis of rows of he originl mrix. 4 () () Since we wn sis for he row spce consising of rows of he originl mrix, we will ke A T nd p i ino redced row echelon form. The posiion of he s in he redced row echelon form indice he rows of A h form sis for he row spce. We omi he deils of he row redcion process. A T From he redced form ove, he following is sis for he row spce: { 4, 7 } Since we wn sis for he row spce consising of rows of he originl mrix, we will ke A T nd p i ino redced row echelon form. The posiion of he s in he redced row echelon form indice he rows of A h form sis for he row spce. We omi he deils of he row redcion process. A T From he redced form ove, he following is sis for he row spce: { 4, 4, }. Find sis for he colmn spce of ech of he mrices from prolem. Yor sis shold consis of colmns of he originl mrix. () Since we wn sis for he colmn spce consising of colmns of he originl mrix, we will ke A nd p i ino redced row echelon form. The posiion of he s in he redced row echelon form indice he colmns of A h form sis for he row spce. We omi he deils of he row redcion process. A From he redced form ove, he following is sis for he colmn spce: () Since we wn sis for he colmn spce consising of colmns of he originl mrix, we will ke A nd p i ino redced row echelon form. The posiion of he s in he redced row echelon form indice he colmns of A h form sis for he row spce. We omi he deils of he row redcion process. A , 7
2 From he redced form ove, he following is sis for he colmn spce:, 4, 4 4. Find he rnk of ech of he mrices from prolem. () From he work done ove, we see h rnka. () From he work done ove, we see h rnka. 4. Find sis for he nll spce of ech of he mrices from prolem. () To find sis for he nll spce, we need o row redce he mrix represening he reled homogeneos sysem of eqions. Since we lredy p A in redced row echelon form, we cn js dd n gmened colmn of zeros o he redced row echelon form of A: 8 From his, we see h x x 8 x 4 nd x x + x 4. Le x s nd x 4. s 8 Then elemens of he nll spce hve he form: s+ s. Therefore, he following se is sis for he nll 8 spce:, () find sis for he nll spce, we need o row redce he mrix represening he reled homogeneos sysem of eqions. Since we lredy p A in redced row echelon form, we cn js dd n gmened colmn of zeros o he redced row echelon form of A: 6 6 From his, we see h x 6x 4 nd x x 4 nd x 6x 4. Le x 4. 6 Then elemens of he nll spce hve he form: 6. Therefore, he following se is sis for he nll 6 spce: 6. Use he deerminn o deermine wheher or no ech of he following mrices hs rnk : () 6 Noice h de(a) ++ ( ) ( 4) Since de(a), A is inverile, nd hence redces o I. Ths rnka.
3 () Noice h de(a) (8) ( 4) Since de(a), A is singlr nd is redced form hs les one row of zeros. Ths rnka <. 6. Use he rnk of he coefficien mrix o deermine wheher or no ech of he following homogeneos sysems hve nonrivil solion. 4 () A x if A 7 We find he rnk of A y ping A ino redced row echelon form: 4 A 7 Noice h A hs rnk 4, so A hs nlliy. Th is, he reled homogeneos sysem of eqions hs only he rivil solion. 4 () A x if A We find he rnk of A y ping A ino redced row echelon form: A Noice h A hs rnk, so A hs nlliy. Th is, here re nonrivil solions o he reled homogeneos sysem of eqions. 7. Sppose h A is 4 7 mrix. () Wh is he mximm rnk of A? Since A hs 4 rows nd 7 colmns, he lrges ideniy mrix h fis inside A is I 4. Therefore, he mximm rnk of A is 4. () Cold he colmns of A e linerly independen? Jsify yor nswer. No. Since here re 7 colmns nd he mximm rnk of A is 4, some of he colmns ms e liner cominions of he oher colmns. (c) Cold he rows of A e linerly independen? Jsify yor nswer. Yes. If A ins is mximm rnk of 4, hen he rows wold e linerly independen. (d) If he rnk of A is, find he nlliy of A. If he rnk of A is, hen he redced row echelon form of A hs hree s. Then here ms e 4 free vriles, so he nlliy of A is 4. (e) If he rnk of A is, find he nlliy of A T. If he rnk of A is, hen he rnk of A T is lso. Then he redced row echelon form of A T lso hs hree s. Since A T hs 7 rows nd 4 colmns, here is one free vrile, so A T hs nlliy.
4 8. Le A e n n n mrix. Show h rnk A n if nd only if he colmns of A re linerly independen. Le A e n n n mrix nd firs sppose h rnka n. Then A is row eqivlen o I n. Therefore, every colmn of A is vecor in sis for he colmn spce of A. Ths he colmns of A re linerly independen. Now, le A e n n n mrix nd sppose h he colmns of A re linerly independen. By definiion, he colmns of A spn he colmn spce of A. Since hese colmns re linerly independen, he n colmns of A form sis for he colmn spce of A. Then A hs colmn rnk n. However, he colmn rnk of A is eql o he rnk of A. Ths A hs rnk n. 9. Consider he following fncions from R R : L L L L 4 () Deermine wheher or no ech of hese fncions is liner rnsformion. Jsify yor nswer. i. L v NoicehL ( )+L ( v) L +L v v v + +v v v v +v ( +v ) ( +v ) ( +v ) +v Similrly, L ( + v) L +v ( +v ) ( +v ) L ( +v +v ) ( )+L ( v). Hence propery () holds. Nex, noice h L (c ) L c c c c c c () lso holds. Therefore, L is liner rnsformion. ii. L NoicehL ( )+L ( v) L ( +v ) ( +v ) +L v v v c v v + c L ( ). Ths propery +v v + Similrly, L ( + v) L holds. +v +v +v Nex, noice h L (c ) L ( +v ) ( +v ) L ( )+L ( v). Hence propery () c c c c c () lso holds. Therefore, L is liner rnsformion. iii. L c c L ( ). Ths propery
5 Noice h L By Theorem 6.(), he imge of nder ny liner rnsformion ms e he zero vecor. Hence L is no liner rnsformion. iv. L 4 Noice h L 4 On he oher hnd, L 4 L 4 9 Therefore, L 4 is no liner rnsformion. () For ech L i his is liner rnsformion, find mrix represening he liner rnsformion. i. L Noice h L, L, nd L Then,kingheseimgesofhecolmnsoformrix,hefollowingmrixrepresensL : A ii. L Noice h L, L, nd L Then,kingheseimgesofhecolmnsoformrix,hefollowingmrixrepresensL : A (c) For ech L i his is liner rnsformion, find he kernel nd rnge of he liner rnsformion. i. L Noice h o e in kerl, we ms hve nd. Therefore, kerl To find he rnge, noice h if R, hen if we ke, nd, hen L. Hence L is ono. Th is, rngel R. ii. L : R. NoicehoeinkerL,wemshve,nd cnenyhing. Therefore,kerL s s : s, R.
6 Alhogh his ws no sked, one shold noice h he kernel of L hs dimension, while he kernel of L hs dimension. To find he rnge, noice h if R, hen we ms hve. If we ke, nd, { } hen L. Hence rngel : R.. Sppose h L is liner rnsformion wih: L, L, nd L We will clly sr y doing pr (): () Find he sndrd mrix represening his liner rnsformion. Noice h L Similrly, L L L From his, he mrix represening his liner rnsformion is: A () Find L. Using he mrix we fond in pr (), L (c) Deermine wheher or no L is oneoone. Noice h de(a) Since de(a), hen kera { }. Hence L is oneoone. (Alhogh his ws no sked, yo shold noice h since L is rnsformion from R o R, since L is oneoone, i is lso ono). 6. Le L + + () Prove h L is liner rnsformion. v Noice h L( )+L( v) L +L v + + v +v + +v v + + +v +v +v ( +v )+( +v )+( +v ) + +v Similrly, L( + v) L +v ( +v )+( +v )+( +v ) L( ) + L( v). Hence propery +v () holds.
7 Nex, noice h L(c ) L c c c holds. Therefore, L is liner rnsformion. c +c +c + c +. Ths propery () lso () Show h L is no oneoone nd find sis for kerl. To e in kerl, we need + +. Th is, we ms hve. Le s nd. s Then kerl s : s, R. Therefore, he following se is sis for kerl:,. Noe h since ker,l { }, L is no oneoone. (c) Deermine wheher or no L is ono. + Noice h every elemen or he rnge of L is of he form + From his, we see h is no in he rnge of L. Hence L is no ono.. Le L + () Prove h L is liner rnsformion. Noice h L( )+L( v) L +L ( +v )+( +v ) ( +v ) ( v ) +v Similrly, L( + v) L +v +v Nex, noice h L(c ) L c c c Therefore, L is liner rnsformion. v v v + ( +v )+( +v ) ( +v ) ( +v ) () Show h L is no oneoone nd find sis for kerl. v +v + v v c +c + c c c + +v +v +v v L( )+L( v). Hence propery () holds.. Ths propery () lso holds. To e in kerl, we need + nd. Th is, we ms hve nd. Le. Then kerl : R. Therefore, he following se is sis for kerl:. Noe h since ker,l { }, L is no oneoone. (c) Deermine wheher or no L is ono.
8 Le Hence L is ono. R. Noice h if, nd, hen L.. Prove Theorem 6.() Theorem 6.(): Le L : V W e liner rnsformion. Then L( V ) W. Proof: Recll h v + V V. Then L( V ) L( V + V ) L( V )+L( V ), where he ls eqliy ses propery () of liner rnsformions. Then L( V ) L( V ) L( V )+L( V ) L( V ). So W L( V ).. 4. Prove Theorem 6.4() Theorem 6.4(): Le L : V W e liner rnsformion. L is oneoone if nd only if kerl { V }. Proof: Firs, sppose h L is oneoone. Le v kerl. Then L( v) W. Recll from Theorem 6.() ove h L( v ) W. Therefore, since L is oneoone, we ms hve v v. Hence kerl { V }. Nex, sppose h kerl { V }. Sppose L( v ) L( v ) for some v,v V. Then, srcing, L( v ) L( v ) w. Th is, sing Theorem 6.(), L( v v ) w. Therefore, v v is n elemen of kerl. Hence, since kerl { V }, we ms hve v v V, so v v. Ths L is oneoone... Given liner rnsformion L : V W, sppose h dimv n nd dimw m wih m > n. () Cold L e oneoone? Jsify yor nswer. Yes. Recll h L is oneoone if nd only if kerl { V }. Also, dim(kerl) + dim(rngel) dimv n. Since dimw m > n, i is possile o hve dim(rngel) n. This wold mke dim(kerl), nd hence L wold e oneoone. () Cold L e ono? Jsify yor nswer. No. Recll h L is ono if nd only if rngel W. Then we ms hve dim(rngel) dimw m. However, he mximm possile vle for dim(rngel is n. Since dimw m > n, i is no possile o hve dim(rngel) n. Hence L cnno e ono. (c) Cold L e inverile? Jsify yor nswer. No. Recll h L is inverile if nd only if L is oh oneoone nd ono. We showed in pr () ove h L cnno e ono. Hence L is no inverile. 6. Consider he mrix: 4 () Find he chrcerisic polynomil for ech of his mrix. 4 λ 4 Noice h if A, hen λi A λ. Then de(λi A) (λ 4)λ+ λ 4λ+. Ths p(λ) λ 4λ+ is he chrcerisic polynomil for his mrix. () Find he eigenvles for his mrix. Noice h p(λ) λ 4λ+ (λ )(λ ). Therefore, he eigenvles re he solions o he chrcerisic eqion p(λ). Th is, solions o (λ )(λ ) Hence he eigenvles re λ nd λ.
9 (c) For ech eigenvle, find n ssocied eigenvecor. To find eigenvecors for ech eigenvle, we ssie ech eigenvle ino he mrix λi A nd hen find he nll spce of he ssied version of he mrix. If we ke λ, hen λi A Afer dding n gmened row of zeros nd hen ping ino redced row echelon form, we oin: Hence, we hve x x nd x is free, so le x. Then he eigenvecors ssocied wih his eigenvle re ll of he form: one possile eigenvecor is x If we ke λ, hen λi A {. }. For exmple, if we se, Afer dding n gmened row of zeros nd hen ping ino redced row echelon form, we oin: Hence, we hve x x nd x is free, so le x. Then he eigenvecors ssocied wih his eigenvle re ll of he form: one possile eigenvecor is x {. }. For exmple, if we se, 7. Consider he mrix: () Find he chrcerisic polynomil for ech of his mrix. Noice h if A λ+, hen λi A λ λ. Then (sing Mple o sve lile work) de(λi A) λ λ λ+. Ths p(λ) λ λ λ+ is he chrcerisic polynomil for his mrix. () Find he eigenvles for his mrix. Noice h p(λ) λ λ λ+ (λ+)(λ λ+). Therefore, he eigenvles re he solions o he chrcerisic eqion p(λ). Hence he eigenvles re λ nd λ ±, where he ls wo come from pplying o he qdric porion of he fcored form of p(λ). (c) For ech eigenvle, find n ssocied eigenvecor. To find eigenvecors for ech eigenvle, we ssie ech eigenvle ino he mrix λi A nd hen find he nll spce of he ssied version of he mrix. If we ke λ, hen λi A 4 Afer dding n gmened row of zeros nd hen ping ino redced row echelon form, we oin:. Hence, x is free, x, nd x. so le x.
10 Then he eigenvecors ssocied wih his eigenvle re ll of he form: one possile eigenvecor is x If we ke λ +, hen λi + A + +. For exmple, if we se, + Afer dding n gmened row of zeros nd hen ping ino redced row echelon form, we oin: +. Hence, x is free, x +, nd x +. so le x. + Then he eigenvecors ssocied wih his eigenvle re ll of he form: +. For exmple, if we se, one possile eigenvecor is x If we ke λ, hen λi A Afer dding n gmened row of zeros nd hen ping ino redced row echelon form, we oin: +. Hence, x is free, x +, nd x +. so le x. Then he eigenvecors ssocied wih his eigenvle re ll of he form: se, one possile eigenvecor is x For exmple, if we
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