# MTH 5102 Linear Algebra Practice Exam 1 - Solutions Feb. 9, 2016

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1 Nme (Lst nme, First nme): MTH 502 Liner Algebr Prctice Exm - Solutions Feb 9, 206 Exm Instructions: You hve hour & 0 minutes to complete the exm There re totl of 6 problems You must show your work Prtil credit my be given even for incomplete problems s long s you show your work () Give the de nition of vector spce V over eld F ; (b) Write out explicitly the de nitions of ddition nd sclr multipliction for the vector spces (i) f0g; (ii) F n ; (iii) M mn (F ); (iv) F (S; F ), where S is ny nonempty set; (v) P (F ) nd P n (F ); (vi) (R) nd n (R); (vii) The set of symmetric nn mtrices over eld F ; (viii) n s vector spce over the complex numbers ; (ix) n s vector spce over the rel numbers R; (c) Prove tht P n (F ) is subspce of P (F ) nd both re subspce of F (F; F ); (d) Prove tht n (R) is subspce of (R) nd both re subspces of F (R; R); (e) Prove tht the set in (vii) is subspce of M nn (F ) Solution () See the book; (b): We will denote c 2 F then (i) = 0 nd c0 = 0; (ii) ( ; : : : n )+(b ; : : : ; b n ) = ( + b ; : : : ; n + b n ) nd c ( ; : : : n ) = (c ; ; c n ); (iii) n b b A m mn b m b mn nd 0 n m mn A = A + b n + b n m + b m mn + b mn 0 c c c m c mn (iv) f + g nd cf for f; g 2 F (S; F ) re the functions (f + g) (s) = f (s) + g (s) nd (cf) (s) = c [f (s)] for ll s 2 S; (v) sme de nition s in F (F; F ); (iv) sme de nition s F (R; R); (vii) sme de nition s M nn (F ); (viii) sme de nition s F n with F = ; (ix) sme de nition s F n with F = but sclr multipliction is restricted to sub eld R of the eld ; (c) It is not true tht P n (F ) nd P (F ) cn both cn be treted s subsets of F (F; F ) by f (x) = n x n when treting x s n independent vrible in domin F is function from F into F In fct, consider the eld F = f0; g of chrcteristic 2 then x = x 2 in F (F; F ), but, x, x 2 re linerly independent in P 2 (F ) = spn ; x; x 2 The correct identi ction of P n (F ) nd P (F ) is with the subspce of F (N; F ) in Exmple 5 Sec 2 of the book by identitfying f (x) = m x m with the sequence f n g with n = 0 if n > m; (d) n (R) is the set of ll functions from R into R whose derivtives up to the nth order (with A ; A

2 n 0) re continuous with 0 (R) = (R) the set of ll continuous functions, in prticulr, this implies n (R) F (R; R) nd since if f; g 2 n (R) nd ; b 2 R then f + bg hs ll its derivtives up to the nth order (with n 0) re continuous by elementry results in single-vrible clculus (ie, sum of continuous functions is continuous, the sum of the derivtives is the derivtives of the sum, etc) implying f + bg 2 n (R) implying n (R) is subspce of (R) which is subspce of F (R; R) for ll n 0; (e) if A nd re symmetric nn mtrices over F nd c 2 F then in prticulr A t = A, t = 2 M nn (F ) nd (ca + ) t = ca t + t = ca+ (since the trnspose opertion on M nn (F ) into M nn (F ) is liner trnsformtion) implying ca + is symmetric n n mtrix over F implying the set in (vii) is subspce of M nn (F ), s desired This completes the proof 2

3 2 () Prove the intersection W \ W 2 of ny two subspces W nd W 2 of vector spce V over eld F is subspce of V ; (b) Find necessry nd su cient conditions for the union W [ W 2 to be subspce of V ; (c) Provide counterexmple to the sttement: If W nd W 2 re subspces of vector spce V over eld F then W [ W 2 is subspce of V over F Proof () First, W \W 2 V Next, if ; b 2 F nd u; v 2 W \W 2 then, since W nd W 2 re vector spces over F contining both u nd v nd hence re closed under liner combintions of u nd v, we must hve u + bv 2 W \ W 2 Therefore, this proves W \ W 2 is subspce of V (b) if W nd W 2 re two subspces of vector spce V over eld F then W [ W 2 is subspce of V if nd only if W i = W [ W 2 for i = or i = 2 Proof Suppose tht W nd W 2 re two subspces of vector spce V over eld F If W i = W [ W 2 for i = or i = 2 then it is subspce of V by hypothesis tht W i is subspce Let us now prove the converse Suppose W [W 2 is subspce of V If W = W [W 2 or W 2 = W [W 2 then we re done Thus, suppose tht W ; W 2 6= W [ W 2 Then we cn nd w 2 W n W 2 nd w 2 2 W 2 n W Hence since by hypothesis W [ W 2 is subspce of V then we must hve w + w 2 2 W [ W 2 implying tht w + w 2 is in W or W 2 ut then either w 2 = (w + w 2 ) w is in W or w = (w + w 2 ) w 2 is in W 2 contrdicting tht w 2 W n W 2 nd w 2 2 W 2 n W Therefore, this contrdiction proves tht W = W [ W 2 or W 2 = W [ W 2 This completes the proof ounterexmple (c) Let W = f(; 0) : 2 Rg nd W 2 = f(0; b) : b 2 Rg Then these re subspces of R 2, but W [ W 2 = f(; b) : ; b 2 R nd either = 0 or b = 0g is not subspce since if 2 R nd 6= 0 then (; 0); (0; ) 2 W [ W 2 but (; 0) + (0; ) = (; ) 62 W [ W 2 so tht W [W 2 is not closed under the opertion of ddition in R 2 nd therefore cnnot be subspce of R 2 3

4 3 () Suppose V is vector spce over eld F nd let S be subset of F Give the de nition of: (i) the spn of S; (ii) S is linerly dependent/independent; (iii) S genertes V ; (iv) S is bsis for V ; (v) V is nite-dimensionl; (vi) V is in nite-dimensionl Solution 2 See book 4

5 4 () Prove, for the following exmples of vector spces, tht they re nitedimensionl by nding bsis for them nd nd their dimensions: (i) f0g; (ii) F n ; (iii) M mn (F ); (iv) F (S; F ), where S is ny set with n elements; (v) P n (F ); (vi) n over ; (vi) n over R; (vii) the set of symmetric nn mtrices over eld F ; (b) Prove, for the following exmples of vector spces, tht they re in nite-dimensionl: (iv) F (S; F ), where S = fv ; v 2 ; : : :g is n in nite set with distinct elements v, v 2, ; (v) P (F ); (vi) (R) nd n (R) Proof (): (i) A bsis for f0g is the empty set ; nd so dim (f0g) = 0; (ii) A bsis for F n is the stndrd bsis f(; 0; : : : ; 0) ; : : : ; (0; 0; : : : ; )g nd since there re n vectors in this bsis then dim (F n ) = n; (iii) A bsis for M mn (F ) is E ij : i m, j n, where E ij is the m n mtrix with zeros in ll entries except for in the ith row nd jth column entry nd so dim (M mn (F )) = mn; (iv) Let S = fx ; : : : ; x n g be set of n distinct elements x ; : : : ; x n De ne the functions i : S! F by i (x) = 0 (the dditive identity element of the eld F ) if x 6= x i nd i (x i ) = (the multiplictive identity element of F ) It follows tht f i : i ng is bsis for F (S; F ) nd so dim (F (S; F )) = n; (v) The stndrd bsis f; x; : : : ; x n g nd so dim (6 P n (F )) = n + ; (vi) Sme s in (ii) but with F = ; (vii) A bsis is f(; 0; : : : ; 0) ; : : : ; (0; 0; : : : ; )g [ f(i; 0; : : : ; 0) ; : : : ; (0; 0; : : : ; i)g, where i = p so dim ( n ) = 2n where n is the vector spce over eld R; (vii) Using (iii), bsis is E ij + E ji : i < j n [ E ii : i n which hs 2n (n + ) elements implying this is the dimension of the set of symmetric n n mtrices over eld F Proof (b): (iv) Use prt ()(iv) with y : S! F for ech y 2 S de ned by y (x) = 0 if x 6= y nd y (y) = The set f y : y 2 Sg is linerly independent in F (S; F ) nd not nite since S is not; (v) P n (F ) P (F ) for ll n with dim P n (F ) = n + which implies P (F ) is in nite-dimensionl; (vi) The functions f m (x) = x m, m re linerly independent in n (R) (for ny n 0) for if X n if i = 0 i= for some i 2 R, i = ; : : : ; n then this mens 0 = X n i= if i (x) = X n i= ix i for ll x 2 R implying the polynomil p (x) = P n i= ix i with rel coe cients hs in nitely mny zeros in R ut fundmentl theorem of polynomils over the eld tells us tht polynomil with coe cients in of degree n 0 hs n zeros in (counting multiplicities) Hence the degree of p (x) must be nd so p (x) is the zero polynomil implying i = 0 for i = ; : : : ; n This proves tht the set of functions ff m : m g is linerly independent in n (R) nd hence n (R) must be in nite-dimensionl This completes the proof One my wish to consider problem (c) in conjunction with the proof of 4(b)(vi) to understnd it better 5

6 5 Using the method to prove Theorem 9 (p 44): Prove tht the set S = f(2; 3; 5) ; (8; 2; 20); g (; 0; 2); (0; 2; ); (7; 2; 0)g genertes R 3 Also, nd bsis for R 3 which is subset of S Proof See exmple 6 in Sec 6 of the book 6

7 6 () Stte nd prove the Dimension Theorem; (b) Let T : P 3 (R)! P 2 (R) nd T 2 : P 2 (R)! R be the functions T (f (x)) = f 0 (x) ; T 2 (f (x)) = Z b f (x) dx (for b > ): Prove tht these functions re: (i) well-de ned; (ii) liner trnsformtions; (c) Find bsis for R (T j ) nd N (T j ), for j = ; 2 (d) Is the mp T 3 : P 2 (R)! P 3 (R) de ned by T 3 (f) (x) = Z x f (t) dt; x 2 R liner trnsformtion? (e) Is the mp T 4 : P 2 (R)! (R) de ned by T 4 (f) (x) = Z x f (t) dt; x 2 R liner trnsformtion? If so, nd bsis for R (T 4 ) nd N (T 4 ) Proof () see the book; (b): (i) Let p (x) 2 P 3 (R) then p (x) = 3 x x 2 + x+ 0 for some i 2 R, i = 0; ; 2; 3 Then p 0 (x) = 3 3 x x+ 2 P 2 (R) Thus, T : P 2 (R)! P 2 (R) is well-de ned function Similrly since ll the elements of P (R) treted s functions re continuous nd hence R b q (x) dx is well-de ned rel number for ny q (x) 2 P 2 (R) Thus, T 2 : P 2 (R)! R is well-de ned function; (ii) The derivtive nd integrls re liner is obvious from results from single-vrible clculus for functions tht re di erentible nd continuous, respectively, such s ll the elements of P (R) (c) A bsis for P 3 (R) nd P 2 (R) is ; x; x 2 ; x 3 nd ; x; x 2, respectively And T () = 0, T (x) =, T x 2 = 2x, T x 3 = 3x 2 nd hence T (x) ; T x 2 ; T x 3 is bsis for R (T ) nd fg is bsis for N (T ) Similrly, T 2 () = b > 0 nd R is one-dimensionl so tht bsis for R (T 2 ) is ft 2 ()g nd N (T 2 ) = 2 x 2 + x P 2 (R) : 2 3 b b (b ) = 0 y the Dimension Theorem we must hve implying 3 = dim (P 2 (R)) = nullity (T 2 ) + rnk (T 2 ) = nullity (T 2 ) + dim (N (T 2 )) = nullity (T 2 ) = 2 Now p (x) = x + b, where b = b b is in N (T 2 ) s is p 2 (x) = x 2 + c, where c = b b Thus since fp (x), p 2 (x)g is linerly independent subset of N (T 2 ) nd dim (N (T 2 )) = 2 this implies tht fp (x), p 2 (x)g is bsis for N (T 2 ) (d) Yes (e) Yes Now ; x; x 2 is bsis for P 2 (R) nd T 4 () = x, T 4 (x) = 2 x2 2, T 4 x 2 = 3 x3 3 which re linerly independent in (R) so T 4 () ; T 4 (x) ; T 4 x 2 is bsis for R (T 4 ) nd N (T 4 ) = f0g hs bsis ; This completes the proof/solution 7

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