# UCLA MATH 131A - midterm1 solution (5 pages)

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- University of California, Los Angeles
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- Math 131a - Analysis

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Linear Algebra Solution to Midterm 1 Problem 1 Let V a1 a2 a1 a2 R Define addition of elements of V coordinatewise and for a1 a2 in V and c R define n 0 0 if c 0 c a1 a2 ca1 ac2 if c6 0 Is V a vector space over R with these operations Justify your answer 5 points Solution No If c d R c d 6 0 c 6 0 d 6 0 then c d a1 a2 c d a1 a2 c d usually is not equal to c a1 a2 d a1 a2 ca1 da1 a1 a2 c d VS8 does not hold Problem 2 Let W1 and W2 be subspaces of a vector space V Prove that W1 W2 is a subspace of V if and only if W1 W2 or W2 W1 9 points Proof that W1 W2 or W2 W1 then W1 W2 W1 or W2 Since W1 and W2 are subspaces V we have W1 W2 is also a subspace of V Suppose that W1 W2 is a subspace of V Also suppose that W1 6 W2 and W2 6 W1 then there exist u v V such that u W1 W2 v W2 W1 u v W1 W2 u v W1 W2 If u v W1 then u u v W1 v W1 If u v W2 then u v v W2 u W2 Hence W1 W2 or W2 W1 Problem 3 Show that if S1 and S2 are arbitrary subsets of a vector space V then span S1 S2 span S1 span S2 9 points 1 P Pn Proof Let u span S1 S2 then u m i 1 ai vi j 1 bj wj for some scalars ai i 1 m bj j 1 n where vi i 1 m are in S1 and wj j 1 n are in P Pn S2 Since m a v is in span S and i i 1 i 1 j 1 bj wi is in span S2 we have u span S1 span S2 Hence span S1 S2 span S1 span S2 Now let v x y span S1 span S2 where x span S1 and y span S2 We Pm can write x i 1 ai vi for some scalars ai i 1 m and vi S1 i 1 m Pn and y j 1 bj wj for some scalars bj j 1 n and wi S2 j 1 n Then Pm Pn we can see that v x y a v i i i 1 j 1 bj wj is in span S1 S2 since vi i 1 m wj j 1 n are in S1 S2 Hence span S1 span S2 span S1 S2 Therefore span S1 S2 span S1 span S2 Problem 4 Prove that a set S is linear dependent if and only if S 0 or there exist distinct vectors v u1 u2 un in S such that v is a linear combination of u1 u2 un 9 points Proof If S is linearly dependent and S 6 0 then there exist distinct vectors u0 u1 un S such that a0 u0 a1 u1 an un 0 with at least one of the scalars a0 a1 an is not zero say a0 6 0 Then we have a2 an a1 u1 u2 un u0 a0 a0 a0 Hence v u0 is a linear combination of u1 u2 un If S 0 then it s clear that S is linearly dependent Assume that there exist distinct vectors v u1 u2 un S such that v is a linear combination of u1 u2 un say v a1 u 1 a2 u 2 an u n for some scalars a1 a2 an Then we have 0 1 v a1 u1 a2 u2 an un Hence S is linearly dependent Problem 5 Prove that if W1 is any subspace of a finite dimensional vector space V then there exists a subspace W2 of V such that V W1 W2 9 points Proof Let u1 un be a basis for W1 Since W1 is a subspace of V By Replacement Theorem we can extend to a basis for V say u1 un un 1 um Let W2 span un 1 um Claim that V W1 W2 1 V W1 W2 If v V then v m X i 1 ai ui n X i 1 ai ui m X ai ui W1 W2 for some scalars ai i 1 m i n 1 This implies that V W1 W2 But by the definition of W1 W2 we also know that W1 W2 V Hence V W1 W2 2 W1 W2 0 P P Let u W1 W2 Then u ni 1 bi ui m i n 1 ci ui for some scalars b1 bn cn 1 cm Then we have n m X X bi ui ci ui 0 i 1 i n 1 But is linearly independent since is a basis Hence b1 bn cn 1 cm 0 This implies that u 0 That is W1 W2 0 By 1 and 2 we have V W1 W2 We are done Problem 6 Prove that if W1 and W2 are finite dimensional subspaces of a vector space V then the subspace W1 W2 is finite dimensional and dim W1 W2 dim W1 dim W2 dim W1 W2 9 points Hint Start with a basis u1 u2 uk for W1 W2 and extend this set to a basis u1 u2 uk v1 v2 vm for W1 and to a basis u1 u2 uk w1 w2 wp for W2 Proof dim W1 W2 dim V W1 W2 has a finite basis u1 u2 uk We can extend to a basis 1 u1 u2 uk v1 v2 vm for W1 and to a basis 2 u1 u2 uk k1 k2 kp for W2 Let u1 u2 uk v1 v2 vm w1 w2 wp We claim that is a basis for W1 W2 To prove the claim we need to check that 1 is linearly independent Let a1 u1 ak uk b1 v1 bm vm c1 w1 cp wp 0 for some scalars a1 ak b1 bm c1 cp Then b1 v1 bm vm a1 u1 ak uk c1 w1 cp wp W1 W2 Since is a basis for W1 W2 we have b1 v1 bm vm d1 u1 dk uk for some scalars d1 dk d1 u1 dk uk b1 v1 bm vm 0 d1 dk b1 bm 0 since 1 is a basis for W1 a1 u1 ak uk c1 w1 cp wp 0 a1 ak c1 cp 0 since 2 is a basis for W2 Hence is linearly independent 2 W1 W2 span Let u v w W1 W2 where v W1 and w W2 be any vector in W1 W2 Since 1 is a basis for W1 and 2 is a basis for W2 we can find some scalars x1 xk y1 ym z1 zk such that u x1 u1 xk uk y1 v1 ym vm z1 u1 zk uk t1 w1 tp wp x1 z1 u1 xk zk uk y1 v1 ym vm t1 w1 tp wp …

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