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1 Section 1.7. Linear Independence De nition 1 Asetofpvectors~u ;~u ;:::;~u in Rm is called linearly independent if the vector 1 2 p equation ~ x1~u1 + x2~u2 + ::: + xp~up = 0 has only the trivial solution x = 0; i = 1;2;:::;p: Otherwise, the set is called linearly depen- i dent; the coe¢ cients x1;:::;xn are called a linear relation. Example 2 Anyonesingle vector ~u is always independent. two vectors ~u ;~u are dependent 1 2 i¤ ~u1 = ~u2. The notation of linear dependence (or independence) is closely related to homogeneous systems. In fact, let A = [~a ;~a ;:::;~a ] be a m n matrix with column vectors ~a ; then we 1 2 n i have the following relation: ~ Claim 3 A~x = 0 has NO non-trivial solution i¤ its column vectors ~a ;~a ;:::;~a are linearly 1 2 n independent. Example 4 (1) Determine whether the set of the following three vectors is dependent: 2 1 3 2 2 3 2 4 3 ~u =425; ~u =4 1 5; ~u =4 5 5: 1 2 3 3 0 6 (2) Find a linear relation. ~ Solution. (1) We need to determine if A~x = 0 has a non-trivial solution, where A = [~u ; ~u ; ~u ] 1 2 3 2 3 1 2 4 A=42 1 5 5: 3 0 6 h ~i To this end, we perform row operation on the augmented matrix A;0 :For simplicity, we only work on A : 2 1 2 4 3 h ~i 2 1 2 4 0 3 A!40 3 3 5; since A;0 !4 0 3 3 0 5: (1) 0 0 0 0 0 0 0 ~ Apparently, x3 is a free variable. Thus, A~x = 0 has non-trivial solution, and the set f~u ;~u ;~u g is linearly dependent. 1 2 3 Solution. (2) Finding a linear relation ~u ;~u ;~u , i.e., 1 2 3 ~ x1~u1 + x2~u2 + x3~u3 = 0 1 means to nd a non-trivial solution of ~ A~x = 0: ~ From (1), we nd that A~x = 0 reduces to x1 +2x2 +4x3 = 0 3x2 x3 =0: Since we only need one non-zero solution, we take x3 = 3 (we could have choose any number here. The only reason I chose 3 is to avoid fraction.) Then x2 = 1; x1 = (2x2 +4x3) = 10; and ~ 10~u1 ~u2 +3~u3 = 0: From the above equation, we see that, for instance, ~u2 = 10~u1 + 3~u3; i.e., when f~u ;~u ;~u g is linearly dependent, ~u is a linear combination of f~u ;~u g: In general, 1 2 3 2 1 3 Theorem 5 (Characterization of linear dependence) Vectors ~u ;~u ;:::;~u are linearly de- 1 2 p pendent i¤ at least one is a linear combination of the rest. Proof. Suppose that one is a linear combination of the rest. Without loss of generality, we say that ~u is a linear combination of ~u ;:::;~u ; i.e., there exist ;:::; such that 1 2 p 2 p ~u =~u +:::+ ~u : 1 2 2 p p It follows that ~u + ~u +:::+ ~u = 0: 1 2 2 p p This implies that ~u ;~u ;:::;~u are linearly dependent. On the other hand, suppose that 1 2 p ~u ;~u ; :::; ~u are linearly dependent. Then, we have 1 2 p ~u + ~u +:::+ ~u = 0; 1 1 2 2 p p where at least one of p constants f ; ;:::; g is not zero. Without loss of generality, we 1 2 p say 6= 0: Then we can solve for ~u as 1 1 ~u = 2 ~u +:::+ p ~u : 1 2 p 1 1 Hence, ~u is a linear combination of ~u ;:::;~u ; the rest. 1 2 p Notethatwhen~u ;~u ;:::;~u are linearly dependent, it does NOT mean ANY one member 1 2 p is a linear combination of the rest. For instance, 1 ; 1 ; 1 are linearly dependent. 2 2 0 2 But 1 is not a linear combination of 1 ; 1 : 0 2 2 In summary, to determine whether ~u ;~u ;:::;~u in Rm are linearly independent, we need 1 2 p ~ to see whether the system A~x = 0 has only trivial solution, where A is the m p column matrix [~u ;~u ;:::;~u ]: 1 2 p De nition 6 We call the number of pivots of a mp matrix A the RANK of A; and denote it by r (A): Obviously, the rank cannot exceed the number of rows or columns, i.e., r(A) m; r(A) p: The following theorem follows directly from the last conclusion in Section 1.5. Theorem 7 A set of vectors ~u ;~u ;:::;~u in Rm is linearly dependent i¤ p > r(A); where 1 2 p Ais the mp column matrix [~u ;~u ;:::;~u ]: In particular, it is linearly dependent if p > m; 1 2 p i.e., the number of vectors is more than the dimension. Example 8 Determine whether the following set is linearly dependent. If it is linearly dependent nd a set of linearly independent vectors ~v1;~v2;::: such that Spanf~v1;~v2;::g = Spanf~u ;~u ;::g: 1 2 2 1 3 2 4 3 2 2 3 2 1 3 ~u =425; ~u =4 5 5; ~u =4 1 5; ~u =4 1 5: 1 2 3 4 3 6 0 1 Solution. The answer is yes, it is linearly dependent because the number of vectors is great than the dimension. ~ Wenext describe all vectors b 2 Spanf~u ;~u ;::g using parametric vector representation. 1 2 ~ ~ AswedidinExample??,weneedtodescribebsuchthat[~u ;~u ;~u ;~u ] ~x = b has a solution. h i1 2 3 4 ~ To this end, we perform row operation on ~u ;~u ;~u ;~u ;b to arrive at a Echelon form: 1 2 3 4 2 3 2 3 1 4 2 1 b 1 4 2 1 b 1 R 2R !R 1 4 5 2 1 2 4 5 2 5 1 1 b 0 3 3 1 b 2b 2 R 3R !R 2 1 3 6 0 1 b 3 1 3 0 6 6 2 b 3b 3 ! 3 1 2 3 1 4 2 1 b 1 4 5 R 2R !R 0 3 3 1 b 2b : 3 2 3 2 1 ! 0 0 0 0 b 2b +b 1 2 3 The system is consistent i¤ b 2b +b = 0; or 1 2 3 b =2b b : 1 2 3 3 Therefore, 2 3 2 3 2 3 2 3 b 2b b 2 1 1 2 3 ~ 4 5 4 5 4 5 4 5 b = b = b =b 1 +b 0 ; 2 2 2 3 b b 0 1 3 3 and the set of 2 3 2 3 2 1 4 5 4 5 ~v1 = 1 ; ~v2 = 0 0 1 is linearly independent, and Spanf~v ;~v g = Spanf~u ;~u ;~u ;~u g: Homework 1.7: 1 2 1 2 3 4 #5,11,21,23,27,33,37 4
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