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1 section 1 7 linear independence denition 1 asetofpvectors u u u in rm is called linearly independent if the vector 1 2 p equation x1 u1 x2 u2 xp ...

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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
                                                                                     3x2x3 =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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