Last lecture: Examples and the column space of a matrix
        Suppose that A is an n × m matrix.
        DefinitionThecolumnspaceofAisthevectorsubspace
        Col(A)ofRn whichisspannedbythecolumnsofA.
                            £             ¤
            That is, if A =  a ,a ,...,a    then Col(A) =
             ¡              ¢ 1 2       m
        Span a ,a ,...,a     .
                1  2      m
       Linear dependence and independence (chapter. 4)
            • If V is any vector space then V = Span(V ).
            • Clearly, we can find smaller sets of vectors which
              span V.
            • This lecture we will use the notions of linear
              independence and linear dependence to find the
              smallest sets of vectors which span V .
            • It turns out that there are many “smallest sets” of
              vectors which span V , and that the number of
              vectors in these sets is always the same.
              This number is the dimension of V .
                                0-0
          Linear dependence—motivation Let lecture we saw that
                                   · ¸ · ¸ ·       ¸
                                 n 1       3     5   o
          the two sets of vectors    2 , 5 ,     9        and
                                     3     7     13
           · ¸ · ¸ · ¸
          n 1      3     0 o                 3
             2 , 5 , 1        donotspanR .
             3     7     2
             • Theproblemisthat
                · 5 ¸      ·1¸    ·3¸
                  9    =2 2 + 5           and
                  13        3       7
                ·0¸       ·1¸    ·3¸
                  1  =3 2 − 5 .
                  2        3       7
             • Therefore,
                      ³·1¸ ·3¸ · 5 ¸´               ³·1¸ ·3¸´
                Span      2 , 5 ,     9    =Span        2 , 5
                          3    7     13                 3    7
                and ³· ¸ · ¸ · ¸´                  ³· ¸ · ¸´
                          1    3     0                1     3
                Span      2 , 5 , 1       =Span       2 , 5      .
                          3    7     2                3     7
             • Notice that we can rewrite the two equations
                above in the following form:
                 ·1¸     ·3¸     · 5 ¸    ·0¸
                2 2 + 5 − 9 = 0                   and
                   3       7       13       0
                 ·1¸     ·3¸     ·0¸     ·0¸
                3 2 − 5 − 1 = 0
                   3       7       2       0
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            This is the key observation about spanning sets.
        Definition
       Suppose that V is a vector space and that x ,x ,...,x
                                        1  2     k
       are vectors in V .
       Thesetofvectors {x ,x ,...,x } is linearly dependent
                       1  2     k
       if
                 r x +r x +···+r x =0
                 1 1   2 2        k k
       forsomer ,r ,...,r ∈ Rwhereatleastoneofr ,r ,...,r
               1  2    k                     1 2     k
       is non–zero.
       Example
        ·1¸   ·3¸  · 5 ¸  ·0¸
       2 2 + 5 − 9 = 0 and
         3     7     13    0
        ·1¸   ·3¸  ·0¸   ·0¸
       3 2 − 5 − 1 = 0
         3     7     2    0 ·  ¸ · ¸ · ¸
                           n 5    1   3 o
       Sothetwosetsofvectors  9 , 2 , 5   and
        · ¸ · ¸ · ¸          13   3   7
       n 0    1   3 o
          1 , 2 , 5   are linearly dependent.
          2   3   7
       Question Suppose that x,y ∈ V. When are x and y
       linearly dependent?
                           0-2
        Question What do linearly dependent vectors look like
            2      3
        in R and R ?
         Example
                 "1#       "3#          "0#
        Let x =    2  y = 2 andz = 4 . Is {x ,x ,x }
                   3        1             8        1  2  3
        linearly dependent?
        We have to determine whether or not we can find real
        numbers r,s,t, which are not all zero, such that
        rx +sy+tz =0.
        Tofindallpossibler,s,t wehavetosolvetheaugmented
        matrix equation:
        " 1 3 0 0 # R2:=R2−2R1 " 1 3             0  0 #
           2  2   4 0    −−−−−−−−−→      0  −4 4 0
           3  1   8 0     R3:=R3−3R1     0  −8 8 0
         R3:=R3−2R2 " 1 3       0   0 #
        −−−−−−−−−→      0  −4 4 0
                        0  0    0   0
        Sothis set of equations has a non–zero solution.
        Therefore,{x,y,z}isalinearlydependentsetofvectors.
                       "1# "3# "0# "0#
        Tobeexplicit, 3 2   − 2 − 4 = 0 .
                         3      1      8      0
                                0-3