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linear independence linear independence denition let v v v rn be a set of vectors 1 2 k the vectors are linearly dependent if there exist scalars 1 2 k ...

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                               Linear Independence
                               Linear Independence.
                               Definition. Let v ,v ,...,v ∈ Rn be a set of vectors.
                                               1  2      k
                                  • The vectors are linearly dependent if there exist scalars λ1,λ2,...,λk ∈ R, not
                                     all zero, such that
                                                          λ v +λ v +···+λ v =0.
                                                           1 1     2 2         k k
                                  • The vectors are linearly independent if λ v + ··· + λ v   = 0 for scalars
                                                                            1 1          k k
                                     λ1,...,λk implies that
                                                             λ1 = λ2 = ··· = λk = 0.                       N
                               Example1.
                                  1.   • Thevectors                                    
                                                             1             0                 0
                                                           0           1               0 
                                                     e = ,e = ,...,e = 
                                                      1    .  2        .          n    . 
                                                           .           .               . 
                                                              .            .                 .
                                                             0             0                 1
                                          in Rn are linearly independent.
                                       • Toshowthis, consider any scalars λ ,...,λ such that λ e + λ e ··· +
                                                                          1       n          1 1     2 2
                                          λ e =0,i.e.suchthat
                                           n n
                                               1          0               0        λ1        0 
                                               0          1               0        λ2        0 
                                           λ       +λ  +···+λ  =                        = 
                                             1  .       2  .            n .        .   . 
                                               .          .               .        .   . 
                                                  .           .                  .          .          .
                                                 0            0                  1         λn          0
                                       • Thelast equation implies that λ1 = λ2 = ··· = λn = 0.
                                  2.   • Thevectors                                     
                                                             1             4                  7
                                                    w = 2 ,w = 5 , andw = 8 
                                                      1             2                  n
                                                             3             6                  9
                                          in R3 are linearly dependent, since w1 − 2w2 + w3 = 0.
                                       • Howdidwecomeupwiththis? Startwiththeequation in the definition of
                                          linear dependence
                                                        1          4          7       0 
                                                    λ  2 +λ  5 +λ  8 = 0 
                                                      1           2            3
                                                          3            6            9         0
                                          and solve this system for possible values of λ ,λ ,λ .
                                                                                   1  2  3
                                     2.    • Wecanwritethis system in matrix form as
                                                                 1 4 7  λ1   0 
                                                                 2 5 8  λ2 = 0 .
                                                                    3  6   9        λ3           0
                                           • The coefficient matrix A is simply the matrix with columns w ,w ,w
                                                                                                               1   2    3
                                              which we can write as A = (w1w2w3). If we reduce A to row echelon
                                              form we get
                                                                          1     4     7 
                                                                          0 −3 −6 .
                                                                            0    0     0
                                           • We have found that rankA is not maximal, and so the system is singular
                                              and has a nonzero solution (many in fact).                               
                                      • Thefollowing theorem generalizes the last example.
                                  Theorem1(S&B11.1). Vectorsv ,...,v ∈ Rn arelinearly dependent iff the linear
                                                                     1       k
                                  system                                      
                                                                           λ
                                                                            1
                                                                        λ 
                                                                     A 2 =0,
                                                                        . 
                                                                        . 
                                                                            .
                                                                           λk
                                  where A = (v v ···v ), has a nonzero solution (λ ,...,λ ).
                                                1 2      k                            1       k
                                      • The following theorem gives us a way of checking that a set of n vectors in Rn
                                        are linearly independent.
                                  Theorem2(S&B11.2). Vectorsv1,...,vn ∈ Rn are linearly independent iff
                                                                   det(v v ···v ) 6= 0.
                                                                        1 2      n
                                  Theorem3(S&B11.3). Ifk > n,anysetofk vectors in Rn is linearly dependent.
                                  Proof. Let v1,...,vk be k vectors in Rn with k > n. By theorem 1, this set of vectors
                                  is linearly dependent if and only if the system
                                                                            λ1 
                                                                            λ2 
                                                                  Aλ=A           =0
                                                                            . 
                                                                            . 
                                                                               .
                                                                              λk
                                  where A = (v v ···v ), has a nonzero solution (λ ,...,λ ).
                                                1 2      k                            1       k
                                      • But, since the vectors belong to Rn, the matrix A has n rows and k columns with
                                        k > n.
                                                                             2
                             • Since A has more columns than rows, there must be some column, say column
                               j, in the row echelon forms of A without a pivot.
                             • But then λj is a free variable, and since Aλ = 0 has a solution, it must have
                               infinitely many solutions all but one of which is nonzero.
                             • Hencethevectors are linearly dependent.                    
                             • Instead of the above, we could have used fact 7.11(a)(i) of S&B which says
                               that if there are more unknowns than equations, the system Aλ = 0 must have
                               infinitely many solutions.
                          SpanningSets.
                             • Asubset of V of the vector space Rn is a subspace of Rn if it is
                                 – closed under addition (the sum of any two elements in V is also in V ), and
                                 – closed under scalar multiplication (any scalar multiple of an element in V
                                   is also in V ).
                               Example2.
                                 – Theset{(1,1,1)}isnotasubspaceofR3,becauseforexample(0,0,0) ∈/
                                   {(1,1,1)}.
                                 – ThesetV = L[(1,1,1)] = {λ(1,1,1) | λ ∈ R} is a subspace of R3 as you
                                   can check.
                          Definition. Let v ,...,v ∈ Rn be a set of vectors and let V ⊆ Rn be a subspace.
                                       1     k
                                 • Theset of all linear combinations of these vectors
                                          L[v ,...,v ] = {λ v +···+λ v |λ ,...,λ ∈ R}
                                            1     k     1 1        k k  1     k
                                   is called the set generated or spanned by v ,...,v .
                                                                   1     k
                                 • Equivalently, the span of {v ,...,v } is the set L[v ,...,v ] (sometimes
                                                        1     k           1     k
                                   written span(v ,...,v )).
                                              1     k
                                 • If                   V =L[v ,...,v ],
                                                              1     k
                                   wesayv ,...,v spanV.                                   N
                                         1     k
                                 • A set of vectors spans (or is a spanning set of) a subspace V of Rn if we
                                   can write any element in V as a linear combination of the vectors.
                          Example3.
                            1. The n-dimensional Euclidean space is spanned by the vectors e1,...,en of our
                               earlier example. Take any vector a = (a ,...,a ) ∈ Rn. Then a = a e +
                                                              1     n                 1 1
                               · · · + a e .
                                    n n
                            2. There are many sets of vectors that span the same space. Each of the following
                               sets spans the space R2.
                                                          3
                                            (a)  1   0                                  1   0   56            
                                                   0    ,   1    .                    (c)     0    ,   1    ,    −24    .
                                                 −1   0                                 6       31 
                                           (b)     0      ,   1    .                  (d)     −6     ,   31    .
                                   Theorem4(S&B11.4). Letv ,...v ∈ Rn beasetofvectors. LetA = (v v ···v )
                                                                   1      k                                       1 2      k
                                   andletb ∈ Rn beavector. ThenbliesinthespaceL[v ...,v ]spannedbyv ,...,v
                                                                                            1       k               1       k
                                   iff the system Aλ = b has a solution λ.
                                       • The first corollary to this theorem provides a way of checking whether or not a
                                          set of vectors spans all of Rn, while the second gives the minimum number of
                                          vectors needed in a set spanning Rn.
                                   Corollary1(S&B11.5). Letv ,...v ∈ Rnbeasetofvectors. LetA = (v v ···v ).
                                                                   1      k                                       1 2      k
                                   Then v ,...,v span Rn iff the system of equations Ax = b has a solution x for every
                                           1       k
                                   b ∈ Rn.
                                   Corollary2(S&B11.6). AsetofvectorsthatspansRn mustcontainatleastnvectors.
                                   Basis and Dimension.
                                       • Once we have a spanning set of vectors, we can always make the spanning set
                                          larger by including more vectors.
                                       • We want to find the smallest set of vectors that spans a subset V of the n-
                                          dimensional Euclidean space.
                                   Definition. Let v ,...v      ∈ V be a set of vectors and let V ⊆ Rn. Then v ,...v
                                                      1      k                                                       1      k
                                   forms a basis of V if
                                       1. v ,...v span V, and
                                           1      k
                                       2. v ,...v are linearly independent.                                                N
                                           1      k
                                   Example4.        1. Fromourpreviousexamples,wecanseethattheunitvectorse1,...,en
                                          form a basis of Rn. This natural basis, is called the canonical basis of Rn.
                                                                                                        2
                                       2. In a previous example (2), we listed four spanning sets of R . The set (2c) is not
                                          linearly independent, since it contains more than two vectors. The collections
                                          (2a), (2b), and (2d) are linearly independent and so each forms a basis for R2.   
                                   Theorem5(S&B11.7). EverybasisofRn containsexactly n vectors.
                                       • Using theorems 1, 2 and corollary 1, together with the fact that a square matrix
                                          is nonsingular iff its determinant is nonzero gives the following theorem.
                                   Theorem6(S&B11.8). Letv ,...,v ∈ RnbeasetofvectorsandletA = (v v ···v ).
                                                                   1       n                                         1 2      n
                                   Thenthe following statements are equivalent.
                                                                                4
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