Linear Independence and Span   Span We have seen in the last discussion that the span of vectors v1, v2, ... , vn is the set of linear combinations         c1v1 + c2v2 + ... + cnvn   and that this is a vector space.   We now take this idea further.  If V is a vector space and S  =  {v1, v2, ... , vn) is a subset of V, then is Span(S) equal to V? Definition Let V be a vector space and let S  =  {v1, v2, ... , vn) be a subset of V.  We say that S spans V if every vector v in V can be written as a linear combination of vectors in S.         v  =  c1v1 + c2v2 + ... + cnvn   Example Show that the set          S =  {(0,1,1), (1,0,1), (1,1,0)}  spans R3 and write the vector (2,4,8) as a linear combination of vectors in S.   Solution A vector in R3 has the form          v  =  (x, y, z) Hence we need to show that every such v can be written as         (x,y,z)  =  c1(0, 1, 1) + c2(1, 0, 1) + c3(1, 1, 0)         =  (c2 + c3, c1 + c3, c1 + c2) This corresponds to the system of equations               c2 + c3  =  x         c1 +       c3  =  y         c1 + c2         =  z which can be written in matrix form We can write this as          Ac  =  b Notice that          det(A)  =  2 Hence A is nonsingular and         c  =  A-1b So that a nontrivial solution exists.  To write (2,4,8) as a linear combination of vectors in S, we find that so that We have         (2,4,8)  =  5(0,1,1) + 3(1,0,1) + (-1)(1,1,0) Example Show that if         v1  =  t + 2        and         v2  =  t2 + 1 and  S  =  {v1, v2} then          S does not span P2   Solution A general element of P2 is of the form          v   =  at2 + bt + c We set          v  =  c1v1 + c2v2  or         at2 + bt + c  =  c1(t + 2) + c2(t2 + 1)  =  c2t2 + c1t + c1 + c2  Equating coefficients gives         a  =  c2          b  =  c1          c  =  c1 + c2  Notice that if          a  =  1        b  =  1        c  =  1 there is no solution to this.  Hence S does not span V. Example Let Find a spanning set for the null space of A.   Solution We want the set of all vectors x with          Ax  =  0 We find that the rref of A is The parametric equations are         x1  =  7s + 6t         x2  =  -4s - 5t         x3  =  s         x4  =  t We can get the span in the following way.  We first let         s  =  1        and         t  =  0 to get          v1  =  (7,-4,1,0)    and let         s  =  0        and     t  =  1 to get         v2  =  (6,-5,0,1) If we let S  =  {v1,v2} then S spans the null space of A. Linear Independence We now know how to find out if a collection of vectors span a vector space.  It should be clear that if S  =  {v1, v2, ... , vn) then Span(S) is spanned by S.  The question that we next ask is are there any redundancies.  That is, is there a smaller subset of S that also span Span(S).  If so, then one of the vectors can be written as a linear combination of the others.          vi  =  c1v1 + c2v2 + ... + ci -1vi -1 + ci+1vi+1 + ... + cnvn  If this is the case then we call S a linearly dependent set.  Otherwise, we say that S is linearly independent.  There is another way of checking that a set of vectors are linearly dependent.   Theorem Let S  =  {v1, v2, ... , vn) be a set of vectors, then S is linearly dependent if and only if 0 is a nontrivial linear combination of vectors in S.  That is, there are constants c1, ..., cn with at least one of the constants nonzero with          c1v1 + c2v2 + ... + cnvn  =  0  Proof Suppose that S is linearly dependent, then         vi  =  c1v1 + c2v2 + ... + ci -1vi -1 + ci+1vi+1 + ... + cnvn  Subtracting vi from both sides, we get          c1v1 + c2v2 + ... + ci -1vi -1 + vi  + ci+1vi+1 + ... + cnvn  =  0 In the above equation ci  =  1 which is nonzero, so that 0 is a nontrivial linear combination of vectors in S.   Now let           c1v1 + c2v2 + ... + ci -1vi -1 + civi  + ci+1vi+1 + ... + cnvn  =  0 with ci nonzero.  Divide both sides of the equation by ci and let aj  =  -cj / ci to get          -a1v1 - a2v2 - ... - ai -1vi -1 + vi - ai+1vi+1 - ... - anvn  =  0 finally move all the terms to the other right side of the equation to get         vi  =  a1v1 + a2v2 + ... + ai -1vi -1 + ai+1vi+1 + ... + anvn  Example Show that the set of vectors          S  =  {(1, 1, 3, 4),  (0, 2, 3, 1),  (4, 0, 0, 2)} are linearly independent.   Solution We write         c1(1, 1, 3, 4) + c2(0, 2, 3, 1) + c3(4, 0, 0, 2)  =  0 We get four equations         c1 + 4c3  =  0         c1 + 2c2  =  0         3c1 + 3c2  =  0         4c1 + c2 + 2c3  =  0 The matrix corresponding to this homogeneous system is and Hence          c1  =  c2  =  c3  =  0 and we can conclude that the vectors are linearly independent. Example Let            S  =  {cos2 t, sin2 t, 4) then S is a linearly dependent set of vectors since         4  =  4cos2t + 4sin2t   Back to the Linear Algebra Home Page