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MATH 244 -- Linear Algebra 

Problem Set 7 Solutions 

March 30, 2007 

Section 4.3 

24.   Let A  be the n x n matrix with columns  

       By the Invertible Matrix Theorem (section 2.3) 

       A  is invertible.  Hence for all the  

       system of linear equations   has a solution. 

       But this implies  Span(ℬ) Therefore,  ℬ 

       is a basis for  

       (Note:  this also follows from the reasoning given 

       in the proof of Section 4.5/26 below.) 

26.    The set S = is linearly 

        dependent because of a trigonometric identity: 

        for all t.  Hence  

         The set is linearly independent 

         since if for all t, then 

         setting shows    hence  

         But then as well, since is not the zero  

         function.   

31.     If is linearly dependent, then there exist 

        scalars    not all equal to zero such that 0 =  

        Since  T  is assumed to be a linear mapping, applying it 

        to both sides of the last equation shows 

         But  for all linear mappings so 

            0 =  

          Since the are not all zero (they are the same scalars as 

          in the original linear dependence on the this shows that 

          is also linearly dependent.   

32.     Assume that there are scalars not all equal to zero such that  

          = Since is linear, this implies 

         Since T  is one-to-one, 

         this implies that = Since are not all 

         equal to zero, this implies that is linearly dependent. 

34.    ``By inspection'',  As in 24 above, this  

        relation shows  or the span of  

        any of the other two-element subsets).  The set is linearly 

        independent since as a polynomial 

        implies It is easy to check that 

        the only solution of this 2 x 2  homogeneous system of linear  

        equations is Thus,  is one basis for 

Section 4.4 

10.  By the standard ``recipe'' 

      (The columns of this matrix are the coordinate vectors of the 

      basis ℬ  with respect to the standard basis.) 

12.   so   

14.   We can follow the method used in problem 12.  Take 

       the "standard basis"   Then 

       (columns are the coordinate vectors of the  

       polynomials in ℬ).   

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    Then  

     Check:  OK! 

18.  Follow the standard ``recipe'' for getting the coordinate 

      vectors.  The unique linear combination of the vectors in ℬ 

      that equals  is 

      Picking off the scalars here and placing them into a vector 

      in we get  the ith standard basis vector, which is the  

      same as the ith column of the n x n  identity matrix.  

22.   The matrix that does this is the change of coordinates matrix 

        This is the inverse of the matrix whose 

       columns are the (standard coordinates) of the vectors in the basis 

       ℬ. 

28.    With respect to the basis for the coordinate  

        vectors of the polynomials are 

        The mapping is an isomorphism of vector spaces 

        (linear, 1-1, and onto).  This implies that the polynomials are 

        linearly independent if and only if the coordinate vectors are.   

        To check linear dependence or independence, we apply our 

        standard methods.  Reducing the 4 x 3  matrix with these  

        columns to echelon form we find: 

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        Since there is a free variable in the corresponding homogeneous 

        system, there is a nontrivial solution, so the vectors (and polynomials) 

        are linearly dependent. 

Section 4.5 

26.     Let ℬ =  be a basis for  H.   

        We will argue by contradiction.  Suppose 

        H  is strictly contained in  V.  Then there is some   

        such that We claim that the set  =  

        must be linearly independent.   

         To see this, consider a linear combination  

         If  then we could use this equation to solve for 

         x, and that would say  So   

         But then the remaining terms only involve vectors in  

         ℬ and we are assuming ℬ is a basis, hence linearly  

         independent.  Therefore for all i.    But this 

         leads to a contradiction because  also. 

         In an dimensional vector space, any set of   

         vectors is linearly dependent (Theorem 9 in this section 

         of the text).  Therefore   

28.    The subset for all   

        spans the subspace of consisting of all polynomial 

        functions.  This set is linearly independent because 

        any nonzero polynomial has only finitely many real roots. 

        (That is, the only polynomial that is the zero function  

        is the zero polynomial, with all coefficients equal to zero.) 

        This implies that is not a finite-dimensional vector 

        space (Theorem 9 again -- in a vector space of dimension 

        n, the largest linearly independent sets have n elements). 

32.     To prove it suffices to show that if 

        is a basis for then   

         is a basis for do you see why?) 

         First, note that the results of problems 31 and 32 from section  

         4.3 above say that when  T  is 1-1 (as assumed here),  

         is linearly dependent if and only if 

         is linearly dependent.  Thus, it follows by logic that 

is linearly independent if and only if 

         is linearly independent  too.    

         Next, since for all we have  

          for some scalars Since T  is linear this implies 

          This shows  is contained in  

          On the other hand, we can  

          take any linear combination as on the right of the last equation, 

          use the linearity of T  ``in reverse'', and see that it equals 

           for some This shows that  

          is equal to  Since spans and 

          is linearly independent, it is a basis for