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MATH 244 -- Linear Algebra 

Problem Set 6 Solutions 

March 16, 2007 

Section 3.3 

4.  By Cramer's Rule, the solution is 

6.  By Cramer's Rule, 

8.  There is a unique solution when    

This is true for all real s.    By Cramer's Rule, the solution is 

20.   The area of the parallelogram is 

24.   The volume is  

29.   Since the diagonal of the parallelogram with  

       vertices and  

       divides the parallelogram into two congruent triangles,  

       the area of the triangle is one half of the area of the  

       parallelogram:   

       where is the matrix with columns  

30.   If we translate the triangle to place at the origin, 

      the other vertices are  

      Applying the formula from problem 

      29, the area is  

      Now on the other hand, if we consider the matrix given 

      in the problem, we can apply row operations to evaluate 

      the determinant as follows -- replace row 2 by row 2 - row 1 

      and replace row 3 by row 3 - row 1: 

      Expanding the determinant along the third column gives  

                       =  

      which is the same as the result above, which shows the equality 

32.  a)  There are several such mappings, but the "obvious" one 

           is the  defined by the matrix  A   

           whose columns are the vectors  (in that order). 

           Then for each  

      b)  By Theorem 10,  where is the matrix 

           described in part a. 

Section 4.1 

2.  a)  Yes:  If then  So the vector  

         satisfies  Hence  

    b)  If then  

4.  The figure should show a line  L  not passing through (0,0), 

    two vectors from the origin to points on  L,  and their 

    vector sum constructed by the parallelogram law 

    for vector addition in  The head of the vector sum is 

    not on the line  L,  so L  is not closed under vector sums. 

5.  The set W of all polynomialsis a vector subspace of  

    because:  (a) If we take  

    (b)  If then  

    (c)  If and then  

6.  The set of polynomials is not  a vector 

    subspace of because 0 W.  (The other two properties 

    also fail, but one is sufficient to say W  is not a subspace.) 

8.  The set of polynomials such that  

    is a vector subspace of for each .  (a)  The 

    zero polynomial satisfies  

    (b)  If then  

    (c)  If then  

20. a)  We need to know that the zero function is continuous, 

         that sums of continuous functions are continuous, and 

         that a constant times a continuous function is continuous. 

    b)  Let The zero function z is in  

         because If   then  

         so Finally, ifthen  

         so   

21.  is a vector subspace of the vector space because 

     (a) the zero matrix is in H (take in the general 

          form). 

     (b) If then for some  

          Hence  

          This matrix is in H  (because of the zero in the second row, 

          first column).   

     (c) If and then  

22.  H is a vector subspace of because 

     (a) so   

     (b) If  then   

          This shows   

     (c) If then  

          Hence  

31.  If is a subspace and then by property c in  

     the definition of a subspace, for all scalars  

     Then, by property b in the definition, 

     The set consists of all the  

     linear combinations so we have shown  

32.  If are subspaces, they satisfy all three properties in  

      the definition.  Hence Therefore,  

      If then Therefore,  

      But this says   

      Finally, if then so  

      Similarly, so This implies  

      Since  satisfies the three properties for a subspace 

      it is itself a subspace.   

      For the last part, consider   

      and These are vector subspaces  

      of  The union is not a vector subspace, though,  

      because, for instance, but  

33.   

      a.  Since are assumed to be subspaces themselves, 

          and Hence  

          Next, let Then by definition,  

          and where  

          But then commutativity 

          and associativity of vector addition).  Since are subspaces, 

          and This shows  

          Finally if then  

          Since are subspaces,  

          This shows  

      b. This is immediate since and they satisfy 

          the three parts of the definition of subspaces. 

34.  Since given  

      any we have   for 

      some scalars Similarly,  for some 

      scalars But then  

      which is in This shows 

      that Conversely, 

      if then for some scalars 

            =  

       This shows  

       Because of the two inclusions, the sets are equal. 

Section 4.2 

30.   We assume is a linear transformation and 

       consider for some
       (a) so (b)  Let  

       Then for some Then 

       because  is linear,   

       which shows (c)  Finally, let  

       and Then  for some Hence 

       This shows that  

       All three properties in the definition are satisfied, so  

       is a subspace of  

31.  a)  Let Then by the definition of T 

  This shows T  is a linear mapping. 

  b)  The kernel of is, by definition, the set of all  

       such that This says is a quadratic 

       polynomial with roots at Any such polynomial 

       is for some So 

       The image (range) of is all of since for any there is  

       a polynomial with and  

       fact, we can do this with a linear polynomial (the coefficient 

       of  

33.  a)  If  then  

          On the other hand,  

                               =  

                               =  

           Therefore, This can  

           also be proved by using properties of transposes.) 

           If then  

           Therefore, T  is linear. 

     b)  if  since  

since we assumed  

     c)  If  B  is any matrix with then part b shows 

          Conversely, if is any matrix  

          then  

          Hence the image of  T  equals the set of such matrices 

     d)  the set of all matrices with 

          a = 0, d = 0, b + c = 0.  These can be written as 

34.  By the second part of the Fundamental Theorem of Calculus, 

      if  then Hence 

      our mapping can be described as  

      By properties of integrals,  

      and 

      Hence  T  is a linear mapping. 

      The kernel of T  is the set of all functions with  

      (the zero function).  If we differentiate both sides of this, 

      we get  So the only element of 

      is the zero function in this case.   

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