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MATH 244 -- Linear Algebra Section 1 

Problem Set 9 Solutions 

April 20, 2007 

Section 5.2 

6.  The characteristic polynomial is  

    The eigenvalues are the roots of 0 = so by the quadratic 

    formula:  

     Neither root is real, so there are no real eigenvalues. 

10.  The characteristic polynomial is  

      (after expanding out and simplifying). 

14.  The characteristic polynomial is (expanding determinant along row 2): 

18.  The eigenspace for is: 

      If we consider reducing this to row-reduced 

      echelon form, then there is only one free variable in the system  

      unless For h = 6, we have rref =  

      dimension 2.  In all other cases   

23.   If then This shows that and  

      are similar since the invertible matrix satisfies  

      (see the definition at bottom of page 314 in the text).   

25.   a)  By direct calculation,  

           so the given vector is an eigenvector with eigenvalue λ = 1.  The characteristic 

           polynomial is so the other eigenvalue is 

           Since the eigenvalues 

           1 and .3 are distinct, the set is linearly independent (Theorem 2 on 

            page 307 of the text, or direct check), hence a basis for  

       b)   

       c)  This is similar to the computations you did on Discussion 3: 

            Similarly, for all , by mathematical induction, 

            In the limit as so   

Section 5.3 

12.   Given that the eigenvalues are we have  

       Then, ℬ = is a basis for consisting of eigenvectors 

       of A.  This shows that  A  is diagonalizable, with  for  

       Check: 

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14.   Given the eigenvalues are (characteristic polynomial factors as 

       Theorem 7 on page 326 implies that in order for  

       A  to be diagonalizable, we must have  

       which is row-equivalent to    so there are 2 free 

       variables in the homogeneous system   

Then  

       This shows that A  is diagonalizable with  

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24.   No -- the hypotheses of Theorem 7 on page 324 cannot be satisfied in  

       this case.  The largest linearly independent set of eigenvectors will have 

       size 2, so there is no basis consisting of consisting of eigenvectors of A. 

26.   Yes -- if the remaining eigenspace has dimension only 1, then the matrix 

       is not diagonalizable (see Theorem7).   The following matrix gives an example  

       The eigenvalues are  (with multiplicity 2),  

      with multiplicity 3) , and (with multiplicity 2).  By the form of the 

      matrix  has dimension 2,   

      has dimension 3, but  

32.  Since A  is not invertible,    must be one of the eigenvalues (see Theorem on page 

      312 -- continuation of the Invertible Matrix Theorem).  By Theorem 6 on page 323,  

      if the other eigenvalue is (or anything other than 0), then  A  will be diagonalizable. 

      This means that A  should be similar to the diagonal matrix  To get  

      not diagonal we can try taking an arbitrary invertible matrix like with 

      Then   

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     is a matrix that satisfies all of the properties we want.   

Section 6.2 

6.   Call the three vectors  We have  

     So the set is not  orthogonal.  (It would be enough just to show  

     of course).   

10.  It is clear that here.  This implies that 

     is an orthogonal set.  By Theorem 4 in this section, it 

     follows that is linearly independent.  (This can also be checked  

     directly:  If  for some scalars then 

     we can take dot products with   

          0 =  

     Since  Similarly, dotting with  shows  and 

     dotting with  shows  It follows that is a basis for    

     Then because of the orthogonality, there is a shortcut  we can use for 

     computing the weights in a linear combination: 

     For each  i,  i = 1, 2, 3,  we have 

              (since the dot product of  with the other two terms 

                                         is zero). 

     Hence  for each  i = 1, 2, 3. 

12.   The orthogonal projection on the line spanned by is defined by  

       With  we get  

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