MATLAB problems from Midterm Problem Set

  II D.  Finding the 2-norm of $B$ by the formula (3):

       >> B = bucky;
       >> max(eig(B'*B))
       ans =
              9.0000

  Hence ||B||_2 = 3.  Check with MATLAB's norm
  function:


       >> norm(full(B),2)
       
       ans =

              3

  II F.  MATLAB can compute Frobenius norms directly 
  like this:

      >> norm(B,'fro')

      ans =

         13.4164

  It's not hard to see that this can also be computed
  directly as sqrt(sum(diag(B'*B))) (i.e. the square root
  of the trace of the matrix B'*B -- the sum of the diagonal
  entries).  

  The inequality (3) is satisfied since  

             3 <= 13.4164 <= 3 sqrt{60} =~ 23.2379.

  III A.  X is the following 2 x 7 matrix:

      >> X = [1.001 1; 1.002 1; 1.003 1; 1.004 1; 1.005 1; 1.006 1; 1.007 1]

       X =

          1.0010    1.0000
          1.0020    1.0000
          1.0030    1.0000
          1.0040    1.0000
          1.0050    1.0000
          1.0060    1.0000
          1.0070    1.0000

      >> cond(X'*X,inf)

         ans =

            1.0121e+06

      This is rather large, which indicates that the normal equations
      are relatively ill-conditioned.  We can see this from the form
      of the matrix X'*X also:

      >> X'*X

         ans =

             7.0561    7.0280
             7.0280    7.0000

       We can see that the columns are quite close to being proportional,
       so this matrix is "nearly singular".  (This condition number is
       not large enough that MATLAB's default double precision floating point
       number system gives an inaccurate result, however.)

  III B.  The shifted and rescaled system is

      >> Xprime = [-.9 1; -.6 1; -.3 1; 0 1; .3 1; .6 1; .9 1]

         Xprime =

               -0.9000    1.0000
               -0.6000    1.0000
               -0.3000    1.0000
                     0    1.0000
                0.3000    1.0000
                0.6000    1.0000
                0.9000    1.0000

      >> cond(Xprime'*Xprime,inf)

      ans =

          2.7778

      This is much smaller, so the modified normal equations will
      be much better conditioned.  The solution is 

      >> Y = [3.664; 3.789; 3.891; 4.022; 4.233; 5.200; 5.329] 

          Y =

             3.6640
             3.7890
             3.8910
             4.0220
             4.2330
             5.2000
             5.3290

      >> Xprime'*Xprime \ Xprime'*Y

      ans =

            0.9713
            4.3040

      This says the best-fit line is given by 

            y = (.9713)(300(x - 1.004)) + 4.3040

              = 291.39 x - 288.25

      (agrees with solving the usual normal equations).