------------- Problem 4 ----------------
Applying Rayleih quotient algorithm to
With A = [2 1 1; 1 2 1; 1 1 2];
x0 = [1; -1; 2];
and printing at the end of the loop (see p63 of notes) the eigenvalue and
eigenvector and the difference between two succesive eigenvalue iterates
----------------------------------------
 i= 0 la=1.6666667e+00 err=           
   x=   0.408248290463863  -0.408248290463863   0.816496580927726

 i= 1 la=1.0683761e+00 err=5.9829060e-01
    x=  -0.065372045046061   0.849836585598797  -0.522976360368491

 i= 2 la=1.0000381e+00 err=6.8338005e-02
    x=   0.156358879152925  -0.769455347348119   0.619265992403447

 i= 3 la=1.0000000e+00 err=3.8063029e-05
Done in 3 iterations
eigenvalue = 1.0000000e+00
eigenvector =  -0.154303323868882   0.771516775903129  -0.617213373755831

------------- Problem 5 ----------------
Applying QR algorithm with deflation as stated in the HW to the matrix
A = [2 -1 0; -1 2 -1; 0 -1 2];
and printing the updated A at every iteration after the update 
A(1:m,1:m) = RQ + mu*I
________________________________________

**** iteration 1
   1.000000000000000   0.707106781186547  -0.000000000000000
   0.707106781186547   2.000000000000000  -0.707106781186547
                   0  -0.707106781186547   3.000000000000000

**** iteration 2
   0.693997690565051   0.375974484550621   0.000000000000000
   0.375974484550621   1.892417374123494  -0.030396964933976
                   0  -0.030396964933976   3.413584935311455

**** iteration 3
   0.614479590481565   0.199386431066825  -0.000000000000000
   0.199386431066825   1.971306847145480  -0.000000447976474
                   0  -0.000000447976474   3.414213562372954

**** iteration 4
   0.585786437626907   0.000000000000000  -0.000000000000000
  -0.000000000000000   2.000000000000139                   0
                   0                   0   3.414213562372954

done in 4 iterations
The approximate eigenvalues are
   0.585786437626907   2.000000000000139   3.414213562372954