function [normPR,normQR]=hw3_2_2(n)

% kinda covers problems 2.2 from hw3 and 2.3 from hw4.
% the size of the matrix, n, is the input.
% normPR=norm(P-R) and normQR=norm(Q-R) and the outputs.
% you can write another script to run through different values of n
% just supress the the other disp commands in this file by commenting them out

format short g
A=hilb(n)
P=inv(A'*A)
Q=inv(A)*inv(A)'
disp('P and Q look the same but...Q-P is:')
disp(Q-P)

% R is the exact inverse of A*A', i.e., computed algebraicly not numerically
R=invhilb(n)*invhilb(n)'

fprintf('Distance from P to R: %6.2f \nand the distance from Q to R is: %6.2f \n',norm(P-R),norm(Q-R))

% so Q is "closer" to R than P is, but there is still some error in the 
% computation...why? Let's look at condition numbers

fprintf('condition number of A: %6.2f \n',cond(A))
fprintf('condition number of A''*A: %6.2f',cond(A'*A))
format short

% so it's MUCH more difficult so solve the system A'*A*x=b than it is to solve 
% the system A*x=b. i.e. it's harder to compute inv(A'*A) accurately than it is
% to just compute inv(A). So Q has less error than P since Q only inverts A and 
% P directly inverts A'*A