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# 04_linear_algebra | |
HW4 ME3255 | |
#Problem 2 | |
#Part A | |
Outputs of taking norms of matrices | |
4x4: (2 norm = 1.5002) (frobenius-norm = 1.5097) (1 norm = 2.0833) (infinity-norm = 2.0833) | |
5x5: (2 norm = 1.5671) (frobenius-norm = 1.5809) (1-norm = 2.2833) (infinity-norm = 2.2833) | |
#Part B | |
Outputs of taking norms of inverse Hilbert matrices | |
4x4: (2 norm = 1.0341e+04) (frobenius-norm = 1.0342e+04) (1 norm = 13620) (infinity-norm = 13620) | |
5x5: (2 norm = 3.0414e+05) (frobenius-norm = 3.0416e+05) (1 norm = 413280) (infinity-norm = 413280) | |
#Part C | |
Outputs of condition numbers on 4x4 and 5x5 Hilbert matrices | |
4x4: (2-norm = 1.5514e+04) (frobenius-norm = 1.5614e+04) (1 norm = 2.8375e+04) (infinity-norm = 2.8375e+04) | |
5x5: (2 norm = 4.7661e+05) (frobenius-norm = 4.8085e+05) (1 norm = 9.4366e+05) (infinity-norm = 9.4366e+05) | |
#Problem 3 and 4 | |
``` | |
function [d,u]=chol_tridiag(e,f); | |
% chol_tridiag is a function that takes 2 vectors as inputs and calculates | |
%the Cholseky factorization of a tridiagonal matrix | |
% given e, the off-diagonal vector | |
% and f, the diagonal vector | |
% output = [d,u] | |
% d is the diagonal of the Upper matrix | |
% u isthe off-diagonal of the Upper matrix | |
d = zeros(length(f),1); | |
u = zeros(length(f)-1,1); | |
d(1) = sqrt(f(1)); | |
u(1) = e(1)/d(1); | |
l = 2; | |
while l <= length(f)-1 | |
d(l) = sqrt(f(l)- (u(l-1))^2); | |
u(l) = e(l)/d(l); | |
l = l+1; | |
end | |
d(4) = sqrt(f(4)-(u(3))^2); | |
end | |
``` | |
``` | |
function [x] = solve_tridiag(u,d,b) | |
% provides solution of Ax=b | |
% d = diagonal of upper matrix of cholesky factorization | |
% u = off-diagonal of upper matrix of cholesky factorization | |
% b = the vector | |
x = zeros(1,length(b)); | |
y = zeros(1,length(b)); | |
y(1) = b(1)/d(1); | |
for i=2:length(b) | |
y(i)=(b(i)-u(i-1)*y(i-1))/d(i); | |
end | |
x(length(b)) = y(length(b))/d(length(b)); | |
for i = n-1:-1:1 | |
x(i) = ((y(i)-u(i))*x(i+1))/d(i); | |
end | |
end | |
``` | |
#Problem 5 | |
for k = 1000, x = 29.28 and error = 29.28 | |
for k2 = 1000e12 x = (2.9e10) and error = 1.13e12 | |
for k2 = 1000e-12 x = (3e-10) and error = 9.001e12 | |
#Problem 6 | |
x = [0.039, 0.069, 0.083, 0.098] | |
#Problem 7 | |
answer that I got was =[40.522, 14.409, 2.569] | |
#Problem 8 | |
| number of segments | Largest Load | Smallest Load | num of eigenvalues | | |
| ------------- | ------------- |---------------|--------------------| | |
| 5 | 1.1756 | 0.6180 | 2 | | |
| 6 | 1.2 | 0.6212 | 3 | | |
| 10 | 1.2361 | 0.6257 | 7 | | |
When the segment length approaches 0, the number of eigenvalues also approaches zero. |