04_linear_algebra

Linear Algebra HW 4

Problem 2

Part A

H4 = [1,1/2,1/3,1/4;1/2,1/3,1/4,1/5;1/3,1/4,1/5,1/6;1/4,1/5,1/6,1/7];
norm(H4,2)
norm(H4,'fro')
norm(H4,1)
norm(H4,inf)

H5 = [1,1/2,1/3,1/4,1/5;1/2,1/3,1/4,1/5,1/6;1/3,1/4,1/5,1/6,1/7;1/4,1/5,1/6,1/7,1/8;1/5,1/6,1/7,1/8,1/9]
norm(H5,2)
norm(H5,'fro')
norm(H5,1)
norm(H5,inf)

Output (4x4):

• 2-norm: 1.5002
• frobenius-norm: 1.5097
• 1-norm: 2.0833
• inf-norm: 2.0833

Output (5x5):

• 2-norm: 1.5671
• frobenius-norm: 1.5809
• 1-norm: 2.2833
• inf-norm: 2.2833

Part B

H4 = [1,1/2,1/3,1/4;1/2,1/3,1/4,1/5;1/3,1/4,1/5,1/6;1/4,1/5,1/6,1/7];
iH4 = inv(H4)
norm(iH4,2)
norm(iH4,'fro')
norm(iH4,1)
norm(iH4,inf)

H5 = [1,1/2,1/3,1/4,1/5;1/2,1/3,1/4,1/5,1/6;1/3,1/4,1/5,1/6,1/7;1/4,1/5,1/6,1/7,1/8;1/5,1/6,1/7,1/8,1/9];
iH5 = inv(H5)
norm(iH5,2)
norm(iH5,'fro')
norm(iH5,1)
norm(iH5,inf)

Output (4x4):

• 2-norm: 1.0341e+04
• frobenius-norm: 1.0342e+04
• 1-norm: 1.3620e+04
• inf-norm: 1.3620e+04

Output (5x5):

• 2-norm: 3.0414e+05
• frobenius-norm: 3.0416e+05
• 1-norm: 4.1328e+05
• inf-norm: 4.1328e+05

Part C

H4 = [1,1/2,1/3,1/4;1/2,1/3,1/4,1/5;1/3,1/4,1/5,1/6;1/4,1/5,1/6,1/7];
cond(H4,2)
cond(H4,'fro')
cond(H4,1)
cond(H4,inf)

H5 = [1,1/2,1/3,1/4,1/5;1/2,1/3,1/4,1/5,1/6;1/3,1/4,1/5,1/6,1/7;1/4,1/5,1/6,1/7,1/8;1/5,1/6,1/7,1/8,1/9];
cond(H5,2)
cond(H5,'fro')
cond(H5,1)
cond(H5,inf)

Output (4x4):

• 2-norm: 1.5514e+04
• frobenius-norm: 1.5614e+04
• 1-norm: 2.8375e+04
• inf-norm: 2.8375e+04

Output (5x5):

• 2-norm: 4.7661e+05
• frobenius-norm: 4.8085e+05
• 1-norm: 9.4366e+05
• inf-norm: 9.4366e+05

Problem 3

function [d,u]=chol_tridiag(e,f);
  % chol_tridiag calculates the components for Cholesky factorization of a symmetric
  % positive definite tridiagonal matrix
  % given its off-diagonal vector, e  
  % and diagonal vector f
  % the output is
  % the diagonal of the Upper matrix, d
  % the off-diagonal of the Upper matrix, u

 d = zeros(length(f),1);
 u = zeros(length(f)-1,1);

 d(1) = sqrt(f(1));
 u(1) = e(1)/d(1);

 k = 2;

 while k <= length(f)-1
     d(k) = sqrt(f(k)- (u(k-1))^2);
     u(k) = e(k)/d(k);
     k = k+1;
 end

 d(4) = sqrt(f(4)-(u(3))^2);

end

Problem 4

function [x] = solve_tridiag(u,d,b)
  % Function that provides the solution of Ax=b
  % given:
  % diagonal of upper matrix of cholesky factorization, d
  % off-diagonal of upper matrix of cholesky factorization, u
  % the vector, b

  x = zeros(1,length(b));
  y = zeros(1,length(b));
  % forward substitution
  y(1) = b(1)/d(1);

  for i=2:length(b)
      y(i)=(b(i)-u(i-1)*y(i-1))/d(i);
  end
  % back substitution
  x(length(b)) = y(length(b))/d(length(b));
  for i = length(b)-1:-1:1
      x(i) = ((y(i)-u(i))*x(i+1))/d(i);
  end
end

Problem 5

Part A

k = 1000;
k1 = k;
k2 = k1;
k3 = k2;
k4 = k3;
K = [2*k,-k,0,0;-k,2*k,-k,0;0,-k,2*k,-k;0,0,-k,k];
x = cond(K)
c = log10(x);
t_k = 0; % accuracy of coefficients of K: 10^-t (accuracy is 1 so t = 0)
e = 10^(c-t_k)

Condition:

• 29.2841

Error:

• 29.2841

Part B

k = 1000;
k1 = k;
k2 = 1000*10^12;
k3 = k1;
k4 = k3;
K = [k1+k2,-k2,0,0;-k2,k2+k3,-k3,0;0,-k3,k3+k4,-k4;0,0,-k4,k4];
x = cond(K)
c = log10(x);
t_k = 9; % accuracy of coefficients of K: 10^-t
e = 10^(c-t_k)

Condition:

• 1.1291e13

Error:

• 1.1291e22

Part C

k = 1000;
k1 = k;
k2 = 1000*10^-12;
k3 = k1;
k4 = k3;
K = [k1+k2,-k2,0,0;-k2,k2+k3,-k3,0;0,-k3,k3+k4,-k4;0,0,-k4,k4];
x = cond(K)
c = log10(x);
t_k = 9; % accuracy of coefficients of K: 10^-t
e = 10^(c-t_k)

Condition:

• 9.001e12

Error:

• 9.001e3

Problem 6

Part A

k = 1000;
k1 = k;
k2 = k1;
k3 = k2;
k4 = k3;
e = [-k2,-k3,-k4];
f = [k1+k2,k2+k3,k3+k4,k4];
[d,u] = chol_tridiag(e,f)
b = [1;1;1;1];
[x] = solve_tridiag(u,d,b)

Displacements:

• mass 1: 0.0025 m
• mass 2: 0.0050 m
• mass 3: 0.0075 m
• mass 4: 0.0100 m

Part B

k = 1000;
k1 = k;
k2 = 1000e12;
k3 = k1;
k4 = k3;
e = [-k2,-k3,-k4];
f = [k1+k2,k2+k3,k3+k4,k4];
[d,u] = chol_tridiag(e,f)
b = [1;1;1;1];
[x] = solve_tridiag(u,d,b)

Displacements:

• mass 1: 0.0023 m
• mass 2: 0.0023 m
• mass 3: 0.0047 m
• mass 4: 0.0070 m

Part C

k = 1000;
k1 = k;
k2 = 1000e-12;
k3 = k1;
k4 = k3;
e = [-k2,-k3,-k4];
f = [k1+k2,k2+k3,k3+k4,k4];
[d,u] = chol_tridiag(e,f)
b = [1;1;1;1];
[x] = solve_tridiag(u,d,b)

Displacements:

• mass 1: 3008695.77 m
• mass 2: 3008695769.04 m
• mass 3: 3005690078.97 m
• mass 4: 2999690697.57 m

Problem 7

function  [v,w] = mass_spring_vibrate(k1,k2,k3,k4)
    % Function that finds the vibration modes (v) and natural frequencies (w) of a
    % spring - mass system given four parameters k1,k2,k3,k4
    m1 = 1; %kg
    m2 = 2;%kg
    m3 = 4; %kg
    % x1 = A1*sin(w*t);
    % x2 = A2*sin(w*t);
    % x3 = A3*sin(w*t);
    %((k2+k1)/m1-w^2)*x1 - (k2/m1)*x2 = 0; differential equation 1
    %((k2+k3)/m2)*x2 - (k2/m2)*x1 - (k3/m2)*x3 = 0; differential equation 2
    %((k3+k4)/m3)*x3 - (k4/m3)*x2 = 0; differential equation 3
    K = [(k2+k1)/m1, -k2/m1, 0; -k2/m2,(k2+k3)/m2, -k3/m2; 0, -k4/m3, (k2+k1)/m3];
    [v,lambda] = eig(K);
    w = sqrt(lambda);
end

Outputs:

• Mode 1: Natural Frequency = 6.345 rad/sec
  • Mass 1 and 3 in phase with each other, mass 2 out of phase
• Mode 2: Natural Frequency = 3.594 rad/sec
  • Mass 1 and 2 in phase with each other, mass 3 out of phase
• Mode 3: Natural Frequency = 2.08 rad/sec
  • Mass 1,2 and 3 in phase with each other

Problem 8

# of segments	largest load (N)	smallest load (N)	# of eigenvalues
5	10998.82	1161.18	4
6	16337.43	1172.97	5
10	47449.69	1190.31	9

As the number of segments approaches zero, the eigenvalues will decrease until there is only one