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//Example 11.11
//Generalized Eigenvalue Problem
//Page no. 365
clc;close;clear;
A=[1,1,0.5;1,1,0.25;0.5,0.25,2]
B=[2,2,2;2,5,5;2,5,11]
disp(B,"B =",A,"A =")
for i=1:3
G(i,i)=sqrt(B(i,i))
end
G=[B;eye(3,3)];
//transformation to frobenius matrix
for k=3:-1:2
g(k)=0;
for j=1:k-1
if(g(k)<G(k,j))
g(k)=G(k,j)
p=j;
end
end
if(g(k)~=0)
for j=1:3
r(1,j)=G(k,j)
end
for i=1:6
G(i,k-1)=G(i,k-1)/g(k)
end
for j=1:3
if(j~=k-1)
l=G(k,j)
for i=1:6
G(i,j)=G(i,j)-l*G(i,k-1)
end
end
end
end
for j=1:3
for i=1:3
c(i,1)=G(i,j)
end
G(k-1,j)=0
for i=1:3
G(k-1,j)=G(k-1,j)+r(1,i)*c(i,1)
end
end
end
//partition g
for i=4:6
for j=1:3
T(i-3,j)=G(i,j)
end
end
//eigenvalues computation
p=poly(B,'x')
a=roots(p)
printf('\n\nDiagonalized Matrix B = \n\n')
for i=1:3
for j=1:3
if i~=j then
B(i,j)=0
else
B(i,j)=a(i)
end
end
end
disp(B)
//eigenvectors computation
for k=1:3
m=2
for l=1:3
y(l,k)=a(k)^(m)
m=m-1;
end
end
printf('\n\n')
for k=1:3
for l=1:3
y1(l,1)=y(l,1)
y2(l,1)=y(l,2)
y3(l,1)=y(l,3)
end
x1=T*y3;
x2=T*y2;
x3=T*y1;
end
printf('\n\nEigenvectors of B are :\n\n')
for i=1:3
printf('|%.5f|\t\t|%.5f|\t\t|%.5f|',x3(i,1),x2(i,1),x1(i,1))
printf('\n')
end
x=[x3,x2,x1]
B=[2,2,2;2,5,5;2,5,11]
G=0
for i=1:3
for j=1:3
if i==j then
G(i,j)=sqrt(B(i,j))
else
G(i,j)=0;
end
end
end
B=inv(G)*x'*A*x*inv(G)
disp(B,"Eigenvectors of A =")
printf('\n\n\nNote : Computation Error in book in caculation of eigenvector of B thus for A')
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