Hilbert matrix come on
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n=100;
for i=1:n
for j=1:n
A(i,j)=1/(i+j-1);
end
end
[L,U,P]=lu(A),xex=ones(n,1),
b=A*xex,R=L*U-P*A,y=L\P*b;
x=U\y ,r=max(abs(R))
for k=1:n; er(k)=norm(xex-x,2)/norm(xex,2) ; end
semilogy((1:n),er,'r--',(1:n),r,'g--'),
there is a problem with er when i draw the curve i find it stable (red)
2 个评论
David Goodmanson
2021-1-14
Hi Karim
You can save some lines of code with the hilb function.
The big problem here is that you are always operating on the same size matrix, so of course er does not change. You need the matrix to change size for each iteration of the for loop.
Walter Roberson
2021-1-14
for k=1:n; er(k)=norm(xex-x,2)/norm(xex,2) ; end
is always doing the same calculation so it always has the same result.
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