Reduced Echelon form is wrong

Respuestas (1)

John D'Errico
John D'Errico el 9 de En. de 2021
Editada: John D'Errico el 9 de En. de 2021

0 votos

A = [-2 2;-.4 .2];
[V,D] = eig(A)
V =
-0.974588432346814 -0.754384720454159
-0.224003097156669 -0.656432550644239
D =
-1.54031242374329 0
0 -0.259687576256715
B = A - D(1,1)*eye(2)
B =
-0.459687576256715 2
-0.4 1.74031242374329
Now, B is a rank 1 matrix. We can approximate the matrix quite well as an outer product of two vectors, as we can see here:
rank(B)
ans =
1
[U,S,V ] = svd(B)
norm(U(:,1)*S(1,1)*V(:,1)' - B)
ans =
1.90253315901036e-15
That rref seems to miss this is just an issue of tolerances.
[R,jb] = rref(A - D(1,1)*eye(2))
R =
1 0
0 1
jb =
1 2
However, if we increase the tolerance by a little beyond the default applied in rref, we see:
[R,jb] = rref(A - D(1,1)*eye(2),1e-14)
R =
1 -4.35078105935822
0 0
jb =
1
Now rref is able to agree with the finding of the other tools.
Remember that all numerical computations are subject to trash in the least significant bits, and that you need to use tolerances properly and carefully to resolve any problem. Understanding the computations done is a huge part of resolving those problems.

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R2020b

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Preguntada:

el 9 de En. de 2021

Editada:

el 9 de En. de 2021

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