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Reduced Echelon form is wrong

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Diana Dawoud
Diana Dawoud on 9 Jan 2021
Edited: John D'Errico on 9 Jan 2021
As you can see the reduced echelon form is wrong, it should be [1 -4.35; 0 0]

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Answers (1)

John D'Errico
John D'Errico on 9 Jan 2021
Edited: John D'Errico on 9 Jan 2021
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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