# Different results for the same equation

2 views (last 30 days)
Diana on 14 Jan 2021
Commented: Walter Roberson on 14 Jan 2021
Why E1 is different than E3, though lambda(1)= lambda(3)

Steven Lord on 14 Jan 2021
lambda(1) is displayed the same as lambda(3) but its value is not identical to the value of lambda(3).
A = [-10 10 -15; 10 5 -30; -5 -10 0];
lambda = eig(A)
lambda = 3×1
-15.0000 25.0000 -15.0000
lambda(1) == lambda(3) % false
ans = logical
0
lambda(1)-lambda(3) % very small but not 0
ans = 3.5527e-15
##### 2 CommentsShowHide 1 older comment
Walter Roberson on 14 Jan 2021
A = [-10 10 -15; 10 5 -30; -5 -10 0];
lambda = eig(A)
lambda = 3×1
-15.0000 25.0000 -15.0000
lambda(1) + 15
ans = 3.5527e-15
lambda(3) + 15
ans = 0
The third lambda is an exact integer. When you calculate A-lambda(3)*eye(3) you get exact integers, and rref() is able to calculate exact integer outputs.
When you use format short (which is the default) and all of the outputs to be displayed are exact integers, then no decimal points are shown. When any of the outputs are not exact integers, then decimal points are shown in all of the outputs.

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