Different results for the same equation

Why E1 is different than E3, though lambda(1)= lambda(3)

Answer 1

[Steven Lord]

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

[Diana]

but that should not effect the final result

[Walter Roberson]

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.