Different results for the same equation

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Diana
Diana 2021-1-14
评论： Walter Roberson 2021-1-14
Why E1 is different than E3, though lambda(1)= lambda(3)

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回答（1 个）

Steven Lord
Steven Lord 2021-1-14
Ran in:
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 个评论
Diana
Diana 2021-1-14
but that should not effect the final result
Walter Roberson
Walter Roberson 2021-1-14
Ran in:
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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