Checking whether two complex matrices are equal

For matrix A=[3 -3 4;2 3 4;0 -1 1]; I have to prove that eigenvalues of A^2 is equal to square of eigenvalues of A. After obtaining eigenvales I used == to compare results. Both matrix were same but using == gives me result [0 1 1;1 0 1; 1 1 0]. Why?

Answer 1

[James Tursa]

You can't rely on floating point calculations to give exact results. See this link:

https://ch.mathworks.com/matlabcentral/answers/57444-faq-why-is-0-3-0-2-0-1-not-equal-to-zero?s_tid=srchtitle

You need to use tolerances for these types of floating point comparisons. E.g.,

>> A = [3 -3 4;2 3 4;0 -1 1];
>> A2 = A*A;
>> eig(A).^2
ans =
 -0.999999999999988 +18.973665961010287i
 -0.999999999999988 -18.973665961010287i
  1.000000000000000 + 0.000000000000000i
>> eig(A2)
ans =
 -1.000000000000000 +18.973665961010266i
 -1.000000000000000 -18.973665961010266i
  1.000000000000001 + 0.000000000000000i
>> eig(A).^2 == eig(A2)
ans =
  3×1 logical array
   0
   0
   0
>> abs(eig(A).^2 - eig(A2)) < 1e-10
ans =
  3×1 logical array
   1
   1
   1

But, this isn't really a "proof" of course ... just a demonstration within numerical tolerances.

A proof for this case would be to use the Symbolic Toolbox. E.g., start with this:

   As = sym(A);

And then work everything with As instead of A. Then see how the results compare ... (hint: you may need to use the simplify( ) function and you may need to reorder the eigenvalues before the comparison)

A general proof for an arbitraty matrix A would be starting with A*x = lambda*x on paper, multiplying both sides by A, and then seeing what you get as a result using hand calculations and simplifying.

[Amrutha Rajan]

Thank you so much.