Loser66
  • Loser66
Let A be a real 2 x 2 matrix If A has 2 distinct eigenvalues, then A^2 has 2 distinct eigenvalues, True of False. why? Please, help
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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Loser66
  • Loser66
@Empty
Loser66
  • Loser66
@phi
Loser66
  • Loser66
@cwrw238

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phi
  • phi
I think you can write the matrix as S^-1 L S where L has the eigenvalues on the diagonal, and S is the eigenvectors A^2 = S^-1 L S S^-1 L S= S^-1 L^2 S if the values are distinct say -1 and +1 , after squaring you would get +1 +1 no longer distinct
Loser66
  • Loser66
Since it is False, we can find out a counter example to prove it False What if it is True, how to prove? Like: A is 2 x2 matrix, the determinant of A^2 is nonnegative,. How to prove?
phi
  • phi
Let the eigenvalues be -1 1 make up two eigenvectors and compute A = S^-1 L S
phi
  • phi
example A= -3 -4 2 3
phi
  • phi
eigs(A) ans = 1 -1 eigs(A^2) ans = 1 1
Loser66
  • Loser66
@phi I got you. One more question, if I not lucky enough to pick the right one to manipulate, (I meant if I pick the wrong one which gives me the eigenvalues of A^2 distinct,) I fail the test, right?
Empty
  • Empty
If the matrix has two eigenvalues then you can diagonalize it: \[A=PDP^{-1}\] where \[D=\left[ \begin{array}c \lambda_1 & 0\\0 & \lambda_2\\\end{array} \right]\] so you can square A to get: \[A^2= (PDP^{-1})( PDP^{-1}) = PD^2P^{-1}\] and D^2 has the form: \[D^2=\left[ \begin{array}c \lambda_1 ^2 & 0\\0 & \lambda_2^2\\\end{array} \right]\] which are the eigenvalues of D, so like @phi is saying, \[\lambda_1 = - \lambda_2 \] would be a counter example. I realizeI probably just wasted my time repeating everything that was just said because I'm a slow typer lol oh well
phi
  • phi
In general, if the values are distinct , then the squares are distinct. So it is a special case that they become "not distinct"
phi
  • phi
@Empty but it looks nicer
Loser66
  • Loser66
Hey!! At the end, it is T or F?
phi
  • phi
It only takes one counter-example to make it false. The statement is false
Loser66
  • Loser66
From @Empty we have lambda1 , lambda 2 (distinct) after then, the eigenvalues of A^2 are lambda1 = - lambda2, (distinct also) ??? I confused
phi
  • phi
He is saying if you start with L1 = - L2 (thus distinct) for matrix A then for matrix A^2 you get L1^2 = L2^2 (no longer distinct)
Loser66
  • Loser66
Got you. Thanks a lot.

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