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Loser66
 one year ago
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
Loser66
 one year ago
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

This Question is Closed

phi
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.0Since 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
 one year ago
Best ResponseYou've already chosen the best response.2Let the eigenvalues be 1 1 make up two eigenvectors and compute A = S^1 L S

phi
 one year ago
Best ResponseYou've already chosen the best response.2eigs(A) ans = 1 1 eigs(A^2) ans = 1 1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.2In general, if the values are distinct , then the squares are distinct. So it is a special case that they become "not distinct"

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Hey!! At the end, it is T or F?

phi
 one year ago
Best ResponseYou've already chosen the best response.2It only takes one counterexample to make it false. The statement is false

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0From @Empty we have lambda1 , lambda 2 (distinct) after then, the eigenvalues of A^2 are lambda1 =  lambda2, (distinct also) ??? I confused

phi
 one year ago
Best ResponseYou've already chosen the best response.2He 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
 one year ago
Best ResponseYou've already chosen the best response.0Got you. Thanks a lot.
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