## 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

1. Loser66

@Empty

2. Loser66

@phi

3. Loser66

@cwrw238

4. 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

5. 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?

6. phi

Let the eigenvalues be -1 1 make up two eigenvectors and compute A = S^-1 L S

7. phi

example A= -3 -4 2 3

8. phi

eigs(A) ans = 1 -1 eigs(A^2) ans = 1 1

9. 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?

10. 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

11. phi

In general, if the values are distinct , then the squares are distinct. So it is a special case that they become "not distinct"

12. phi

@Empty but it looks nicer

13. Loser66

Hey!! At the end, it is T or F?

14. phi

It only takes one counter-example to make it false. The statement is false

15. Loser66

From @Empty we have lambda1 , lambda 2 (distinct) after then, the eigenvalues of A^2 are lambda1 = - lambda2, (distinct also) ??? I confused

16. 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)

17. Loser66

Got you. Thanks a lot.