Question

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

Answer 1

@Empty

Answer 2

@phi

Answer 3

@cwrw238

Answer 4

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

Answer 5

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?

Answer 6

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

Answer 7

example
A=

-3 -4
2 3

Answer 8

eigs(A)

ans =

1
-1

eigs(A^2)

ans =

1
1

Answer 9

@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?

Answer 10

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

Answer 11

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

Answer 12

@Empty but it looks nicer

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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)

Answer 17

Got you. Thanks a lot.