Question

Suppose A is n*n square matrice on R that A^2=-I
proof : 1. A has no Eigenvalues on R .
2. n is even

Answer 1

hey friend, A is a matrix how A^2 =-1? is there something missed?

Answer 2

I am studying this part, too. so that's why I concern. We can do the stuff together.

Answer 3

that is not -1 is -I (identity matrice)

Answer 4

I mean A^2+I=0

Answer 5

oh, ok, I misunderstand, hihi, sorry, I'm thinking.

Answer 6

hi friend, I think we can start at A^2 +I = 0, since I is identity matrix, so A^2 must have the form of -1 in main diagonal line and any other entries =0. and then A^2 = A *A it's mean -1= something^2 it's mean...hehehe...what? you know and I know , too

Answer 7

A^2 =\[\left[\begin{matrix}-1 & 0 &0 &..... \\ 0 & -1 & 0 & ..........\\0 & 0 & -1 & .....\end{matrix}\right]\]

Answer 8

do you get what I mean? if my English is toooo complicated, and you don't understand, just speak out. I'm ok

Answer 9

I think do not understand

Answer 10

@Amirsd , you there? sorry, my computer lost connection with the site a while

Answer 11

Yes I am here friend

Answer 12

ok, I explain you now, you have A^2 + I = 0 right?

Answer 13

yes

Answer 14

\[\left[\begin{matrix}-1 & 0 &0&......\\ 0 & -1&0&.........\\0&0&-1&......\end{matrix}]+\left[\begin{matrix}1 & 0&0&.....\\ 0 & 1&0&...\\0&0&1&...\end{matrix}\right]\right]\] =0

Answer 15

the first matrix is A^2

Answer 16

got it?

Answer 17

wait

Answer 18

the second one is I , sum of them is 0

Answer 19

Ok

Answer 20

so, A^2 = A * A what it mean? it mean A is a complex matrix with entries in main diagonal line is . Why? because i^2 = -1

Answer 21

entries = i

Answer 22

yes that is right

Answer 23

ok, can you make up the characteristic equation of \[\left[\begin{matrix}i & 0&0&.... \\ 0& i&0&...\\0&0&i&.....\end{matrix}\right]\]

Answer 24

and get a real eigenvalue?

Answer 25

no

Answer 26

that the proof of your a) question. no real eigenvalue for A

Answer 27

so what about 2 ?

Answer 28

you mean question b) n must be even?

Answer 29

yes

Answer 30

because when you diagonalize a matrix to get A^n and L^n (lamda) to the theorem, you have L^n is eigenvalue of A^n , since -1 =i^2 , that means n =2 if n =3 i^3 is not -1, it's is 1 ---> A^2 + I != 0 (not equal )

Answer 31

got it? hehehe... thanks for your question, I gain too much

Answer 32

so \[AX=\lambda A\]
why whe diagonalize ?

Answer 33

I mean for get A^n?

Answer 34

our matrice was A^2 is't it ?

Answer 35

hey, friend, your problem relate to that, A^2 if you don't diagonalize how to come up with A^2 by D and no way to deal with Lamda.

Answer 36

so what about A^n why you write that ?

Answer 37

you see your A is a diagonal matrix, it mean A^2 = A *A = {diagonal entries ^2} and A^n = {diagonal entries ^n} and the eigenvalue is counted by L ^2, L ^n .... respectively,. If n= even , i ^ (even) number = -1, if n = odd, i^(odd) =1 which not satisfy the condition of A^n + I =0

Answer 38

I have to go to my friend' post to help her.

Answer 39

thank you for helping friend I took your time

Answer 40

it's ok, I help you to learn too.