acidhoony

[math]A^2= \-I[/math]

 

Note: [math](det \ A)^2=det(-I)=(-1)^n \ \Rightarrow \ n[/math] is even.

 

Note also that the conclusuin that n is even and [math] det \ A \ = \ \pm 1 [/math] would not hold over the complex field, so the proof must take advantage of the fact that we are working over the reals.

 

Sketch of proof:

 

1. Using Jordan canonical form A can be assumed to be block diagonal, with 2x2 blocks. This reduces the problem to the case n=2.

 

2. So now [math] A \in (O2)[/math]. [math]A[/math] is thus a composition of a rotation and, if [math]det \ A \ = \ -1[/math], an orientation changing transformation, which can be taken to be

.

 

[math] C= \left ( \begin{array} {cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right )[/math]

 

 

Note that [math]C[/math] does not commute with rotations, but yet that any element in [math]O2[/math] can be written as [math]B[/math] or [math]BC[/math] where [math]B[/math] is some rotation.

 

3. Now examine the possibilities [math]A= BC[/math] where [math]B[/math] is a rotation and show that [math](BC)^2 \ne \ -I[/math] .

 

4. That the shows that [math] A [/math] is a rotation through [math]\frac {\pi}{2}[/math] or [math]\frac {3 \pi}{2}[/math] and hence has determinant 1.

i have another solution

please read me

dr rocket

we haver real matrix A so we have eigen vector which is unreal. and this eigen vector is conjugated

because the polynomier of matrix A is "poly"nomier it self

and every coefficienet is "real"

every eigen value is conjugated!!

and A^2 = - I => A^2X=-X =AAX=AaX=a^2x (here a is eigenvalue of matrix A)

so we know that (eivenvalue of A )^2 = -1

matrix A is 2k*2k matrix either

so we have a1 ................ak, a(k+1),a(k+2),,,,,a2k

and they are a k pair of conjugated eigenvalue

eigenvalue's product is 1

i didn't prove it well but i think you will be understood it.

hello

thank you for reply. DrRocket..

but..

i said that A^2 is not I BUT -I

PLEASEEEGIVE ME ANOTHER PROOF..

khan academy

i think that this video is mainly come from www.khanacademy.org

you may got c1A1 + c2A2 + c3A3 = (a b)

(c d)

which a,b,c,d are arbitary real number maybe~

if you can find c1,2,3for every a,b,c,d

then it A1,2,3 is span all 2 by 2 Matrix.

but i think it will not span.

Because a,b,c,d is fourrr v variable but so we need at least for basis

but we have only A1,2,3 ; only three ;;; i don;t know it is dependent or not , but , whatever it is

it is too little number of matrix to span every two by two matrix~

now i'm using apostol calculus book user,

the problem says that

given n*n matrix A with real entries such that A^2=-I I=UNIT MATRIX

THEN PROVE THAT

det A = 1

i know that (det A)^2=1

so, det A = +1 or -1

but i cannot prove that why -1 is not

how can we prove this thing without using cayley-hamiltion Th. ;because it's proof is either complicated

i think this problem may not using cayley hamiltion Th.

please solve this problem~