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Posts posted by acidhoony

  1. [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.

  2. 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~

  3. 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




    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~



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