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Lepton (1/13)



  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. hello thank you for reply. DrRocket.. but.. i said that A^2 is not I BUT -I PLEASEEEGIVE ME ANOTHER PROOF..
  3. khan academy i think that this video is mainly come from www.khanacademy.org
  4. acidhoony


    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~
  5. 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~
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