acidhoony

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hello
thank you for reply. DrRocket..
but..
i said that A^2 is not I BUT I
PLEASEEEGIVE ME ANOTHER PROOF..
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khan academy
i think that this video is mainly come from www.khanacademy.org
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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~
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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 cayleyhamiltion 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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linear algebra help
in Homework Help
Posted
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.