we know that:
T = (1/3)(TT* - 2I)
so T* = [(1/3)(TT* - 2I)]* = (1/3)(TT*)* - (2I)* = (1/3)((T*)*T*) - 2I*
= (1/3)(TT* - 2I) = T.
ok so we got the T*=T part
now in order o find the eigenvalues
$\displaystyle T(T^{*}-3I)+2I=0$
$\displaystyle T^2-3T+2I=0$
$\displaystyle (T^2-3T+2I)v=0$
$\displaystyle (T-2I)(T-I)v=0$
so k=2 and k=1 are eigen values and they are positive
but i am not sure if i have done it correctly
because by definition k is an eigen value if T(v)=kv
or if k nullifies the caracteristic polinomial
i cant show neither of those here
i just said intuitivly that 1 ,2 are eigenvalues
??
i can replace T=Av
$\displaystyle (Av-2I)(Av-I)v=0$
but it doesnt go anywhere
x^2 - 3x + 2 is a polynomial that T satisfies. so the minimal polynomial m(x) divides this. there are precisely 3 possibilities:
1) m(x) = x - 2
2) m(x) = x - 1
3) m(x) = (x - 2)(x - 1).
1) means T = 2I. 2) means T = I. in both of these cases, T is positive-definite.
if we have case 3), we know that the minimal polynomial for T has the same roots as the characteristic polynomial for T. thus T has eigenvalues 1,2 > 0, so T is positive-definite.
you are correct when you say we don't know what the characteristic polynomial of T is, we aren't given enough information. and you can easily verify that I and 2I both satisfy the given equation, so T might be one of those 2 matrices. but even if it isn't, we still know it HAS to be positive-definite, because we know all the roots of the characteristic polynomial, even though we don't know what the polynomial actually is.