Linear Operators: Proving Determinant Property

[MathematicalPhysicist]

i need to prove the next statement:
let S and T be linear operators on a vector space V, then det(SoT)=det(S)det(T).

my way is this:
let v belong to V, and {e_i} be a basis of V
v=e1u1+...+e_nu_n
then T(v)=e1T(u1)+...+enT(un)
(SoT)(v)=S(T(v))=S(e1T(u1)+...+enT(un))=e1S(T(u1))+...+enS(T(un))
but i don't know how to proceed from here.

 

[matt grime]

Wasn't there a far too long thread on this recently somewhere round here.

How have you defined the determinant, and what are you allowed to assume about linear operators?

 

[MathematicalPhysicist]

the determinant is the sum on all permutations t=j1,...jn in Sn such that detA=
sum (sgn t) a1j1...anjn
t

and about linear operators:
i used linear combinations of the operator T because i learned it, you can also assume that i know ((-:) the algebra of linear operators i.e if F,G,H are operators then:
1)F(G+H)=FG+FH
2)(G+H)F=GF+HF
3)k(GF)=(kG)F=G(kF)
where k is scalar.

basically if you know schaum series book for linear algebra then i went through chapters from 1-8.

 

[matt grime]

I don't know Schaum from personal experience; from what I can tell from other people's it's a waste of money, and I have no desire to buy a copy, if it's even avaliable in the UK.

So you just need to show that the two expressions you have, one for Det(ST) and one for Det(S)Det(T), agree. It is not hard but it is tedious; bookkeeping some people call it.

Det(S)Det(T) is gong to be a big horrible mess, and it will contain the expression Det(ST), it just remains to show that the rest cancels, or is zero. Remember that often be the best way to show X is zero is to show X=-X (I can't recall if that is exactly what happens here; it is some time since I proved this fact).