My question asks:
Suppose that V is finite dimensional and S,T belong to L(V). Prove that ST=1 if and only if TS=1.
Could someone get me started on this please? Thanks for any help.
I think what you need is: if V is a fin. dimensional v.s. then a linear transformation is 1-1 iff it is onto, and thus:
A lin. transf. in a fin. dim. v.s. is invertible iff it is 1-1.
Finally, for any two functions $\displaystyle f,\,\,g$: $\displaystyle fg=id. \Longrightarrow g \,\,is\,\,1-1$
Tonio
Okay, so I went and tackled this question a different way and came up with:
Assume ST=I. Let v be an element of V, such that Tv is an element of null S. v=Iv=STv=S(Tv)=0. Since V is finite dimensional S must be invertible since it is injective. Thus if ST=I then TS=I.
Assume TS=I. Let v' be an element of V such that Sv' is in the null space of T. v'=Iv'=TSv'=0. Since V is finite dimensional T must be invertible since it is injective. Thus if TS=I then ST=I.
Therefore ST=I if and only if TS=I.
Does this make sense?
So could itb e changed to:
Assume ST=I. Let v be an element of V, such that Tv is an element of null S. v=Iv=STv=S(Tv)=0. Since V is finite dimensional T must be invertible since it is injective. Thus if ST=I then TS=I.
and also best to add:
ST = I
TSTT' = TIT' (where T' is the inverse of T)
TS = I
or am i still on the wrong track?
This just follows from the fact that composition of invertible linear transformations is associative (function composition in general is associative). If $\displaystyle ab=1$ then
$\displaystyle (ab)a=1a=a$
but by associativity
$\displaystyle (ab)a=a(ba)$
So $\displaystyle a(ba)=a$. Multiplying on the left by $\displaystyle a^{-1}$ on both sides yields $\displaystyle ba=1$.
The above obviously holds for any group.