- #1
gravenewworld
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Hi I was wondering if someone could tell me if this is a valid proof for this question.
Suppose that V is finite dimensional and S,T elements of L(V). Prove that ST=I if and only if TS=I.
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.
Suppose that V is finite dimensional and S,T elements of L(V). Prove that ST=I if and only if TS=I.
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.