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Is this a valid proof?

  1. Nov 7, 2004 #1
    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.
     
  2. jcsd
  3. Nov 7, 2004 #2

    AKG

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    Have you proven S is injective? If S(v) = 0 only if v = 0, then it's injective. Have you proven that null(S) = {0}, or just shown that the only v in V such that T(v) is in null(S) is 0? You've shown that R(T) intersect N(S) = {0}, not that N(S) = {0}, as far as I can tell.
     
  4. Nov 7, 2004 #3

    AKG

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    You were on the right track I believe:

    ST(v) = v. If v is not zero, then T(v) cannot be in null S, and so T(v) is not zero (since zero is in null S) if v is not zero. So it is T which is injective, which is thus invertible, so:

    ST = I
    TSTT' = TIT' (where T' is the inverse of T)
    TS = I

    The converse proof is similar. So, I think you had it right except on the first line you concluded that S was injective, when it should have been that T was injective, and vice versa for the second line. Maybe I'm wrong and your first line also proves that S is injective, but as far as I can see, it only proves that T is, not S.
     
  5. Nov 8, 2004 #4
    Doesn't This line prove that null S is ={0}? STv=0 only if v=0. Thus Tv=T(0)=0. STv=S(0)=0. so the only Tv that are in null S is the 0 vector, so S must be injective.
     
  6. Nov 8, 2004 #5

    AKG

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    I don't think that's true in general. Like I said, all you've proven is that "the only Tv that are in null S is the 0 vector", but there might be some vectors x such that S(x) = 0, but there is no v such that T(v) = x. In other words, it's like I said, you've proven that N(S) intersection R(T) = {0}, where N(S) is the null space of S, and R(T) is the range of T. Say, for example, V = R² and S and T were both the projection onto the x-axis. Then the null space of S would be all the points on the the y-axis, but the only element in the range of T which is in the null space of S (i.e. which is on the y-axis) is the origin, since only the origin is both on the x-axis (in the range of T) and on the y-axis (in the null space of T). So the only Tv that are in null S is the zero vector, i.e. R(T) intersect N(S) = {0}, but null S includes more than just the zero vector, and S is not injective.
     
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