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Math Help - Prove ST=1 iff TS=1

  1. #1
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    Prove ST=1 iff TS=1

    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.
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  2. #2
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    Quote Originally Posted by GreenDay14 View Post
    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 f,\,\,g: fg=id. \Longrightarrow g \,\,is\,\,1-1

    Tonio
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    Quote Originally Posted by tonio View Post
    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 f,\,\,g: fg=id. \Longrightarrow g \,\,is\,\,1-1

    Tonio
    Thank you very much Tonio, if i may ask tho, what do you mean by 1-1?
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    Quote Originally Posted by GreenDay14 View Post
    Thank you very much Tonio, if i may ask tho, what do you mean by 1-1?

    One to one, injective: f(x)\,\,is\,\,injective\,\,iff\,\,f(x)=f(y) \Longrightarrow x=y

    Tonio
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  5. #5
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    Quote Originally Posted by GreenDay14 View Post

    Suppose that V is finite dimensional and S,T belong to L(V). Prove that ST=1 if and only if TS=1.
    Sorry, this should read:


    Suppose that V is finite dimensional and S,T belong to L(V). Prove that ST=I if and only if TS=I.

    Notice that the 1's were changed to the letter I. Sorry for any mix up.
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  6. #6
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    Quote Originally Posted by GreenDay14 View Post
    Sorry, this should read:


    Suppose that V is finite dimensional and S,T belong to L(V). Prove that ST=I if and only if TS=I.

    Notice that the 1's were changed to the letter I. Sorry for any mix up.

    It never minds: anyone understanding this stuff won't get confused and, in fact, many times 1 is used in place of I.

    Tonio
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  7. #7
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    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?
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  8. #8
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    Quote Originally Posted by GreenDay14 View Post
    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?

    You didn't show S is injective: you only showed that if Tv is in Null(S) then v = 0. You should show that v in Null(S) ==> v = 0.

    Tonio
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    Quote Originally Posted by tonio View Post
    You didn't show S is injective: you only showed that if Tv is in Null(S) then v = 0. You should show that v in Null(S) ==> v = 0.

    Tonio
    So are you saying that in my first line i am actually showing that t is injective, rather than S? or is neither shown. Also does this mean that my second line is garbage?
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  10. #10
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    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?
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  11. #11
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    This just follows from the fact that composition of invertible linear transformations is associative (function composition in general is associative). If ab=1 then

    (ab)a=1a=a

    but by associativity

    (ab)a=a(ba)

    So a(ba)=a. Multiplying on the left by a^{-1} on both sides yields ba=1.

    The above obviously holds for any group.
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  12. #12
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    Quote Originally Posted by GreenDay14 View Post
    So are you saying that in my first line i am actually showing that t is injective, rather than S? or is neither shown. Also does this mean that my second line is garbage?

    You're showing neither, the same as in your second line. You could try to prove the following easy lemma from basic set theory:

    ST=I \Longrightarrow T\,\,is\,\,\,1-1\,\,\,and\,\,S\,\,is\,\,\,onto

    Tonio
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    Quote Originally Posted by tonio View Post
    You're showing neither, the same as in your second line. You could try to prove the following easy lemma from basic set theory:

    ST=I \Longrightarrow T\,\,is\,\,\,1-1\,\,\,and\,\,S\,\,is\,\,\,onto

    Tonio
    I apologize Tonio, but I am unsure as to what you mean by onto? onto what?
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