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Linear Algebra Proof

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Let V be a vector space, and let T:V->V be linear. Prove that T2=T0 if and only if R(T) is a subset of N(T)


    2. Relevant equations
    I brainstormed everything I know while looking through my text book and compiled the following which I use within my proof.

    I'm letting beta be a basis for V and beta be composed of {x1,...,xn}

    T2(x)=TT(x)=T(T(x)) /forall X /in V
    T0(x)=0 and since T is linear, T(0)=0

    N(T)={xi \in V : T(xi)=0} (1<i<n)
    R(T)={T(xi): xi \in V} (1<i<n)

    3. The attempt at a solution
    TT(x1,...,xn)
    = T(T(x1,...,xn))
    = T( T(x1),...,T(xn))
    =T(R(T))
    =0 when R(T)={0}

    so R(T) must be a subset of N(T)

    So my question.... I am worried that I have made too many leaps or assumptions that I am not allowed. This is my first semester writing proofs so I would not appreciate a full proof from someone else (which is against the rules anyways right?) but rather, I think I would benefit if people could point out flaws in my "proof," point out any steps that are illogical, etc.

    So basically, point out what I can't do or what is vague so I can scour my book and notes and fix it.

    thanks ahead of time to anyone that can help.
     
  2. jcsd
  3. Feb 21, 2009 #2
    First, we won't need to consider a basis...

    First direction: Suppose [tex]T^2(x) = T_0(x) = 0[/tex] for all [tex]x\in V[/tex]. We want to show [tex]R(T) \subseteq N(T)[/tex].

    To show that [tex]R(T) \subseteq N(T)[/tex], since an element of R(T) is of the form T(x) for some x in V, we let [tex]T(x)\in R(T)[/tex] and we wish to show that [tex]T(T(x)) = 0[/tex], so that [tex]T(x) \in N(T)[/tex] (cf. T(v)=0 means [tex]v\in N(T)[/tex]).

    Continue...

    For the second direction: begin by writing what we want to show (as we did for the first direction). Then see how you go!

    You will notice that after writing out what is required of us to show, the actual effort to carry it out is almost trivial!
     
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