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Math Help - Infinite Demension Question

  1. #1
    Member RedBarchetta's Avatar
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    Infinite Demension Question

    Let T:V \rightarrow W be a linear transformation of some linear space V to a linear space W. Let V be infinite-dimensional. Prove that at least T(V) or N(T) is infinite-dimensional.

    Intuitively it makes sense by the rank plus nullity theorem since for a finite-dimensional linear space V we have dim \,N(T) + dim\, T(V) = dim\, V. Clearly, if V is infinite-dimensional then N(T) or T(V) must be infinite dimensional as well. Although, I can't seem to show this. Contradiction seems the best approach.

    Okay, lets assume that dim\, N(T)=k and dim\, T(V)=r. Let e_1,e_2,\ldots,e_k be a basis for N(T). Also, let T(e_{k+1}), T(e_{k+2}),\ldots, T(e_{k+r}) be a basis for T(V).

    Now let e_1,\ldots,e_k, e_{k+1},\ldots, e_{k+r},\ldots,e_{k+n} be linear independent vectors in V where n>r. Now applying T to a linear combination of these elements we see that

    y=T(\displaystyle\sum_{i=1}^{k+n}c_i e_i)=\displaystyle\sum_{i=1}^{k+n}c_i T(e_i)=\displaystyle\sum_{i=k+1}^{k+n}c_i T(e_i) since T(e_1)=T(e_2)=\cdots=T(e_k)=O

    But y\in\, T(V) such that y=\displaystyle\sum_{i=k+1}^{k+r}d_i T(e_i). From this we see that \displaystyle\sum_{i=k+1}^{k+r}d_i T(e_i)=\displaystyle\sum_{i=k+1}^{k+n}c_i T(e_i). Which is saying a linear combination of linear independent elements is equal to a linear combination of linearly dependent elements....which is a contradiction?....I don't know what else to do from here.

    Any help would be appreciated! Thank you.
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  2. #2
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    Quote Originally Posted by RedBarchetta View Post
    Let T:V \rightarrow W be a linear transformation of some linear space V to a linear space W. Let V be infinite-dimensional. Prove that at least T(V) or N(T) is infinite-dimensional.

    Intuitively it makes sense by the rank plus nullity theorem since for a finite-dimensional linear space V we have dim \,N(T) + dim\, T(V) = dim\, V. Clearly, if V is infinite-dimensional then N(T) or T(V) must be infinite dimensional as well. Although, I can't seem to show this. Contradiction seems the best approach.

    Okay, lets assume that dim\, N(T)=k and dim\, T(V)=r. Let e_1,e_2,\ldots,e_k be a basis for N(T). Also, let T(e_{k+1}), T(e_{k+2}),\ldots, T(e_{k+r}) be a basis for T(V).

    Now let e_1,\ldots,e_k, e_{k+1},\ldots, e_{k+r},\ldots,e_{k+n} be linear independent vectors in V where n>r.


    You can't do that: you ALREADY chose e_1,...,e_k,e_{k+1},...,e_{k+r} : the first k to be a basis of N(T) and the next ones such that their images under T are
    a basis of T(V)
    What you can say (and show!) is : since e_1,...,e_k,e_{k+1},...,e_{k+r} are lin. ind. (proof?) in V and since V is infinite dim., we can add lin. ind. elements such that...etc.

    Now applying T to a linear combination of these elements we see that

    y=T(\displaystyle\sum_{i=1}^{k+n}c_i e_i)=\displaystyle\sum_{i=1}^{k+n}c_i T(e_i)=\displaystyle\sum_{i=k+1}^{k+n}c_i T(e_i) since T(e_1)=T(e_2)=\cdots=T(e_k)=O

    But y\in\, T(V) such that y=\displaystyle\sum_{i=k+1}^{k+r}d_i T(e_i). From this we see that \displaystyle\sum_{i=k+1}^{k+r}d_i T(e_i)=\displaystyle\sum_{i=k+1}^{k+n}c_i T(e_i). Which is saying a linear combination of linear independent elements is equal to a linear combination of linearly dependent elements....which is a contradiction?....I don't know what else to do from here.

    Any help would be appreciated! Thank you.

    I think you were close: indeed e_1,...,e_k,e_{k+1},...,e_{k+r} are k+r lin. ind. elements in V; since V is inf. dim. there's an element v which isn't a lin. combination of e_1,...,e_k,e_{k+1},...,e_{k+r}, but nevertheless:

    Since\,\, Tv\in T(V)\,,\,then\,\,Tv=\sum\limits_{i=k+1}^ra_iTe_i=\  sum\limits_{i=k+1}^rT(a_ie_i)\Longrightarrow T\left(v-\sum\limits_{i=k+1}^ra_ie_i\right)=0 \Longrightarrow v-\sum\limits_{i=k+1}^ra_ie_i\in N(T) \Longrightarrow v-\sum\limits_{i=k+1}^ra_ie_i=\sum\limits_{j=1}^kb_j  e_i...and there: v is a lin. comb. of the e_i's , contradiction.

    Tonio
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