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Thread: Infinite Demension Question

  1. #1
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    Infinite Demension Question

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

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

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

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

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

    But $\displaystyle y\in\, T(V)$ such that $\displaystyle y=\displaystyle\sum_{i=k+1}^{k+r}d_i T(e_i)$. From this we see that $\displaystyle \displaystyle\sum_{i=k+1}^{k+r}d_i T(e_i)=\displaystyle\sum_{i=k+1}^{k+n}c_i T(e_i)$. 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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    Quote Originally Posted by RedBarchetta View Post
    Let $\displaystyle T:V \rightarrow W $ be a linear transformation of some linear space $\displaystyle V$ to a linear space $\displaystyle W$. Let $\displaystyle V$ be infinite-dimensional. Prove that at least $\displaystyle T(V) $ or $\displaystyle N(T)$ is infinite-dimensional.

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

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

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


    You can't do that: you ALREADY chose $\displaystyle e_1,...,e_k,e_{k+1},...,e_{k+r}$ : 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 $\displaystyle e_1,...,e_k,e_{k+1},...,e_{k+r}$ are lin. ind. (proof?) in V and since V is infinite dim., we can add lin. ind. elements such that...etc.

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

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

    But $\displaystyle y\in\, T(V)$ such that $\displaystyle y=\displaystyle\sum_{i=k+1}^{k+r}d_i T(e_i)$. From this we see that $\displaystyle \displaystyle\sum_{i=k+1}^{k+r}d_i T(e_i)=\displaystyle\sum_{i=k+1}^{k+n}c_i T(e_i)$. 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 $\displaystyle e_1,...,e_k,e_{k+1},...,e_{k+r}$ 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 $\displaystyle e_1,...,e_k,e_{k+1},...,e_{k+r}$, but nevertheless:

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

    Tonio
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