1. ## 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.

2. Originally Posted by RedBarchetta
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