Is the following Proof Correct?

Theorem. Given that $V$ is an finite-dimensional vector space with $\dim V>0$ and $W$ is infinite-dimensional. $\mathcal{L}(V,W)$ is infinite-dimensional.

Proof. Assume on the contrary that $\mathcal{L}(V,W)$ is finite-dimensional thus for some $T_1,T_2,T_3,...,T_m\in\mathcal{L}(V,W)$ it is the case that $span(T_1,T_2,...,T_m)=\mathcal{L}(V,W)$ consequently given an arbitrary $T\in\mathcal{L}(V,W)$ it is the case that for some $A_1,A_2,...,A_m\in\mathbf{F}$. $$\forall v\in V\left(Tv = \sum_{i=1}^{m}A_iT_iv\right)\tag{1}$$ moreover since $V$ is finite-dimensional with $\dim V = n>0$ it follows that given any $v \in V$ there exist $c_1,c_2,...,c_n\in\mathbf{F}$ such that $$v=\sum_{j=1}^{n}c_jv_j\tag{2}$$ Taking $(1)$ and $(2)$ together we may now state that given any arbitrary $T\in\mathcal{L}(V,W)$ and $v\in V$ $$Tv = T\left(\sum_{j=1}^{n}c_jv_j\right) = \sum_{i=1}^{m}A_i T_i\left(\sum_{j=1}^{n}c_jv_j\right)\tag{3}$$ by changing the order of summation we may express $(3)$ as $$Tv = \sum_{j=1}^{n}c_j\left(\sum_{i=1}^{m}A_iT_iv_j\right)\tag{4}$$ Consider now the following Lemma.

Lemma. There exists vectors $w_1,w_2,...,w_n\in W$ such that $$\forall i\in\{1,2,...,n\}(w_i\not\in U)\tag{5}$$ where $U$ is the span of all $T_iv_j$ in $(4)$, $i=\{1,2,...,m\}$ and $j=\{1,2,...,n\}$.

Proof. We know that $W$ is infinite-dimensional thus there exists $w_1\in W$ such that $w_1\not\in U$.

We can now invoke the existence of $w_2$ where $w_2\not\in U+span(w_1)$ but this implies that $w_2\not\in U$ and via the same reasoning invoke a $w_3$ where $w_3\not\in U+span(w_1,w_2)$ which again implies that $w_3\not\in U$, repeating this process a further $m-3$ times will yield all the required vectors.


Using the above Lemma we can invoke the existence of the vectors $w_1,w_2,..,w_n$ satisfying $(5)$.

Consider now the linear-transformation $S: V\to W$ defined as follows $$\forall i\in\{1,2,...,n\}(Sv_i=w_i)\tag{6}$$ where $v_1,v_2,...,v_n$ are the basis vectors of $V$, using $(2)$ it follows that given any $v\in V$ it is the case that $$Sv = S\left(\sum_{j=1}^{n}c_jv_j\right) = \sum_{j=1}^{n}c_jw_j\tag{7}$$ and using $(4)$ we may deduce that $$Sv = \sum_{j=1}^{n}c_j\left(\sum_{i=1}^{m}A_iT_iv_j\right)\tag{8}$$ comparing $(7)$ and $(8)$ we see that $$w_j=\left(\sum_{i=1}^{m}A_iT_iv_j\right)$$ but this results in a contradiction as $w_j\not\in U$ implying that our original assumption is wrong and therefore $\mathcal{L}(V,W)$ is infinite-dimensional.


  • $\begingroup$ The proof works but the end is too long. You can choose a single $w \notin U$ and define $S v_j = w$ for all $j$. $\endgroup$ – Gribouillis Aug 17 '17 at 18:38

As I said in the comments, your proof works but it can be shortened. There are many other possible proofs, here is one:

Let $(w_n)_{n\ge 1}$ be a sequence of independent vectors in $W$, let $W_n = \text{span}(w_1,\ldots,w_n)$. Then ${\cal{L}}(V, W_n) \subset {\cal{L}}(V, W)$, but $\dim({\cal{L}}(V, W_n)) = n \dim(V)$.


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