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In an earlier post Proof of $\mathcal{L}(V,W)$ being infinite dimensional I presented a proof for the proposition that

Theorem. If $V$ is finite dimensional with $\dim V>0$ and $W$ is infinite dimensional then the vector space $\mathcal{L}(V,W)$ is infinite dimensional.

However the proof presented there though correct was preposterously long. I present here another proof which makes use of the following theorem. I would like to know whether this new proof is correct?

Theorem ($2.A.9)$ A vector space $V$ is infinite dimensional if and only if there is a sequence of vectors $v_1,v_2,v_3,v_4,....$ such that for all $n\in\mathbf{Z^+}$ the subsequence $v_1,v_2,....,v_n$ is linearly independent.

Proof. Since $W$ is infinite dimensional we may invoke a sequence of vectors $\psi:\mathbf{Z^+}\to W$ such forall $n\in\mathbf{Z^+}$ the subsequence $\psi(1),\psi(2),...,\psi(n)$ is linearly independent.

Let $v_1,v_2,...,v_m$ be a basis for $V$ and consider the sequence $\phi:\mathbf{Z^+}\to\mathcal{L}(V,W)$ defined as follows $$\phi(n) = T(c_1v_1+c_2v_2+\cdot\cdot\cdot+c_mv_m) = c_1\psi(1)+c_2\psi(2)+\cdot\cdot\cdot+c_m\psi(m+n)\tag{1}$$

We now show by recourse to Mathematical-Induction that with the above definition for all $n\in\mathbf{Z^+}$ the list $\phi(1),\phi(2),...,\phi(n)$ is linearly independent.

Evidently $\phi(1)$ on its own is linearly independent. Now assume that for some $k\in\mathbf{Z^+}$ the list $\phi(1),\phi(2),...,\phi(k)$ is linearly independent but the list $\phi(1),\phi(2),...,\phi(k),\phi(k+1)$ is not.

Then for some $j\in\{1,2,3,...,k+1\}$ it is the case that $\phi(j)\in\operatorname{span}(\phi(1),\phi(2),...,\phi(j-1))$ but this $j\not\in\{1,2,....,k\}$ since $\phi(1),\phi(2),...,\phi(k)$ is linearly independent, consequently $j = k+1$, which implies that for some $x_1,x_2,...,x_k\in\mathbf{F}$ $$\phi(k+1)= x_1\phi(1)+x_2\phi(2)+\cdot\cdot\cdot+x_k\phi(k)$$ then applying both sides to the vector $v_m$ we have $$\phi(k+1)v_m = x_1\phi(1)v_m+x_2\phi(2)v_m+\cdot\cdot\cdot+x_k\phi(k)v_m\tag{2}$$ $$\psi(m+k+1) = x_1\psi(m+1)+x_2\psi(m+2)+\cdot\cdot\cdot+x_m\psi(m+k)\tag{3}$$

but $(3)$ implies that $\psi(m+k+1)\in\operatorname{span}(\psi(1),\psi(2),...,\psi(m+k))$ contradicting the fact that for all $n\in\mathbf{Z^+}$, $\psi(1),\psi(2),...,\psi(n)$ is linearly independent.

$\blacksquare$

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  • $\begingroup$ Do you know that $\dim(L(V, W)) = \dim(V) * \dim(W)$ ? $\endgroup$ – onurcanbektas Feb 17 '18 at 11:45
  • $\begingroup$ @onurcanbektas I do indeed $\endgroup$ – Atif Farooq Feb 17 '18 at 11:45
  • $\begingroup$ Then why are not using this fact ? $\endgroup$ – onurcanbektas Feb 17 '18 at 11:46
  • $\begingroup$ @onurcanbektas In the text that i am using this fact was establlished using isomorphisms which were not introduced until much later in the text. $\endgroup$ – Atif Farooq Feb 17 '18 at 11:47
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I'd do it differently.

Theorem. A vector space $V$ is finite dimensional if and only if no infinite subset of $V$ is linearly independent.

Proof. ($\Rightarrow$) No subset having more than $\dim V$ elements is linearly independent.

($\Leftarrow$) Suppose that $V$ is not finite dimensional. In particular, $V\ne\{0\}$ so we can pick $v_1\in V$, $v_1\ne0$. By assumption, $\langle v_1\rangle\ne V$, so we can pick $v_2\in V\setminus\langle v_1\rangle$ and $\{v_1,v_2\}$ is linearly independent. This starts a recursion: suppose we have picked $\{v_1,\dots,v_k\}$ linearly independent. Since, by assumption, $\langle v_1,\dots,v_k\rangle\ne V$, we can pick a further vector $v_{k+1}\in V\setminus\langle v_1,\dots,v_k\rangle$ so $\{v_1,\dots,v_k,v_{k+1}\}$ is again linearly independent. This procedure doesn't stop, because we cannot find a finite spanning set for $V$. QED

Note that the proof provides, for a non finite dimensional space, an infinite linearly independent subset, which you seem to take for granted.

Now, for your case, since $W$ is not finite dimensional, the procedure above provides an infinite linearly independent subset of $W$, say $\{w_n:n\ge1\}$. Pick a basis $\{v_1,\dots,v_m\}$ of $V$ and define $f_n\colon V\to W$ by $$ f_n(v_1)=w_n, \qquad f_n(v_k)=0\ (2\le k\le m) $$ It shouldn't be difficult to finish up by proving that $\{f_n:n\ge1\}$ is linearly independent in $\mathcal{L}(V,W)$.

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I did not read you proof, but to show that $L(V, W)$ have infinite dimension, you can argue that the maps of the form $f_{i,j}: V \to W$ s.t

$$f_{i,j}(v_k) = \begin{cases} w_j & k = i\\ 0 & k\not = i\\ \end{cases},$$ - where $\{v_i\}$ and $\{w_j\}$ forms a basis for $V$ and $W$, respectively - generates a subspace of $L(V, W)$.Therefore, since there are infinitely many such maps, and the space generated by such maps is a subspace of $L(V,W)$, we must have $\dim L(V, W)$ has to be infinite.

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  • $\begingroup$ There's no reason why that set should span $\mathcal{L}(V,W)$. $\endgroup$ – José Carlos Santos Feb 17 '18 at 11:54
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    $\begingroup$ @JoséCarlosSantos To be honest, I cannot argue that it does, but I can still argue that such maps forms a subspace of $L(V,W)$, which is infinite dimension, so the space that contains the subspace generated by those maps has to have infinite dimension too. $\endgroup$ – onurcanbektas Feb 17 '18 at 11:57

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