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.
$\blacksquare$
Best Answer
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.