Proving the space of linear maps is infinite-dimensional

Expert intro: The proof uses an infinite linearly independent set in the codomain and the Linear Map Lemma to build infinitely many independent operators from a finite basis of the domain.

Detailed walkthrough

Core strategy

We want to prove that $\mathcal{L}(V,W)$ is infinite-dimensional when $V$ is finite-dimensional with $\dim V>0$ and $W$ is infinite-dimensional. The key observation is that even though the domain is finite-dimensional, the codomain has enough room to create infinitely many distinct linear maps.

Choose a basis $v_1,\dots,v_m$ of $V$, where $m=\dim V\ge 1$. Because $W$ is infinite-dimensional, it contains an infinite linearly independent sequence $w_1,w_2,w_3,\dots$.

Construct many linear maps

For each positive integer $n$, define a linear map $T_n\in\mathcal{L}(V,W)$ by setting

$T_n(v_1)=w_n, \qquad T_n(v_i)=0 \text{ for } i=2,\dots,m.$

This is possible by the Linear Map Lemma: specifying the images of a basis determines a unique linear transformation.

Now consider a finite linear relation

$a_1T_1+a_2T_2+\cdots+a_kT_k=0.$

Apply both sides to $v_1$. We get

$a_1w_1+a_2w_2+\cdots+a_kw_k=0.$

Since $w_1,w_2,\dots,w_k$ are linearly independent, every coefficient must be zero:

$a_1=a_2=\cdots=a_k=0.$

Therefore $\{T_1,T_2,T_3,\dots\}$ is an infinite linearly independent subset of $\mathcal{L}(V,W)$.

Why this proves infinite dimension

A vector space is infinite-dimensional if it contains an infinite linearly independent set. We have just constructed such a set in $\mathcal{L}(V,W)$. Hence

$\mathcal{L}(V,W) \text{ is infinite-dimensional.}$

Common proof pattern

This is a standard operator-space argument: use a basis of the domain, pick infinitely many independent vectors in the codomain, and define maps that send one basis vector to each of those codomain vectors. The finite-dimensionality of $V$ does not restrict the number of different maps enough to make the operator space finite-dimensional when $W$ is infinite-dimensional.

So the conclusion follows directly from the existence of the infinite independent family $\{T_n\}$.

💡 Pitfall guide

A frequent mistake is trying to argue that because $V$ has dimension $m$, the space of all linear maps from $V$ to $W$ should have dimension at most $m$. That is false unless $W$ is also finite-dimensional. Another error is defining only one map $T_1$ and assuming that a single nonzero operator implies infinite dimension. To prove infinite-dimensionality, you must produce infinitely many linearly independent operators, not just infinitely many different operators. The Linear Map Lemma is what makes the construction rigorous.

🔄 Real-world variant

If $V$ were the zero space, then $\mathcal{L}(V,W)$ would actually be the zero space, which is finite-dimensional. So the condition $\dim V>0$ matters. If instead both $V$ and $W$ were finite-dimensional, then $\dim\mathcal{L}(V,W)= (\dim V)(\dim W)$, which is finite. The conclusion changes only because $W$ is infinite-dimensional here. If $W$ is infinite-dimensional but you try to use only one basis vector of $V$, the proof still works as long as $V\neq\{0\}$, because you only need one nonzero vector in the domain to detect linear independence among the maps.

🔍 Related terms

Linear Map Lemma, linearly independent set, operator space dimension