Suppose $V$ is finite-dimensional with dim $V \gt 0$, and $W$ is infinite dimensional. Prove that $\mathcal{L}(V,W)$ is infinite dimensional.

Question

Suppose $V$ is finite-dimensional with dim $V \gt 0$, and $W$ is infinite dimensional. Prove that $\mathcal{L}(V,W)$ is infinite dimensional. (The question is from Linear Algebra Done Right)

My trial (I tried to prove by contradiction. But I am not sure whether this is correct):

Suppose $V$ is finite-dimensional with dim $V \gt 0$, and $W$ is infinite dimensional. Suppose $\mathcal{L}(V,W)$ is finite dimensional. Let $T_1, T_2, ..., T_n$ be the basis of $\mathcal{L}(V,W)$. Then $Tv = span(T_1,...,T_n)$.

Define $T_j(v_j) = w_j $ for $j = 1,...,n$. Then $Tv = W = span(w_1,...,w_n)$. That contradicts with the assumption that $W$ is infinite dimensional.

I am not sure whether this is correct. I am not sure whether $Tv = W$ is correct. Should it be range $T$ but not $W$?

Answer 1

Apart from the choice of words (a basis of $\mathcal{L}(V,W)$, not the basis), the proof is not good.

The idea is good, though. Suppose $\mathcal{L}(V,W)$ is finite dimensional. We want to show that also $W$ is. To this end, choose a basis $\{T_1,\dots,T_n\}$ of $\mathcal{L}(V,W)$. Set $$ W_0=\operatorname{span}\{T_i(v_j)\mid 1\le i\le n, 1\le j\le m\}, $$ where $\{v_1,\dots,v_m\}$ is a basis of $V$. We want to show that $W_0=W$. If not, take $w\in W$, $w\notin W_0$. Then we can define a linear map $T\colon V\to W$ by $$ T(v_j)=w $$ Since $\{T_1,\dots,T_n\}$ is supposed to be a basis of $\mathcal{L}(V,W)$, we have $$ T=a_1T_1+a_2T_2+\dots+a_nT_n $$ which easily leads to a contradiction.

Answer 2

There is no such thing as the basis of $\mathcal{L}(V,W)$ and you don't say what $Tv$ is.

Let $S$ be an infinite linearly independent subset of $W$ and let $\{e_1,\ldots,e_n\}$ be a basis of $V$. For each $v\in S$, let $f_v\colon V\longrightarrow W$ be the linear map such that $f_v(e_j)=v$, for each $j\in\{1,\ldots,n\}$. It follows from the fact that $S$ is lenarly independent that $\{f_v\,|\,v\in S\}$ is linearly independent too. Since $\mathcal{L}(V,W)$ contains an infinite linearly independent subset, it is infinite-dimensional.

Answer 3

Let $n=\dim V\;(>0)$. If $K$ is the base field, we have an isomorphism $\varphi: V \stackrel{\sim\:}\longrightarrow K^n$, whence an isomorphism $\;\mathscr L(K^n,W)\stackrel{\sim\:}{\longrightarrow}\mathscr L(V,W)$.

Now we have canonical isomorphisms: $$\mathscr L(K^n,W)\stackrel{\sim\:}{\longrightarrow}\bigl(\mathscr L(K,W)\bigr)^n\quad\text{and}\quad\mathscr L(K,W)\stackrel{\sim\:}{\longrightarrow} W, $$ so that ultimately $$\mathscr L(V,W)\stackrel{\sim\:}{\longrightarrow} W^n.$$