Proof that a linear map from infinite dimensional vector space to finite dimensional vector space can't be injective

Question

In other words, let $T: V \to W$ be an injective linear map. Show that if $V$ is infinite-dimensional then $W$ must also be infinite-dimensional. I think the way I want to approach this is to show that the kernel of $T$ must be nonempty assuming that $V$, $W$ are infinite dimensional, but I'm not sure how to work that out in a proof.

Answer 1

Lemma. If $\{v_1,\ldots,v_n\}$ span a vector space $V$, and $w_1,\ldots,w_m\in V$ are linearly independent vectors, then $m\le n$. In particular, if $\dim V=k$, then any $k+1$ vectors in $V$ are linearly dependent.

Let now $V$, $W$ be vector spaces over a filed $\mathbb F$, with $V$ infinite dimensional, and let $T:V\to W$ be a linear transformation.

Assume that $W$ is finite dimensional, and $n=\dim W$. We shall show that $T$ is not injective. Choose $v_1,\ldots,v_{n+1}\in V$ linearly independent. Then according to the Lemma above $Tv_1,\ldots,Tv_{n+1}\in W$ are linearly dependent. Hence, there exist $c_1,\ldots,c_{n+1}\in \mathbb F$, not all zero, such that $$ 0=c_1Tv_1+\cdots+c_{n+1}Tv_{n+1}=T\big(c_1v_1+\cdots+c_{n+1}v_{n+1}\big) $$ Hence, $c_1v_1+\cdots+c_{n+1}v_{n+1}\in \ker T$, but $c_1v_1+\cdots+c_{n+1}v_{n+1}\ne 0$, since the vectors $v_1,\ldots,v_{n+1}\in V$ are linearly independent.

Answer 2

Lemma. Let $V$ and $W$ be vector spaces, and let $T\colon V\to W$ be a linear transformation. Then $T$ is one-to-one if and only if for every $S\subseteq V$, if $S$ is linearly independent in $V$, then $T(S)$ is linearly independent.

Proof. If $T$ is not one-to-one, then there exists $v\in V$, $v\neq\mathbf{0}$ such that $T(v)=\mathbf{0}$. Then $S=\{v\}$ is linearly independent, and $T(S)=\{T(v)\} = \{\mathbf{0}\}$ is not linearly independent.

Conversely, suppose that $T$ is one-to-one and $S$ is linearly independent. Let $T(v_1),\ldots,T(v_n)$ be pairwise distinct vectors in $T(S)$, and $\alpha_1,\ldots,\alpha_n$ be scalars such that $$\mathbf{0}=\alpha_1T(v_1)+\cdots + \alpha_n T(v_n) = T(\alpha_1v_1+\cdots+\alpha_nv_n).$$ Since $T$ is one-to-one, this means that $$\alpha_1 v_1+\cdots + \alpha_n v_n = \mathbf{0}.$$ Since $S$ is linearly independent, this implies that $\alpha_1=\cdots=\alpha_n$, as required. Thus, $T(S)$ is linearly independent, as claimed. $\Box$

Now suppose $V$ is infinite dimensional. Then for every positive integer $n$ we know that $V$ has a linearly independent subset $S$ of size $n$. If $T\colon V\to W$ is linear and one-to-one, this means that $T(S)$ is a linearly independent subset of $W$ of size $n$. Thus, $W$ contains linearly independent subsets of arbitrarily large size, and therefore $W$ is infinite dimensional.