91Ó°ÊÓ

Given two finite dimensional vector spaces, $V$ and $W$, there exists a surjective linear map $T: V \to W$ if and only if $\operatorname{dim}(W) \leq \operatorname{dim}(V)$. The proof involves showing two directions: 1. If a surjective linear map exists, then $\operatorname{dim}(W) \leq \operatorname{dim}(V)$: Assume a surjective map $T$ exists and take bases for $V$ and $W$. Since $T$ is surjective, any vector in $W$ can be written as a linear combination of the images of basis vectors in $V$. Thus, $\operatorname{dim}(W) \leq \operatorname{dim}(V)$. 2. If $\operatorname{dim}(W) \leq \operatorname{dim}(V)$, there exists a surjective linear map from $V$ onto $W$: Assume $\operatorname{dim}(W) \leq \operatorname{dim}(V)$ and take bases for $V$ and $W$. Define a linear map $T$ by mapping basis vectors in $V$ to those in $W$ or to zero for the extra vectors. Then, any vector in $W$ can be expressed as a linear combination of the images of basis vectors in $V$, making $T$ surjective. The result is established by showing that a surjective linear map exists from $V$ onto $W$ if and only if $\operatorname{dim}(W) \leq \operatorname{dim}(V)$.