Suppose that $V$ is a finite-dimensional vector space and that $T \in \mathcal L(V,W)$. Prove that there exists a subspace $U$ of $V$ such that $U \cap \text{null } T = \{0\}$ and $\text{range }T = \{Tu : u \in U\}$.

Let $n_1,\dots,n_k$ be a basis for $\text{null }T$ then extend this to a basis $n_1\dots,n_k,u_1,\dots,u_r$ of $V$ using 2.33. Let

Clearly $U \cap \text{null }T = \{0\}$, and $\text{range }T = \{Tu : u \in U\}$ since the nullspace components disappear by linearity

I was tempted to try the set $U' = \{v \in V : Tv \ne 0\} \cup \{0\}$ but this doesn't work as it isn't closed under addition. Specifically notice that $n_1+u_1 \in U$ and $n_1-u_1 \in U$ but their sum $n_1$ isn't. Having $Tv \ne 0$ isn't enough, we need $v$ to contain no nullspace component.