Suppose $U$ is a 3-dimensional subspace of $\mathbf R^8$ and that $T$ is a linear map from $\mathbf R^8$ to $\mathbf R^5$ such that $\text{null }T = U$. Prove that $T$ is surjective.

By 3.22

Since $\text{null }T = U$, $\dim \text{null }T = 3$ implying

Since $\text{range }T$ is a subspace of $\mathbf R^5$ and their dimensions equal, we have $\text{range }T = \mathbf R^5$ (by basis extension like in 3b/13)