Roothaan matrix equation equation orbital coefficients
The $\mathbf{C}$ matrix will always be $K\times K$, where $K$ is the size of the finite basis set you are using. That is to say, with a basis set of size $K$, you will produce exactly $K$ MOs. This means you will never wind up with $\mathbf{C}$ being a vector. In the example you linked, they use a basis set of size 2, so they get 2 MOs for helium and thus the $\mathbf{C}$ matrix is $2\times 2$. While helium has only one occupied spatial orbital, it will have $K-1$ virtual orbitals depending the size of the basis set you use to solve the restricted Roothaan-Hall equations.