Exercise 3.B.29

Proof. Suppose $v\in V$. Since $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->\phi =1$, it follows that $\phi (u)$ is a basis of $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->\phi$. There exists $\lambda \in 𝔽$ such that $\phi (v)=\mathit{\lambda \phi }(u)$, which implies $\phi (v-\mathit{\lambda u})=0$. Hence $v-\mathit{\lambda u}\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\phi$ and $\mathit{\lambda u}\in \mathit{au}:a\in 𝔽$. But $v=(v-\mathit{\lambda u})+\mathit{\lambda u}$, therefore $v\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\phi +\mathit{au}:a\in 𝔽$, proving the inclusion in one direction. The inclusion in the opposite direction is clearly true. Thus $V=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\phi +\mathit{au}:a\in 𝔽$. Since $\phi (u)\ne 0$, it follows that $\phi (\mathit{\lambda u})=0$ if and only $\lambda =0$. Therefore $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\phi \cap \mathit{au}:a\in 𝔽=0$, proving that the sum is a direct one. □