Exercise 3.F.31

Proof. By Exercise 1, all the $\phi$’s are surjective. Consider the following process

• * Step 1.
  Choose $v_1\in V$ such that ${\phi }_1(v_1)=1$.
• * Step j.
  If $j=n+1$, stop the process.
  By the contrapositive of the statement in _Theorem 2_ of Chapter 3 notes, it follows that there is a vector $v_j\in V$ such that $v_j\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{1\le k\le n,k\ne j}$ and $v_j\notin \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->{\phi }_j$.
  Because both these subspaces are closed under scalar multiplication and because ${\phi }_j$ is surjective, we can assume without loss of generality that ${\phi }_j(v_j)=1$.

After step $n$ the process stops and we will have a list $v_1,\dots ,v_n$ such that ${\phi }_j(v_k)=1$ if $j=k$ and ${\phi }_j(v_k)=0$ if $j\ne k$. We but need to prove that $v_1,\dots ,v_n$ is linearly independent, since it already has length $\dim <!--\; FUNCTION\; APPLICATION\; -->V$.

Suppose there are $a_1,\dots ,a_n\in 𝔽$ such that

Applying ${\phi }_j$ to both sides of the equation above gives $a_j=0$, for each $j=1,\dots ,n$. Hence $v_1,\dots ,v_n$ is linearly independent and, therefore, a basis of $V$. □