Proof. The contrapostive of the theorem is: Given that $\sigma_1, \ldots,\sigma_n$ are distinct automorphisms of $E$: If $a_1$, $a_2$, …, $a_n$ are not all zero then there is a $u\in E$ such that $$ a_1\sigma_1(u)+\cdots+a_n\sigma_n(u)\not=0.$$ To prove this it suffices to show that for all $m$, with $\sigma_1,\ldots,\sigma_m$ distinct automorphisms, if $a_1,\ldots,a_m$ are all nonzero, then there is a $u\in E$ such that $$ a_1\sigma_1(u)+\cdots+a_m\sigma_m(u)\not=0.$$ This is because given $a_1$, $a_2$, …, $a_n$ not all zero, we may pick out the $m$ of them that are non-zero, and apply the second result.

So we prove the following statement by induction on $m$: If $\sigma_1,\ldots,\sigma_m$ are any distinct automorphisms of $E$, and $a_1,\ldots,a_m$ are all nonzero then for some $u$, $a_1\sigma_1(u)+\cdots+a_m\sigma_m(u)\not=0$. Call this statement "$S(m)$.''

Base: If $m=1$, pick $u$ so that $\sigma_1(u)\not=0$ ($u=1$, for example). Then $a_1\sigma_1(u)\not=0$ because $E$ has no zero divisors.

Induction: We want to prove that if $S(i)$ is true for $i\le m$, then $S(m+1)$ is true. We prove the contrapositive, namely, that if $S(m+1)$ is false, then $S(i)$ is false for some $i\le m$. So suppose that $a_1,\ldots,a_{m+1}$ are nonzero and for all $u$, $a_1\sigma_1(u)+\cdots+a_{m+1}\sigma_{m+1}(u)=0$. Pick $a\in E$ such that $\sigma_1(a)\not=\sigma_{m+1}(a)$. Now substituting $au$ for $u$ and then using the multiplication-preserving property of isomorphisms we get: $$\eqalignno{ 0&=a_1\sigma_1(au)+\cdots+a_m\sigma_m(au)+a_{m+1}\sigma_{m+1}(au)\cr 0&=a_1\sigma_1(a)\sigma_1(u)+\cdots+a_m\sigma_m(a)\sigma_m(u)+ a_{m+1}\sigma_{m+1}(a)\sigma_{m+1}(u).& (1.1.1)\cr}$$ Also, $$\eqalignno{ 0&=\sigma_{m+1}(a)\bigl(a_1\sigma_1(u)+\cdots+a_m\sigma_m(u)+ a_{m+1}\sigma_{m+1}(u)\bigr)\cr 0&=a_1\sigma_{m+1}(a)\sigma_1(u)+\cdots+a_m\sigma_{m+1}(a)\sigma_m(u)+ a_{m+1}\sigma_{m+1}(a)\sigma_{m+1}(u).& (1.1.2)\cr}$$ Subtracting equation 1.1.2 from equation 1.1.1, we get $$0=a_1(\sigma_1(a)-\sigma_{m+1}(a))\sigma_1(u)+\cdots+ a_m(\sigma_m(a)-\sigma_{m+1}(a))\sigma_m(u).$$ Not all of the coefficients of the $\sigma_i$ are zero in the last equation, because $a_1(\sigma_1(a)-\sigma_{m+1}(a))\not=0$. Choosing just those terms in which $a_j(\sigma_j(a)-\sigma_{m+1}(a))\not=0$, we have shown that for some $i\le m$, $S(i)$ is false, as desired. $\qed$

Proof. Let $u_1,\ldots,u_n$ be a basis for $E$ over $F$. Suppose, for a contradiction, that $G(E/F)$ contains $n+1$ distinct automorphisms, $\sigma_1,\ldots,\sigma_{n+1}$. Consider the system of $n$ simultaneous equations in $n+1$ unknowns: $$\eqalignno{ &\sum_{i=1}^{n+1}\sigma_i(u_j)x_i=0, \qquad j=1\ldots n.& (1.1.3)\cr }$$ By linear algebra, this system has a nontrivial solution $a_1,\ldots,a_{n+1}$. Now for any $u\in E$, $u=\sum_{i=1}^n c_iu_i$, because the $u_i$ form a basis. Then $$\sum_{i=1}^{n+1}a_i\sigma_i(u)= \sum_{i=1}^{n+1}a_i\sum_{j=1}^n c_j\sigma_i(u_j)= \sum_{j=1}^n c_j\sum_{i=1}^{n+1}a_i\sigma_i(u_j)= \sum_{j=1}^n c_j\cdot 0=0,$$ because $a_1,\ldots,a_{n+1}$ is a solution to the system of equations in 1.1.3. This contradicts theorem 1.1.3. $\qed$