I'm having trouble understanding the following proof:
Lemma: Let $K \subseteq L$ be a Galois extension and the Galois group $G(L/K) = \{ \sigma_1, ..., \sigma _n \}$. Elements $x_1, ..., x_n \in L$ form a $K$-basis of $L$ if and only if $\det [\sigma_i(x_j)] \neq 0.$
Proof: The given elements are linearly dependent if and only if there are elements $a_1,...,a_n\in K$ not all equal to $0$ such that $a_1 x_1 + \dots + a_n x_n =0$. Letting all $\sigma _i$ act on this equality, we get \begin{align*} a_1 \sigma_1 (x_1) + \dots + a_n \sigma_1(x_n) &= 0 \\ a_1 \sigma_2 (x_1) + \dots + a_n \sigma_2(x_n) &= 0 \\ \dots \\ a_1 \sigma_n (x_1) + \dots + a_n \sigma_n(x_n) &= 0. \end{align*} The above system of linear equations has a nonzero solution $a_1,...,a_n$ if and only if the determinant of the coefficient matrix (consisting of $\sigma_i(x_j) $) equals $0$.
I understand that if there is a nonzero solution $a_1,...,a_n \in K$, then this determinant must be zero.
Conversely, I do not understand why the determinant being zero implies there is a nonzero solution $a_1,...,a_n \in K$. Since the coefficients are in $L$, the only thing I should be able to deduce is that there is a nonzero solution $a_1,...,a_n \in L$, while in general $K \neq L$.
What am I missing?
My justification might be circular in your context since you are developing Galois theory, but if we forget this context and just view it from the outside (Galois theory can definitely be proved without your theorem), then I can justify it using Galois descent.
If the system of linear equations has a non-trivial $L$-solution, then the corresponding linear transformation has non-trivial $L$-kernel, which forms a vector space $V$. By Galois descent, $V^\Gamma = V \cap K^n$ is non-trivial, so the system of equations has a non-trivial $K$-solution.