Proof of $H^1(G,\operatorname{GL}_n(K))=0$

Question

I have a confusion about the above proof. In the last paragraph, it says let $x_1 \ldots x_n$ such that $y_i=b(x_i)$ are linearly independent over K. How do we know such a $x_1 \ldots x_n$ exists? I feel that we have to use the result of the previous paragraph but I don't get how. Note here $G = \operatorname{Gal}(K/k)$

Answer 1

The previous paragraph is precisely an argument showing that $\mathcal{B} = \{b(x)\mid x\in K^n\}$ spans $K^n$ as a $K$-vector space, so reduce $\mathcal{B}$ to a $K$-basis $B\subseteq\mathcal{B}$. Then $B$ has $n$ elements $B = \{b_i\}_{i = 1}^n$, and each element $b_i$ is of the form $b(x_i)$ for some $x_i\in K^n$ by definition.