I'm looking for some explanations of the proof of Lemma 11.1 of the book Linear Algebraic Groups by Humphreys.
Shortly, in this proof he takes a subspace $M$ of dimension $d$ of $\mathbf{K}^n$ and an element $x \in GL_n$, then he takes a basis of the ambient space $w_1,\ldots,w_n$ and he says that we can choose the basis in such a way that $M$ is spanned by the first $d$ vectors of the basis and $x(M)$ is spanned by $w_{l+1},\ldots, w_{l+d}$ for a certain $l$ (the proof is to show that $l=0$).
Why can this trick work? Why can we find a basis with such a property?
Define $N=M\cap x(M)$, and $l=d-\dim(N)$. Then you can start by taking a basis $w_{l+1},\dots,w_{d}$ of $N$, and since it is a family of linearly independent vectors of $M$ you can complete it by adding $l$ vectors $w_1,\dots,w_l$ to get a basis of $M$. Similarly, you can complete it by adding $l$ vectors $w_{d+1},\dots,w_{d+l}$ to get a basis of $x(M)$. Now the family $w_1,\dots,w_{d+l}$ is a basis of $M+x(M)$, so it can again be completed by adding vectors $w_{d+l+1},\dots,w_{n}$ to get a basis of $\mathbf{K}^n$, which has the required properties.