A question concerning unipotent matrices and a basis choice Let $G$ be a subgroup of $\mathbb{GL}_n$, the group of all invertible $n\times n$ matrices over an algebraically closed field $k$, which consists of unipotent matrices. 
I don't understand a step in a proof. We want to proof that there is an $x\in\mathbb{GL}_n$, such that $xGx^{-1}$ lies in the set of all upper triangular matrices with only $1$'s on the diagonal. The author of the proof states that it suffices to find a $v\in k^n\setminus 0$ such that $gv=v$ for all $g\in G$.
I don't understand why this is sufficient. Given such a $v$, how can I find $x$?
Thank you!
 A: This is indeed a bit quick. Once you've got a fixed vector $v$ for the subgroup $G$, you can take this as first vector in a base, and after base change (which corresponds to a conjugation $g\mapsto xgx^{-1}$) all elements of $G$ will have the form
$$
  \begin{pmatrix}
 1&*&*&\cdots&*\\
 0&*&*&\cdots&*\\
 0&*&*&\cdots&*\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
 0&*&*&\cdots&*\\
  \end{pmatrix}
$$
Now consider the subgroup (check it!) $G'$ of $\mathbb{GL}_{n-1}$ of all matrices that arise from elements of $G$ after dropping the first row and column (this amounts to considering the elements of $G$ as acting on the quotient space $k^n/\langle v\rangle$). Now $G'$ consists of unipotent matrices, so by induction on the dimension we can assume that there exists $y\in\mathbb{GL}_{n-1}$ such that $yG'y^{-1}$ consists of upper unitriangular matrices only. Now lifting $y$ to an element $\hat y\in\mathbb{GL}_n$ by adding a first row and column, with diagonal entry $1$ and all others $0$, one can check that $(\hat yx)G(\hat yx)^{-1}$ consists of upper unitriangular matrices in $\mathbb{GL}_n$ only.
A more geometric version of this argument says that after finding a fixed vector $v_1$ for $G$, we can continue to find a linearly independent vector $v_2$ that is fixed modulo $\langle v_1\rangle$ (meaning that $g(v_2)\in v_2+\langle v_1\rangle$ for all $g\in G$), which is obtained from a fixed vector in $k^n/\langle v_1\rangle$, then a linearly independent vector $v_3$ that is is fixed modulo $\langle v_1,v_2\rangle$ (obtained from a fixed vector in $k^n/\langle v_1,v_2\rangle$) and so on. In the end one obtains a basis $v_1,v_2,\ldots,v_n$ such that each subspace $\langle v_1,\ldots, v_k\rangle$ is $G$-stable (one says this is a $G$-stable flag of subspaces). Then a base change to this basis will make $G$ upper unitriangular.
A: Since $k$ is algebraically closed, we can assume $M$ has a Jordan decomposition. So it suffice to prove the Jordan blocks can be changed into the desired form. But this is not difficult given that the product of the determinant is 1. 
The reason $v$ exists such that $$(g-I)v=0$$ would hold for any $g\in GL_{n}(k)$ follows from dimension consideration. If $v\not=0$, the solution space for $Av=0$ has dimension $n^{2}-1$ by the rank-nullity theorem. And the dimension for $GL_{n}(k)$ is $n^{2}$. Now proceed inductively to find a sequence of basis vectors for $V$ such that $gv_{i}=v_{i}+a_{1}v_{i-1}....+a_{i-1}v_{1}$. This should not be too difficult. Using this basis you get the desired transformation. 
