Let $V$ be an $n-$dimensional vector space and $G \le GL_n$. If I have a complete flag of $G$-invariant subspaces $\langle v_1, \dots v_k \rangle$ $(1 \le k \le n)$, how does a base change to $\{v_1, \dots v_n\}$ make every matrix in $G$ upper triangular?
I don't have a proper understanding of how a base change works, and why it is equivalent to conjugation.
I know that if $T$ is a linear transformation, with eigenvalue $\lambda$, then there exists a basis of $V$ such that the corresponding matrix of the transformation is of the form
$$\begin{pmatrix} \begin{array} \ \lambda & * & \dots &*\\ 0 & * & \dots & * \\ \vdots & \vdots && \vdots\\ 0 & * & \dots &* \end{array} \end{pmatrix},$$
I'm not sure if this is relevant. It feels like there could be a problem as we don't have fixed basis in this way.
Refrence to where the claim is made: A question concerning unipotent matrices and a basis choice