Question:
Let $k$ be a field. Consider the subgroup $B\subset GL_2(k)$ where
$$B=\left\{\begin{bmatrix}a&b\\0&d\end{bmatrix}: a,b,d\in k, ad\neq 0\right\}$$
a) Let $Z$ be the center of $GL_2(k)$. Show that $$\bigcap_{x\in GL_2(k)}x^{-1}Bx=Z.$$
b) Assume $k$ is algebraically closed. Show that $$\bigcup_{x\in GL_2(k)}x^{-1}Bx=GL_2(k)$$
Answer
For the part (a), I know that $Z=\{\lambda I_2:\lambda\in k^* \}$. This trivially yields one inclusion as $Z\subset B$. Also, I managed to show that any element from $\bigcap x^{-1}Bx$ must be diagonal. However, I couldn't show why it must be a scalar matrix.
For the part (b), again one inclusion is obvious. However, I couldn't show the reverse one. Actually, I found a counter example for the case where $k$ is not algebraically closed but I do not see clearly why we need $k$ to be algebraically closed.
Any help/hint would be appreciated. Thanks in advance...
For (a), an element $g$ in the intersection must be upper triangular with respect to every choice of basis. Therefore $g$ preserves (set wise) each line through the origin in $k^2$. This implies, first, your observation that $g$ must be diagonal, and second, that $g$ must be scalar since if the diagonal entries are distinct then $g$ does not preserve the line spanned by the vector $e_1+e_2$ (where $e_1$ and $e_2$ are the standard basis vectors).
For (b), if $k$ is algebraically closed, then any element $g \in \mathrm{GL}_2(k)$ has an eigenvector $v \in k^2$. Choose $w$ so that $v$ and $w$ are linearly independent; then the matrix of $g$ with respect to the basis $(v,w)$ is upper triangular. Hence $g \in x B x^{-1}$ where $x$ is the matrix with columns $v$ and $w$.