## The general linear group over the complex numbers is a union of conjugates of a proper subgroup

Let $G = GL_2(\mathbb{C})$ and let $H = \left\{ \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \ |\ a,b,c \in \mathbb{C}, ac \neq 0 \right\}$. Prove that every element of $G$ is conjugate to some element of $H$. Deduce that $G$ is the union of conjugates of $H$. [Hint: Show that every element of $G$ has an eigenvector.]

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in G$; then $ad-bc \neq 0$.

• If $c = 0$, then $IAI^{-1} = A \in H$.
• If $b = 0$, let $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Note that $\mathsf{det}(B) = 1$, so that $B \in G$; moreover, $B^{-1} = B$. Then $BAB^{-1} = \begin{bmatrix} d & c \\ b & a \end{bmatrix} \in H$.
• If $b,c \neq 0$ and $a = d$, let $B = \begin{bmatrix} c/\sqrt{c} & 0 \\ 1/\sqrt{b} & 1/\sqrt{c} \end{bmatrix}$. Note that $\mathsf{det}(B) = 1$, so that $B \in G$. Evidently $B^{-1} = \begin{bmatrix} 1/\sqrt{c} & 0 \\ -1/\sqrt{b} & c/\sqrt{c} \end{bmatrix}$. Note that $BAB^{-1} = \begin{bmatrix} a-\sqrt{bc} & bc \\ 0 & d+\sqrt{bc} \end{bmatrix} \in H$.
• Suppose now that $b,c \neq 0$ and $a \neq d$. Let $B = \begin{bmatrix} 0 & a-d \\ 1/(d-a) & \omega \end{bmatrix}$, where $\omega = \left(1 + \sqrt{1 + 4bc/(d-a)^2}\right)/2c$. Note that $\mathsf{det}(B) = 1$, so that $B \in G$. Evidently $B^{-1} = \begin{bmatrix} \omega & d-a \\ -1/(d-a) & 0 \end{bmatrix}$. Moreover, we see that $BAB^{-1} = \begin{bmatrix} \omega c(a-d) + d & -c(a-d)^2 \\ 0 & \omega c(d-a) + a \end{bmatrix} \in H$.

Thus every element of $G$ is conjugate to some element of $H$; clearly then $G = \bigcup_{g \in G} gHg^{-1}$.

There is a much easier way- for one thing, it’s enough to note that every invertible matrix has an invertible Jordan canonical form. This is basically the method you mentioned in a more general context. With a repeated eigenvalue, a 2×2 matrix is ‘almost’ diagonalizable, in the sense that we can make the entry $b$ here equal to 1.