Exercise 4.3.25 ($\mathrm{GL}_2(\mathbb{C})$ is the union of conjugates of a proper subgroup)

Proof. Let $M=\left(\begin{array}{cc}\hfill r\hfill & \hfill s\hfill \\ \hfill t\hfill & \hfill u\hfill \end{array}\right)\in G$, where $r,s,t,u$ are complex numbers. Consider the characteristic polynomial

The polynomial $P_M(x)$ of degree $2$ as at least a root $\lambda \in$, which is a proper value of $M$. Then the matrix $M-\mathit{𝜆𝐼}$ is singular, thus there is some eigenvector $v=(a,b)\in {ℂ}^2$, $(a,b)\ne (0,0)$, such that

We can complete $v\ne 0$ into a basis ${ℬ}^{\prime }=(v,w)$ of ${ℂ}^2$ (if $a\ne 0$, take $w=(0,1)$, and if $b\ne 0$, take $w=(1,0)$).

Let $f$ be the endomorphism of ${ℂ}^2$ of matrix $M$ in the natural basis $ℬ=((1,0),(0,1))$ of ${ℂ}^2$. Let $P\in {\mathrm{GL}}_2(ℂ)$ be the change-of-basis matrix from $ℬ$ to ${ℬ}^{\prime }$, so that $M^{\prime }=P^{-1}\mathit{𝑀𝑃}$ is the matrix of $f$ in the base ${ℬ}^{\prime }$. Since $f(v)=\mathit{𝜆𝑣}$,

where $b,c\in ℂ$. This shows that $M^{\prime }\in H$, and so $M=P{M}^{\prime }{P}^{-1}$ is conjugate to some element of the subgroup $H$.

This proves that

so $G$ is the union of conjugates of $H$. □