Conjugates in \(GL_2(\mathbb{C})\) (D&F 4.3.25)

Let \(G = GL_2(\mathbb{C})\) and \(H = \{\begin{bmatrix}a&b\\0&c\\\end{bmatrix} \mid a,b,c \in \mathbb{C}, ac \neq 0\}\). Prove that every element of \(G\) is conjugate to some element of \(H\) and deduce that \(G\) is the union of conjugates of \(H\).

Let \(g = \begin{bmatrix}i&j\\k&l\\\end{bmatrix} \in G\), and \(\beta = \{e_1,e_2\}\) be the standard basis of \(\mathbb{C}^2\). By the Fundamental Theorem of Algebra, the characteristic polynomial of \(G\) has some root \(\lambda \in \mathbb{C}\), which is a complex eigenvalue corresponding to the eigenvector \(v\). Clearly \(v\) is linearly independent from either \(e_1\) or \(e_2\) (or both); we shall assume it is linearly independent from \(e_1\) (the argument for \(e_2\) is similar). Then \(\gamma = \{v,e_1\}\) is another basis for \(\mathbb{C}^2\); now let \(h = [L_g]_\gamma = \begin{bmatrix}\lambda&i'\\0&j'\\\end{bmatrix}\). We know that \(\lambda \neq 0\) since \(h\) is invertible, and also \(j' \neq 0\) because otherwise \(g((\lambda/i')e_1) = (\lambda/i')i'v = \lambda v = gv\) which shows that \(e_1\) and \(v\) are not linearly independent. Thus \(h \in H\), and \(h\) conjugated by the change-of-basis matrix \([I]^\beta_\gamma\) is \(g\).