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}.

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Comments

  • Gobi Ree  On December 12, 2011 at 1:08 am

    Is there any method using eigenvector or diagonalization? I almost forgot linear algebra, so i don’t know if it is possible. If an invertible matrix has a distinct two eigenvalue, it is diagonalizable so it is a conjugate of a diagonal matrix, which is an element of H. But If an invertible matrix has a repeated eigenvalue, is there any way to proceed?

    • nbloomf  On December 12, 2011 at 11:50 am

      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.

      I try to make these solutions refer back only to things already proved in D&F, so I wrote the more explicit solution here.

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