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Let $G = GL_2 (\mathbb {R} $).

Show that $T$ = {$ \begin{bmatrix} a & b \\ 0 & d \\ \end{bmatrix}$ | ad $\neq 0$ } is a subgroup of $G$

My attempt:

det$(TT^{-1})$ = det$(T)$ det($T^{-1})$ = det($T$) $1$/det($T$) = $ad$ $(1/ad)$= $1$. We are done by subgroup test

user26857
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combo student
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  • your attempy is incorrect. the determinant of the product $TT^{-1}$ is always 1, regardless the matrices (of course I'm refereing to invertible matrices). –  Sep 19 '16 at 19:36
  • Do you know how to multiply 2x2 matrices ? –  Sep 19 '16 at 21:54

2 Answers2

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You need to show that with $T$ and $S$ also $TS$ and $S^{-1}$ is again in $G$. So far you only showed that $\det(I_n)=1$, which is clear anyway. In other words, why is $TS$ and $S^{-1}$ also upper-triangular, and nonsingular.

All points have been shown on MSE already, e.g., see here.

Community
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Dietrich Burde
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The subgroup test say that:

a nonempty subset of a group is itself a group if the inverse of any element in the subset multiplied with any other element in the subset is also in the subset.

So you have to prove that $A^{-1}X \in T$ for all $A$ and $X$ in $T$.

It is simpler to show that the inverse of an upper triangular matrix (if it exists) is upper triangular and the product of two upper triangular matrices is upper triangular. And, obviously, the identity matrix is upper triangular.

Emilio Novati
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