I need help on 2a **Transcription for Educational Website:**---**Linear Algebra Exercises on Subgroups**1. **Subgroup Identification** - Let $ K $ be a subgroup of $ \mathbb{R} $. Let $ H = \{ g \in GL(n, \mathbb{R}) : \text{det}(g) \in K \} $. Prove that $ H $ is a subgroup of $ GL(n, \mathbb{R}) $.2. **Subgroup Verification** - Let $ G = GL(2, \mathbb{R}) $. Prove that the following two subsets of $ GL(2, \mathbb{R}) $ are subgroups of $ G $: - (a) $ A = \left\{ \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix} : a > 0 \text{ and } d > 0 \right\} $ - (b) $ N = \left\{ \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} : b \in \mathbb{R} \right\} $3. **Trickier Example of a Subgroup** - Prove that $ K $ is indeed a subgroup of $ GL(2, \mathbb{R}) $: \[ K = \left\{ \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{pmatrix} \right\} \] (You will probably recognize the elements of $ K $ from an earlier homework).4. **Theorem Application** - There is a theorem that says that every element $ g \in GL(2, \mathbb{R}) $ can be written, in a unique way, as $ kan $ for some $ k \in K $, $ a \in A $, and $ n \in N $ (with $ K, A, N $ as in the last two problems). Your job: - (a) If $ g = \begin{pmatrix} 0 & 5 \\ 3 & -12 \end{pmatrix} $, find $ k, a, n $ such that

2a Given set is A=a00d:a>0 and d>0. We have prove that set A is the subgroup of GL2,R. For the…

Answered: 2. Let G = GL(2, R). Prove that the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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I need help on 2a

**Transcription for Educational Website:**

---

**Linear Algebra Exercises on Subgroups**

1. **Subgroup Identification**
- Let $ K $ be a subgroup of $ \mathbb{R} $. Let $ H = \{ g \in GL(n, \mathbb{R}) : \text{det}(g) \in K \} $. Prove that $ H $ is a subgroup of $ GL(n, \mathbb{R}) $.

2. **Subgroup Verification**
- Let $ G = GL(2, \mathbb{R}) $. Prove that the following two subsets of $ GL(2, \mathbb{R}) $ are subgroups of $ G $:
- (a) $ A = \left\{ \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix} : a > 0 \text{ and } d > 0 \right\} $
- (b) $ N = \left\{ \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} : b \in \mathbb{R} \right\} $

3. **Trickier Example of a Subgroup**
- Prove that $ K $ is indeed a subgroup of $ GL(2, \mathbb{R}) $:
\[
K = \left\{ \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{pmatrix} \right\}
\]
(You will probably recognize the elements of $ K $ from an earlier homework).

4. **Theorem Application**
- There is a theorem that says that every element $ g \in GL(2, \mathbb{R}) $ can be written, in a unique way, as $ kan $ for some $ k \in K $, $ a \in A $, and $ n \in N $ (with $ K, A, N $ as in the last two problems). Your job:
- (a) If $ g = \begin{pmatrix} 0 & 5 \\ 3 & -12 \end{pmatrix} $, find $ k, a, n $ such that

Expert Solution

Step 1

$2 (a)$ Given set is $A = \{(\begin{matrix} a & 0 \\ 0 & d \end{matrix}) : a > 0 and d > 0\}$ .

We have prove that set $A$ is the subgroup of $G L (2, R)$ .

For the subgroup, we will prove that set $A$ satisfy all the property of group.

$(1)$ Closure property:- Let $(\begin{matrix} a & 0 \\ 0 & d \end{matrix}) \in A and (\begin{matrix} b & 0 \\ 0 & c \end{matrix}) \in A$ , where, $a, b, c, d > 0$

Then $(\begin{matrix} a & 0 \\ 0 & d \end{matrix}) \cdot (\begin{matrix} b & 0 \\ 0 & c \end{matrix}) = (\begin{matrix} a b & 0 \\ 0 & c d \end{matrix}) \in A$

Hence, closure property satisfy.

$(2)$ Associative property:- Let $x = (\begin{matrix} a & 0 \\ 0 & d \end{matrix}) \in A, y= (\begin{matrix} b & 0 \\ 0 & c \end{matrix}) \in A and z= (\begin{matrix} e & 0 \\ 0 & f \end{matrix}) \in A$ , where, $a, b, c, d, e, f > 0$

We have to show that $(x \cdot y) \cdot z = x \cdot (y \cdot z)$

$\begin{matrix} L H S = (x \cdot y) \cdot z \\ = ((\begin{matrix} a & 0 \\ 0 & d \end{matrix}) \cdot (\begin{matrix} b & 0 \\ 0 & c \end{matrix})) \cdot (\begin{matrix} e & 0 \\ 0 & f \end{matrix}) \\ = (\begin{matrix} a b & 0 \\ 0 & c d \end{matrix}) \cdot (\begin{matrix} e & 0 \\ 0 & f \end{matrix}) \\ = (\begin{matrix} a b e & 0 \\ 0 & c d f \end{matrix}) \end{matrix}$

And

$\begin{matrix} R H S = x \cdot (y \cdot z) \\ = (\begin{matrix} a & 0 \\ 0 & d \end{matrix}) \cdot ((\begin{matrix} b & 0 \\ 0 & c \end{matrix}) \cdot (\begin{matrix} e & 0 \\ 0 & f \end{matrix})) \\ = (\begin{matrix} a & 0 \\ 0 & d \end{matrix}) \cdot (\begin{matrix} b e & 0 \\ 0 & c f \end{matrix}) \\ = (\begin{matrix} a b e & 0 \\ 0 & c d f \end{matrix}) \end{matrix}$