- #1
NasuSama
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Homework Statement
Prove that [itex]SL_{2}(ℝ)[/itex] is generated by the set:
[1 a], [1 0]
[0 1], [b 1], [itex]a,b \in ℝ[/itex]
Homework Equations
- GCD (Greatest common divisor)
- The property of special linear group
- Some basic linear algebra, like determinant
The Attempt at a Solution
[itex]SL_{2}(ℝ)[/itex] is the group consisting of invertible matrices with the determinant 1. Let [itex]A[/itex] be this matrix:
[a b]
[c d]
Then, [itex]det(A) = ad - bc = 1[/itex]
In order for the determinant to be 1, [itex]gcd(a,b) = gcd(a,c) = gcd(d,b) = gcd(d,c) = 1[/itex].
I'm not sure if my reasoning is right. I treat ad - bc = 1 as the linear combination. If 1 divides a and 1 divides b, then 1 divides (ad - bc). But -1 also divides a, b and (ad - bc). Thus a = ±1.
If [itex]a = 1[/itex], then [itex]det(A) = d - bc = 1[/itex]. Here, we need to split into some cases:
If [itex]c = 0[/itex], then [itex]d = 1[/itex], so we are done, having this matrix:
[1 b]
[0 1]
Similar argument shows that if [itex]a = 1[/itex] and [itex]b = 0[/itex], then d = 1, so we are done, leaving off this matrix:
[1 0]
[c 1]
If [itex]a = -1[/itex], then [itex]d = -1[/itex]. Similar arguments show that if either [itex]b = 0[/itex] or [itex]c = 0[/itex], then we obtain these similar matrices that belong to the special linear group:
[-1 0]
[c -1]
[-1 b]
[0 -1]
So these matrices form the set as given.
What if [itex]c ≠ 0[/itex]? Would I need to use Euclidean Algorithm to work out this problem?