# Orthogonal group/linear algebra/group theory

1. Sep 27, 2013

### Computnik

1. The problem statement, all variables and given/known data
This problem has two parts:

i) Determine the range of det: O(n) → ℝ.

ii) Are det-1({1})⊂O(n) and det-1({1})⊂O(n) groups?

2. Relevant equations
AA-1=I & AAT=I

3. The attempt at a solution
i) det(AAT)=det(I)

det(AAT)=1

det(A) det(AT)=1

det(A) det(A)=1

(det(A))2=1

det(A)= +-1

So the range is +-1.

Did I make any mistakes?

ii) The function det: O(n) → ℝ is not injective, so therefore the inverse function does not exist (right?). Which would result in det-1({1})⊂O(n) and det-1({1})⊂O(n) not being groups (right?).

Last edited: Sep 27, 2013
2. Sep 27, 2013

### brmath

It would help if you would define O(n). Is O the set of all unitary matrices? or what? What is the n? And in part 2 you ask the same question twice. Would the second question be re {-1}?

3. Sep 27, 2013

### vela

Staff Emeritus
$\det^{-1}(\{1\})$ is meant to denote the set of matrices in O(n) with determinant 1. You're right that $\det^{-1}$ is not a function, but that has nothing to do with what you're being asked to prove.

4. Sep 28, 2013

### Computnik

O(n) is the orthogonal group.
Ok, I wasn't aware of that. Thanks!

5. Sep 28, 2013

### Computnik

Follow-Up Question:

Is it true that in order to determine whether or not det-1({1})⊂O(n) and det-1({1})⊂O(n) are groups I will have to show that they "hold" for the following?

Let's say that det-1({1})⊂O(n)=G.

i) The following is true x ° (y ° z) = (x ° y) ° z for all x, y, z ∈ G.

ii) There exists some e ∈ G such that x ° e = e ° x = x.

iii) For all x ∈ G, x-1 ∈ G : x ° x-1 = x-1 ° x = e.

Do I make any sense?

Also: What elements does det-1({1}) contain?

6. Sep 28, 2013

### vela

Staff Emeritus
Do you mean $G=\det^{-1}({1})$ or do you mean $G=O(n)$? From what you wrote, it's the latter, but I think you meant the former.

Yes, if you show that $(G,\circ)$ satisfy these three conditions, then you've shown that $(G,\circ)$ is a group.

In general, if you have a function $f: A \to B$, you have by definition that $f^{-1}(Y) = \{x \in A\ \vert\ f(x) \in Y\}$ where $Y \subset B$. In other words, it's everything in the domain that maps onto an element in Y.

In your particular case, $\det^{-1}(\{1\})$ would be the set $\{x \in O(n)\ \vert \ \det(x) \in \{1\}\}$. Since there's only one element in {1}, the elements of the set satisfy det(x)=1.

7. Sep 29, 2013

### Computnik

Yes you were right in assuming that I meant that!

Follow-Up question:

Which elements does O(n) contain and is there a way of knowing which elements (of O(n)) that satisfies det(x)=1 and which elements that satisfies det(y)=-1?

I guess the first step would be to determine if the function det is injective or not? (Since that would indicate if some element of O(n), say z, satisfies only one of following conditions: det(z)=1, det(z)=-1.)

Is it correct to assume that the function det is not injective?

Last edited: Sep 29, 2013
8. Sep 29, 2013

### pasmith

O(n) consists of those nxn matrices $A$ such that $A^{-1} = A^T$. It follows that if $A \in O(n)$ then
$$\det(A)\det(A^T) = \det(AA^T) = \det(I) = 1$$
Since $\det(A) = \det(A^T)$ it follows that $|\det(A)| = 1$ if $A$ is orthogonal.

To return to an earlier point: If you are trying to show that some subset is a subgroup, there's a lot you already know simply because the group operation of a subgroup is exactly the same as the group operation of the full group:

- The group operation will be associative.
- The only way a proposed subgroup can fail to have identity is if the full group identity is not in the proposed subgroup.
- The only way an element of a proposed subgroup can fail to have an inverse is if its inverse in the full group is not in the proposed subgroup.
- The only way two elements of a proposed subgroup can fail to have a product in the proposed subgroup is if their product in the full group is not in the proposed subgroup.

It is implicitly assumed in the definition of a group (as a result of the definition of a binary operation on a set) that the product of two group elements is always in the group, but when you restrict to a subgroup you have actually to check that the product of any two elements in the subgroup will actually be in the subgroup.

9. Sep 29, 2013

### brmath

Now that i know what O(n) is:

The set that mapped into 1 you can show you have a subgroup. The set that mapped into -1 does not look promising as a group. You should show that.

10. Sep 29, 2013

### vela

Staff Emeritus
By definition, O(n) is the set of all nxn orthogonal matrices. That means if you know a matrix A is in O(n), you know that $AA^T=I$ and $A^TA=I$.

If you picked a random element out of O(n), you'd have to calculate its determinant to see which condition it satisfies. If you picked a random element out of $\det^{-1}({1})$, however, you know that its determinant is 1.

I think you mean surjective, right? You showed in part (a) that if a matrix A is in O(n), its determinant is either +1 or -1. It's gotta be one or the other; it can't be anything else.

You're right that det is not injective, but you should be able to prove this. It det were injective, then O(n) could only contain two matrices, one that has determinant +1 and one that has determinant -1. Consider the n=2 case. Can you think of two orthogonal matrices that have determinant 1? That would be enough to prove det is not injective.

11. Sep 29, 2013

### brmath

You know there are lots of unitary functions whose determinants are 1 or -1. So to even talk about an inverse to this operation is confusing. You just want to talk about the set of orthogonal functions whose determinant is 1 or -1. And if a matrix maps into 1, it maps into 1. Talking about injective and surjective is introducing verbiage you don't need. (I'm all for using it when you need it, but you just don't here).

What you need is to look at the set of orthonormal functions with a determinant of 1 and those with a determinant of -1. Is either of these sets a group (under matrix multiplication)? Both? Neither?

12. Oct 1, 2013

### Computnik

I was able to find the solution now. Thanks for all the help!