# How to show that the set of orthogonal n x n matrices forms a group under multiplication

I am studying orthogonal matrices and I am not sure if to show if a set of orthogonal $$n \times n$$ matrices forms a group under multiplication. We must check each of the group axioms.

I found that the axioms are:

1. Closure
2. Associativity
3. Existence of identity matrix
4. Existence of the inverse matrix.

This group is called $$O(n)$$.

To check the four axioms I did:

Let $$A \text{ and } B \in O(n)$$, denoted as orthogonal matrices and assume that $$C=AB$$, then,

### Closure :

To prove that $$C \in O(n)$$ we must prove that $$C$$ is a real $$n \times n$$ orthogonal matrix with uni-modular determinant. Since A and B are real $$n \times n$$ matrices, $$C$$ is also a real $$n \times n$$ matrix so,

$$C^TC=(AB)^T AB=B^T A^T AB = B^TB=I$$

### Associativity :

Matrix multiplication is associative, so the law holds for $$O(n)$$ group elements.

I am not sure if this is enough to prove associativity.

### Identity element :

The $$n \times n$$ identity matrix $$I_{n \times n}$$ represents the identity element.

In this case I am not sure if this is enough to prove the identity element.

### Inverse element :

Let $$A^{-1}$$ be the inverse of $$A$$, then we need to prove that $$A^{-1} \in O(n)$$ since $$(A^{-1})^T=(A^T)^{-1}$$. We have that:

$$(A^{-1})^T A^{-1}=(A^T)^{-1} A ^{-1}=(AA^T)^{-1}=I^{-1}=I$$

Can anyone check if what I did is correct? I also would like to know if I can prove the associativity and the identity element in a better way.

thanks

• You can use the subgroup test instead of going through each axiom if you like (assuming you've already proven that the set of all $n\times n$ invertible matrices is a group under multiplication (i.e. the general linear group)). – user137731 Feb 26 '17 at 3:19
• It woud be useful to know where you got stuck. – Mariano Suárez-Álvarez Feb 26 '17 at 3:32
• HI, I edited my question, I am not showing my answer. – user290335 Feb 26 '17 at 17:09
• @Bye_World for this I would like to use the axioms thank you for your help. – user290335 Feb 26 '17 at 17:10
• $O(n)$ is a subgroup of all matrices, so associativity simply inherited. No proving required! – Varad Mahashabde Oct 20 at 12:14

$(U_1 U_2)^T (U_1 U_2) = I$, hence $U_1 \circ U_2$ is orthogonal.

Associativity follows from associativity of matrix multiplication.

The matrix $I$ is an identity for matrix multiplication.

$U^T U = U U^T = I$, hence $U^{-1} = U^T$ is the required inverse.

• Hi, I edited my question so I could show my work. Is there any other way to prove the associativity and the Identity matrix? Thanks – user290335 Feb 26 '17 at 17:08
• I don't know what you mean by 'and the identity matrix'. The group operation is matrix multiplication, so I am not sure how one would show associativity other than showing associativity of matrix multiplication. – copper.hat Feb 26 '17 at 22:50

First you need to show that $I \in O(n)$ by showing that $$I^TI=II^T=I,~~\text{since}~I^T=I.$$ Then $$AI=IA=A, \forall A\in O(n).$$