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I am studying orthogonal matrices and I am not sure if to show if a set of orthogonal n x 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.

I edited my question, since I was able to find more information about this topic.

This group is called O(n)

To check the four axioms I did:

Let A 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 x n orthogonal matrix with uni-modular determinant. Since A and B are real n x n matrices, C is also real n x n matrix so,

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

Associativity:

Matrix multiplications 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 x n identity matrix I 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

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    You can use the [subgroup test](https://en.wikipedia.org/wiki/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)). – 2017-02-26
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    It woud be useful to know where you got stuck. – 2017-02-26
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    HI, I edited my question, I am not showing my answer. – 2017-02-26
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    @Bye_World for this I would like to use the axioms thank you for your help. – 2017-02-26

2 Answers 2

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$(U_1 U_2)^T (U_1 U_2) = I$, 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.

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    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 – 2017-02-26
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    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. – 2017-02-26
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Your doubt about the identity element is correct.

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