Vtyjkyuk

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

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

Your doubt about the identity element is correct.

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