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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.

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 $A \text{ and } B \in O(n)$, 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

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  • $\begingroup$ 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)). $\endgroup$ – user137731 Feb 26 '17 at 3:19
  • $\begingroup$ It woud be useful to know where you got stuck. $\endgroup$ – Mariano Suárez-Álvarez Feb 26 '17 at 3:32
  • $\begingroup$ HI, I edited my question, I am not showing my answer. $\endgroup$ – user290335 Feb 26 '17 at 17:09
  • $\begingroup$ @Bye_World for this I would like to use the axioms thank you for your help. $\endgroup$ – user290335 Feb 26 '17 at 17:10
  • $\begingroup$ $O(n)$ is a subgroup of all matrices, so associativity simply inherited. No proving required! $\endgroup$ – Varad Mahashabde Oct 20 at 12:14
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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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  • $\begingroup$ 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 $\endgroup$ – user290335 Feb 26 '17 at 17:08
  • $\begingroup$ 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. $\endgroup$ – copper.hat Feb 26 '17 at 22:50
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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 $$AI=IA=A, \forall A\in O(n).$$

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