Center of the Orthogonal Group and Special Orthogonal Group

Question

How can I prove that the center of $\operatorname{O}_n$ is $\pm I_n$?

I have that $AM = MA$, $\forall M \in \operatorname{O}_n$ and $A^{-1} = A^T$, $M^{-1} = M^T$.

Then $M = A^{-1}MA = A^{T}MA$.

I see that since conjugating by $A$ must leave the determinant of $M$ unchanged then the result of conjugation must be a rotation. But now I'm stuck.

How to proceed? Thanks in advance.

Answer 1

Hints:

1. By considering $AD=DA$ for some diagonal matrices $D$ with diagonal entries in $\{-1,1\}$, show that $A$ is a diagonal matrix.
2. By considering $AR=RA$ for some Givens rotation matrices $R$, show that all diagonal entries of $A$ are equal to each other.

Answer 2

Alright, I answered my own question:

Suppose $A$ commutes with every element in $O_n$.

Then $A$ must commute with the elementary orthogonal matrices.

These are

1. the identity

2. the row switching/column switching matrices

3. the matrices that are identical to the identity but with a -1 for one of the entries on the main diagonal.

Then $AE = EA$ implies $A = EAE^{-1}$.

Now, conjugation by Type 2 matrices shows that all the elements on the diagonal must be equal.

And conjugation by Type 3 matrices shows that all the off-diagonal elements must be zero. Suppose we have that the $a_{ii}$ entry of Type 3 is -1 then conjugation leaves $a_{ii}$ unchanged but reverses the signs of all the elements in the same row and column as $a_{ii}$.

Since $A$ must also have determinant $\pm 1$, then the only matrices in the center must be $\pm I$.

The center of $SO_n$ is $\{ \pm I \}$ for $n > 3$ and $SO_2$ for $n=2$.

Suppose A commutes with every element in $SO_n$. Then $A$ must commute with the following matrices,

1. a row switching transformation where one of the switched rows is multiplied by -1.

2. a double row multiplying transformation where the multiplier is -1 in each case.

Now conjugation by Type I, shows that all the elements on the main diagonal must be equal, and that $a_{ij} = -a_{ji}$ for $i \neq j$.

And conjugation by Type 2 matrices shows that for $n > 2$ all the non-main-diagonal elements must be zero.

Since A must also have determinant $1$, then the only matrices in the center must be

1. $SO_2$ if $n=2$. (All matrices in $SO_2$ meet the first condition. This is easily verified by taking arbitrary matrices in $SO_2$ and using sum of angle identities.)

2. $I$ if $n$ is odd.

3. $\pm I$ if $n$ is even.