if $A\in O(n)$ and $\det A=-1$ then $A^2=I$ Any proof without using characteristics polynomial?

Question

I am sorry for this simple and easy question.

Is it true that If $A\in O(n)$ and $\det A=-1$ then $A^2=I$ ?

I know that $\det (A^2-I)=0$ (because $(A-A^t)^t=-(A-A^t)$) then its all eigenvalues are $1$. So it is similar to identity and hence $A^2=I$

1. Is my argument correct?
2. Any proof without using eigenvalues and characteristic polynomial?

Answer 1

For $n=2$, the orthogonal group has elements of two forms: $$ \begin{bmatrix} \cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta) \end{bmatrix}\qquad \begin{bmatrix} -\cos(\theta)&\sin(\theta)\\\sin(\theta)&\cos(\theta) \end{bmatrix} $$ For matrices of the first type, the determinant is $1$, for matrices of the second type, the determinant is $-1$.

Squaring an element of the second type gives the identity, so the claim is true for $n=2$, by explicit calculation.

For $n\geq 3$, @AnginaSeng's answer can be extended via a block matrix. If the matrix in that answer is $A$, then $$ \begin{bmatrix}A&0\\0&I_{n-3}\end{bmatrix} $$ has the same properties (its square is not $I_n$) and is in $O(n)$.

Answer 2

$$\pmatrix{-1&0&0\\ 0&0&1\\ 0&-1&0}$$