So here is the question:
Suppose $A$ is an orthogonal matrix with $\det(A)=-1$, prove $A$ has an eigenvalue $-1$.
Here is my solution, since $$\begin{split} &\hspace{0.75cm} Ax=\lambda x\\&\Rightarrow A^\text TAx=A^\text T\lambda x\\&\Rightarrow(A^\text TA-A^\text T\lambda )x=\theta \end{split}$$ so $$\begin{split} &\hspace{0.75cm} A^\text TA=\lambda A^\text T\\&\Rightarrow \det(A^\text TA)=\det(\lambda A^\text T)=\lambda^n\det(A^\text T)=-\lambda^n \end{split}$$ and given $A$ orthogonal, we have $$\det(A^\text TA)=\det(I)=1$$ so $$-\lambda^n=1\Rightarrow\lambda=i\;\;\text{or}\;\;\lambda=-1$$ and now I'm stuck, how to I prove $A$ also has an eigenvalue $-1$ when $n$ is even?
$$\det(A+I)= \det(A+AA^T)= \det(A(I+A^T))= \det(A) \cdot \det(I+A^T)$$
$$=- \det(I+A)^T= - \det(A+I).$$