TITLE: Is it possible to have a $3 \times 3$ matrix that is both orthogonal and skew-symmetric?
QUESTION [18 upvotes]: Is it possible to have a $3 \times 3$ matrix that is both orthogonal and skew-symmetric?
I know it has something to do with the odd order of the matrix and it is not possible to have such a matrix. But what is the reason?

REPLY [10 votes]: No, it is not possible, for real matrices. Let $A$ be an $n \times n$ real matrix, with $n$ being odd. Suppose it is skew-symmetric, that is, $A^T = -A$. Then, $A^T A = -A^2$. If the matrix is also orthogonal, $I = -A^2$. Now take determinants to get
$$\det(A)^2 + 1 = 0$$
The determinant is imaginary, giving a contradiction, since we assumed $A$ to be a real matrix.