The possible set of eigenvalues of a $4\times 4$ skew symmetric, orthogonal matrix 
The possible set of eigenvalues of a $4\times 4$ Real skew symmetric, orthogonal matrix is
$1.\{\pm i\}$
$2.\{\pm i,\pm 1\}$
$3.\{\pm 1\}$
$4.\{\pm i,0\}$

As it is real skew symmetric so eigenvalues may be $0$ or Purely Imaginary, and as it is orthogonal so  determinant must be $1$ or $-1$. So $1$  may be the possible set. Am I right?
 A: Skew symmetric real matrices (more generally skew-Hermitian complex matrices) have purely imaginary (complex) eigenvalues. Orthogonal real matrices (more generally unitary matrices) have eigenvalues of absolute value$~1$. Either type of matrix is always diagonalisable over$~\Bbb C$. This means that for the given matrix only $\mathbf i$ and $-\mathbf i$ are possible eigenvalues. Moreover for real matrices the multiset of roots of the characteristic polynomial is invariant under complex conjugation; all this implies that $\mathbf i$ and $-\mathbf i$ are both eigenvalues, each with (geometric and algebraic) multiplicity$~2$.
A: Since $det(A−\lambda I)=det(A'−\lambda I)$, A and $A'$ have the same eigenvalues. On the other hand, $A'$ and −A also have the same eigenvalues. Thus if $\lambda$ is an eigenvalue of A, so is $-\lambda$. If n is odd, $\lambda=0$ is an eigenvalue.
If $\lambda$ is an eigenvalue of a matrix A, then $1/\lambda$ is an eigenvalue of $A^{-1}$.
Hence, ${\pm i}$ is the answer.
A: 0 down vote
Since det(A−λI)=det(A′−λI)det(A−λI)=det(A′−λI), A and A′A′ have the same eigenvalues. On the other hand, A′A′ and −A also have the same eigenvalues. Thus if λλ is an eigenvalue of A, so is −λ−λ. If n is odd, λ=0λ=0 is an eigenvalue.
If λλ is an eigenvalue of a matrix A, then 1/λ1/λ is an eigenvalue of A−1A−1.
Hence, ±i±i is the answer.
