Does this matrix have signature 0? Let $A$ be a skew-symmetric $n\times n$ matrix with coefficients in $\mathbb{R}$ or $\mathbb{C}$. Then consider the following $2n\times 2n$ matrix:
$$\begin{pmatrix}0&A\\-A&0 \end{pmatrix}.$$
This matrix is symmetric because $A^t=-A$. Then is the signature of this matrix zero? Equivalently, is the number of positive eigenvalues equal to the number of negative eigenvalues?
 A: Notably, your matrix is equal to 
$$
M = \pmatrix{0&1\\-1&0} \otimes A
$$
where $\otimes$ denotes the Kronecker product.  If $A$ has real coefficients, then both the $2 \times 2$ matrix and $A$ have imaginary eigenvalues in conjugate pairs, so that $M$ has signature zero as a consequence of the spectral properties of the Kronecker product.
A: The answer of @Omnomnomnom is great. 
Let $B$ be your matrix. Computing the characteristic polynomial of $B$, we have
$$
\det (B - \lambda I) = \det(A^2 + \lambda^2I).
$$
Hence, denoting by $\lambda_j^\pm(A) = \pm i \alpha_j$ the eigenvalues of $A$, we have that the eigenvalues of $A^2$ are given by $\lambda_j^{\pm}(A^2) = -\alpha_j^2$. Then, we need that $\lambda^2_j(B) = -\lambda_j^\pm(A^2)$, which means that $\lambda_j^\pm(B) = \pm \alpha_j$.
To conclude, you will have that the e'vals of $B$ will be the imaginary part of the e'vals of $A$, taken with both signs and twice each.
For example, if $A$ is $4\times 4$,you will have the eigenvalues $\lambda(A) = \{i\alpha_1, -i\alpha_1,i\alpha_2, -i\alpha_2\}$. Then the e'vals of $A^2$ are $\lambda(A^2) =\{-\alpha_1^2, -\alpha_1^2, -\alpha_2^2, -\alpha_2^2\}$. For each eigenvalue of $-A^2$, plus or minus the square root will be an eigenvalue of $B$. Hence, $\lambda(B) = \{\pm\alpha_1,\pm\alpha_1, \pm\alpha_2, \pm\alpha_2\}$.
