Skew-symmetric square root of symmetric matrix

Question

Suppose that $A$ and $B$ are real skew-symmetric $4 \times 4$ matrices. Hence $A^2$ and $B^2$ are symmetric matrices. Now we want to find a skew-symmetric 4×4 matrix ($C$) which satisfies

$$A^2+B^2=C^2$$

Thanks!

Answer 1

This is not always possible. E.g. when $$ A=\pmatrix{0&0&0&0\\ 0&0&0&-1\\ 0&0&0&0\\ 0&1&0&0}, \ B=\pmatrix{0&0&0&0\\ 0&0&0&0\\ 0&0&0&-2\\ 0&0&2&0}, $$ we have $S=A^2+B^2=\operatorname{diag}(0,-1,-4,-5)$. Since $S$ has nonzero eigenvalues of odd multiplicities, it is not the square of any real or complex skew-symmetric matrix.