# What is the most general matrix that gives rise to all even characteristic polynomials?

Is there some general form of all matrices which give rise to all even or all odd characteristic polynomial terms?

For skew-symmetric matrices such that $$A^T=-A$$ we necessarily have all even or all odd terms.

For example the matrix $$A=\left(\matrix{0&2&4\\-2&0&-1\\-4&1&0}\right)$$ has the characteristic polynomial $$-\lambda^3-21\lambda$$

However we can also generate characteristic polynomials of this form when the matrix is not skew symmetric.

It appears that we require the trace to be $$0$$, so that the sum of eigenvalues are equal to $$0$$. But this doesn't appear to be a general condition.

An example would be the matrix $$B=\left(\matrix{3&0&2&4&0&0\\0&-3&0&5&-2&-4\\2&0&3&4&0&-5\\4&5&4&7&0&0\\0&-2&0&0&-3&-4\\0&-4&-5&0&-4&-7}\right)$$

which has the characteristic polynomial

$$\lambda^6-189\lambda^4+2472\lambda^2-6084$$