linear algebra - characteristic polynomial of skew Hermitian matrix coefficients are real??

Let $X$ be a skew Hermitian matrix.

Is it true that its characteristic polynomial of $X$ i.e., $\det(\lambda I-X)$ has real coefficients?

Consider the matrix $X=\begin{bmatrix} -i&2+i\\-2+i &0\end{bmatrix}$. This is skew-Hermitian. Its characteristic polynomial is $\lambda^2+\lambda i+5$. So, characteristic polynomial of skew-Hermitian matrix need not have real coefficients.

Consider the matrix $\frac{1}{i}X=\frac{1}{i}\begin{bmatrix} -i&2+i\\-2+i &0\end{bmatrix}$. Its characteristic polynomial is $\lambda^2+\lambda-5 $ whose coefficintes are real.

I have checked some random examples and it turns out that for all of them ($X$ is skew-Hermitian), the characteristic polynomial of $\frac{1}{i}X$ is with real coefficients. Is this true in general? I think it is true. Can not think of a proof in general. Any suggestions are welcome.

The statement is

If $X$ is skew-Hermitian, characteristic polynomial of $\frac{1}{i}X$ is with real coefficients.

One thing is clear. As skew-Hermitian has Eigenvalues purely imaginary, trace is $0$ or $ai$ for some $a\in \mathbb{R}$. So, trace of $\frac{1}{i}X$ is $\frac{1}{i}(ai)=a$ i.e., real. So, one coefficient of characteristic polynomial is real. I can not think of general proof for other coefficients.

Answer

This is true. We have that $\frac{1}{i}=-i$, and if $X$ is skew-Hermitian, $-iX$ is Hermitian. To check this, denoting $X^H$ to be the conjugate transpose, we have
$$
(-iX)^H=\overline{-i}(X^H)=i(-X)=-iX
$$
Now, Hermitian matrices have all real eigenvalues, so the characteristic equation will also have all real coefficients.

Another way to see that all the eigenvalues of $-iX$ are real is to use the fact that skew-Hermitian matrices are diagonalizable, and move the $-i$ coefficient to the diagonal matrix, whose entries are the eigenvalues of $X$.