prove that if $X$ is skew- symmetric matrix of order $n$ and$(X^2+I_n)$ is non singular then $(iX+I_n)^{-1} +I_n$ is unitary matrix I know the basics definition of the unitary, skew-symmetric and invertible matrix but still, I'm not able to prove this statement,
Note: computation is over a field of char $3$.
prove that if $X$ is skew-symmetric matrix of order $n$ and$(X^2+I_n)$ is non-singular then $(iX+I_n)^{-1} +I_n$ is a unitary matrix. Unitary matrix means $UU^T=I$ besides $UU^*=I$.
Any help would be appreciated.
 A: Your question is hard to understand. I suppose that you mean, if $X\in M_n(\mathbb Z_3)$ is skew-symmetric, $X^2+I$ is nonsingular and $i$ is a square root of $-1$ in the algebraic closure of $\mathbb Z_3$, then
$$
U=(iX+I)^{-1}+I\in M_n(\mathbb Z_3[i])
$$
is unitary in the sense that $U^\ast U=I$.
This is true only if $X=0$.
Note that $U^\ast=\left[(iX+I)^{-1}+I\right]^\ast=(-iX^T+I)^{-1}+I=(iX+I)^{-1}+I=U$. Thus
\begin{align}
U^\ast U-I
&=U^2-I\\
&=\left[(iX+I)^{-1}+I\right]^2-I\\
&=(iX+I)^{-2}+2(iX+I)^{-1}\\
&=(iX+I)^{-2}-(iX+I)^{-1}\\
&=(iX+I)^{-1}\left[(iX+I)^{-1}-I\right].\tag{1}\\
\end{align}
Since $X^2+I=(iX+I)(-iX+I)$ is nonsingular, $iX+I$ is nonsingular. Therefore, if $UU^\ast=I$, then from $(1)$ we obtain $(iX+I)^{-1}-I=0$. Hence $iX+I=I$, meaning that $X$ is necessarily zero.
A: Let $U=(\iota X+I_n)^{-1}+I_n$ and for any matrix $(A^{-1})^T=(A^T)^{-1}.$
then  $U^T=(-\iota X+I_n)^{-1}+I_n$.
Now consider
\begin{equation}
    UU^T=[(\iota X+I_n)^{-1}+I_n][(-\iota X+I_n)^{-1}+I_n]
\end{equation}
$$ =(\iota X+I_n)^{-1}(-\iota X+I_n)^{-1}+(\iota X+I_n)^{-1}+(-\iota X+I_n)^{-1}+I$$
$$=[(-\iota X+I_n)(\iota X+I_n)]^{-1}+(\iota X+I_n)^{-1}+(-\iota X+I_n)^{-1}+I$$
as $(AB)^{-1}=B^{-1}A^{-1}$
so, eq.(1) becomes,
\begin{equation}
    =(X^2+I)^{-1}+(\iota X+I_n)^{-1}+(-\iota X+I_n)^{-1}+I
\end{equation}
multiply both side of eq.(2) by $(X^2+I) $ we get RHS as
$=I+(\iota X +I)(-\iota X +I)(\iota X +I)^{-1}+(\iota X +I)(-\iota X +I)(-\iota X +I)^{-1}+(X^2+I)$
then eq.(1)become,
(note that $(\iota X +I)$ and $(-\iota X+ I)$ are both with commuted each other.)
$$(x^2+I_n)UU^T=I_n+(-\iota X+I_n)+(\iota X+I_n)+(X^2+I_n)$$
$$UU^T=(I_n+X^2)^{-1}(4I_n+X^2)=I_n$$ as $4\equiv 1\pmod 3.$
