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

Question

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.

Answer 1

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.

Answer 2

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 $4\equiv 1\pmod 3.$