orthogonal real matrix means characteristic polynomial satsifies $p(t) = t^n p(t^{-1})$ I read somewhere that orthogonal matrix $Q$ with real entries means that the characteristic polynomial $p(t) = \det(tI - Q)$ satisfies $p(t) = t^n p(t^{-1})$. I know that the roots lie on a unit circle due to orthogonality and come in conjugate pairs because $Q$ has real entries. But I can't make sense of the property $p(t) = \det(tI - Q)$
 A: Hint: Let $f(t)=|t-s|$ with  $|s|=1$. Then $tf(\frac 1 1t)=|1-st| =|\frac  1 s -t|=|\overline s -t|=f(t)$ for $t >0$. If $s$ and $\overline s$ are both eigen values then $$(t-s)(t-\overline s)=|t-s|^{2}$$ $$=f(t)^{2}=t^{2}f(\frac 1  t)^{2}$$ $$=t^{2}(\frac 1 t-s) \frac 1 t-\overline s).$$ Apply this to each pair of eigen values and multiply.
This proves that $p(t)=t^{n}p(\frac  1t)$ for $t>0$. Now the identity theorem of Complex Analysis shows that this equation holds for all non-zero complex numbers $t$.
Note that this argument works for conjugate pairs of eigen values. But if $1$ is an eigen value then the argument fails for $f(t)=(t-1)$ since $tf(\frac 1t)=-f(t)$ in this case. So an appropriate adjustment in the sign is needed at the end.
A: You need to add an additional condition $1=(-1)^n {\rm Det}(Q)$.
With this condition, you have almost proven it already:  As the roots of $p(t)$ (other than $1,-1$) come in conjugate pairs and lie in the unit circle, they also come in inverse pairs, as the conjugate of a unit complex number is its inverse: $(x+iy)(x-iy)=x^2+y^2$.  Thus if $p(\lambda)=0$ then $p(\lambda^{-1})=0$.
Thus the polynomials $p(t)$ and $t^np(t^{-1})$ have the same roots - counted with multiplicity.  It remains to check they have the same leading coefficient.  However this follows from our additional condition: $1=(-1)^n {\rm Det}(Q)$.
Without the additional condition we have a $1\times 1$ counterexample: $\left(1\right)$
That is: $t-1\neq t(t^{-1}-1)$.
