Show that if $Q$ is orthogonal transformation matrix, then $Q^t(Q-1)=(1-Q)^t$. Deduce that if $Q$ is also proper, then $\det(1-Q)=0$. Hence show that transformation has nonzero vector that has the same components in both coordinate system.
I tried to solve this problem.I think I got the first part right,
$$Q^t(Q -1)= Q^t Q- Q^t=1- Q^t=(1- Q)^t$$
The second part,
since the orthogonal matrix is proper which means $\det(Q)=1$ and for any matrix, its determinant equals the determinant of its transpose.
So, it's always true for $\det(1-Q)=0$
But that's not what the question asks. I haven't done linear algebra for a while and I am not sure from the concepts I used, so I would be glad if you clarify any mistake I made.