This actually holds for any pair of similar linear mappings! So let $U,V$ be two similar matrices, i.e. there is some invertible matrix $P$ such that $V=PUP^{-1}$. 1. From linear algebra we know that $\det(AB)=\det A\det B$ and therefore: $$ \det V=\det(PUP^{-1})=\det P\det U\det P^{-1}=\det P\det U(\det P)^{-1}=\det U $$ 2. We also know from linear algebra that $\tr(AB)=\tr(BA)$; hence $$ \tr V=\tr(PUP^{-1})=\tr(P^{-1}PU)=\tr U $$