Answer:
Step-by-step explanation:
If A and B are nxn matrices then tr(AB)=tr(BA)... Hypothesis (1)
Prove:
A is similar to B
⇒There exists an invertible n-by-n matrix P such that B=P^{-1}AP.
⇒ tr[B]=tr[P^{-1}AP]
⇒tr[B]=tr[(P^{-1}A)P] Matrix multiplication has the associative property
⇒tr[B]=tr[P(P^{-1}A)] Using hypothesis (1)
⇒tr[B]=tr[(PP^{-1})A] Matrix multiplication has the associative property
⇒tr[B]=tr[IA] I is the identity matrix
⇒tr[B]=tr[A]
