Exercise 2.5.10

Answers

If $A$ and $B$ are similar, we have $A = Q^{- 1} BQ$ for some invertible matrix $Q$ . So we have

tr (A) = tr (Q^{- 1} BQ) = tr (Q Q^{- 1} B) = tr (B)

jephianlin

2011-06-27 00:00