Homepage › Solution manuals › Stephen Friedberg › Linear Algebra › Exercise 2.5.9

Exercise 2.5.9

Answers

We may denote that $A$ is similar to $B$ by $A \sim B$ . First we have $A = I^{- 1} AI$ and hence $A \sim A$ . Second, if $A \sim B$ we have $A = Q^{- 1} BQ$ and $B = {(Q^{- 1})}^{- 1} A Q^{- 1}$ and hence $B \sim A$ . Finally, if $A \sim B$ and $B \sim C$ then we have $A = P^{- 1} BP$ and $B = Q^{- 1} CQ$ . And this means $A = {(QP)}^{- 1} C (QP)$ and hence $A \sim C$ . So it’s a equivalence relation.

jephianlin

2011-06-27 00:00

Exercise 2.5.9

Answers

Comments

Add answer