Exercise 7.2.6

Answers

The fact $rank (M) = rank (M^{t})$ for arbitrary square matrix $M$ and the fact

{(A^{t} - λI)}^{r} = {[{(A - λI)}^{t}]}^{r} = {[{(A - λI)}^{r}]}^{t}

tell us

rank ({(A - λI)}^{r}) = rank ({(A^{t} - λI)}^{r}) .

By Theorem 7.9 we know that $A$ and $At$ have the same dot diagram and the same Jordan canonical form. So $A$ and $A^{t}$ are similar.

jephianlin

2011-06-27 00:00