Homepage › Solution manuals › Gilbert Strang › Introduction to Linear Algebra › Exercise 2.7.37

Exercise 2.7.37

Answers

There are $n!$ permutation matrices of order $n$ . Eventually two powers of $P$ must be the same permutation. And if $P^{r} = P^{s}$ then $P^{r - s} = I$ . Certainly $r - s \leq n!$

P = [\begin{matrix} P_{2} \\ P_{3} \end{matrix}] is 5 by 5 with P_{2} = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] and P_{3} = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] and P^{6} = I .

mathchakra

2022-01-16 12:46

Exercise 2.7.37

Answers

Comments

Add answer