Exercise 5.4.16

Answers

1.: By Theorem 5.21 we know that the characteristic polynomial of the restriction of $T$ to any $T$ -invariant subspace is a factor of a polynomial who splits. So it splits, too.
2.: Any nontrivial $T$ -invariant subspace has dimension not equal to $0$ . So the characteristic polynomial of its restriction has degree greater than or equal to $1$ . So it must contains at least one zero. This means the subspace at least contains one eigenvector.

jephianlin

2011-06-27 00:00