## Chapter 8 Exercise D

1. Solution: By Exercise 11 in section 8B the characteristic polynomial is $z^4$ and by 8.46 this is a polynomial multiple of the minimal polynomial. Since $N^3 \neq 0$, it…

1. Solution: By Exercise 11 in section 8B the characteristic polynomial is $z^4$ and by 8.46 this is a polynomial multiple of the minimal polynomial. Since $N^3 \neq 0$, it…

1. Solution: Because $$ 4 = \operatorname{dim} \mathbb{C}^4 = \operatorname{dim} G(3, T) + \operatorname{dim} G(5, T) + \operatorname{dim} G(8, T), $$ it follows that the multiplicities of the eigenvalues of…

1. Solution: By 8.21 (a), $V = G(0, N)$. Since $G(0, N) = \operatorname{null} N^{\operatorname{dim} V}$ (see 8.11), it follows that $N^{\operatorname{dim} V} = 0$ and so $N$ is nilpotent.…

1. Solution: Since $$ T^2(w, z) = T(z, 0) = (0, 0), $$ it follows that $G(0, T) = V$. Therefore every vector in $\mathbb{C}^2$ is a generalized eigenvector of…