## Chapter 8 Exercise D

Exercise 1 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 follows that the minimal...

Solution to Linear Algebra Done Right

Third Edition

Exercise 1 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 follows that the minimal...

Exercise 1 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 $T$ are at most...

Exercise 1 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. Exercise 2 Define $T...

Exercise 1 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 $T$. Exercise 2 The...