Chapter 9 Exercise B
1. Solution: Choose an orthonormal basis of $\mathbb{R}^3$ that puts the matrix of $S$ in the form given by 9.36. Since $\mathcal{M}(S)$ is a $3$-by-$3$ matrix, one of the diagonal…
Chapter 9 Exercise A
1. Solution: $V_\mathbb{C}$ is clearly closed under addition. We can write each complex number in the form $a + bi$ for some $a, b \in \mathbb{R}$ and we have $$…
Chapter 8 Exercise D
1. Solution: By Exercise 11 in section 8B the characteristic polynomial is $z^4$ and by 8.46 this is a polynomial multiple of the minimal polynomial. Since $N^3 \neq 0$, it…
Chapter 8 Exercise C
1. Solution: Because $$ 4 = \operatorname{dim} \mathbb{C}^4 = \operatorname{dim} G(3, T) + \operatorname{dim} G(5, T) + \operatorname{dim} G(8, T), $$ it follows that the multiplicities of the eigenvalues of…
Chapter 8 Exercise B
1. Solution: By 8.21 (a), $V = G(0, N)$. Since $G(0, N) = \operatorname{null} N^{\operatorname{dim} V}$ (see 8.11), it follows that $N^{\operatorname{dim} V} = 0$ and so $N$ is nilpotent.…
Chapter 8 Exercise A
1. Solution: Since $$ T^2(w, z) = T(z, 0) = (0, 0), $$ it follows that $G(0, T) = V$. Therefore every vector in $\mathbb{C}^2$ is a generalized eigenvector of…
Chapter 7 Exercise D
1. Solution. A quick calculation shows that $T^*Tv = ||x||^2\langle v, u \rangle u$ for every $v \in V$. The map $R \in \mathcal{L}(V)$ defined by $$ Rv = \frac{||x||}{||u||}\langle…
Chapter 7 Exercise C
1. Solution: We give a counterexample. Define $T \in \mathcal{L}(\mathcal{R}^2)$ by $$ \begin{aligned} Te_1 = e_1\\ Te_2 = -e_2 \end{aligned} $$ where $e_1, e_2$ is the standard basis of $\mathbb{R}^2$.…
Chapter 7 Exercise B
1. Solution: It is true. Consider the standard orthonormal basis $e_1,e_2,e_3$ of $\mb R^3$.Define $T\in \ca L(\mb R^3)$ by the rule:\[Te_1=e_1,\quad Te_2=2e_2+e_1,\quad Te_3=3e_3.\]Since we have\[\langle Te_1,e_2\rangle =\langle e_1,e_2\rangle =0,\]\[\langle e_1,Te_2\rangle…
Chapter 7 Exercise A
1. Solution: By definition, we have \[\begin{align*}\langle (z_1,\cdots,z_n),T^*(w_1,\cdots,w_n)\rangle=&\langle T(z_1,\cdots,z_n),(w_1,\cdots,w_n) \rangle \\=& z_1w_2+\cdots+z_{n-1}w_{n}=\langle (z_1,\cdots,z_n),(w_2,\cdots,w_n,0)\rangle.\end{align*}\]Therefore $T^*(w_1,\cdots,w_n)=(w_2,\cdots,w_n,0)$ or $T^*(z_1,\cdots,z_n)=(z_2,\cdots,z_n,0)$. See also Linear Algebra Done Right Solution Manual Chapter 6 Problem 27. 2. Solution:…