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 $$ Rv = \frac{||x||}{||u||}\langle…
1. Solution. A quick calculation shows that $T^*Tv = ||x||^2\langle v, u \rangle u$ for every $v \in V$. The map $R \in \mathcal{L}(V)$ defined by $$ Rv = \frac{||x||}{||u||}\langle…
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 $\mathbb{R}^2$.…
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…
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:…