1. Solution. Because $T_\mathbb{C}$ has no real eigenvalues, if $\lambda$ is an eigenvalue of $T_\mathbb{C}$, then $\overline{\lambda}$ is also an eigenvalue of $T_\mathbb{C}$ with equal multiplicity. The determinant of $T_\mathbb{C}$,…
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…
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: 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: (a) One can easily check that each of the four vectors has norm $\sin^2 \theta + \cos^2 \theta$, which equals $1$. Moreover, we have $$ \begin{aligned} \langle (\cos\theta,…
1. Solution: (a) Note that \[ (I-T)(I+T+\cdots+T^{n-1})=I-T^n=I \]and \[ (I+T+\cdots+T^{n-1})(I-T)=I-T^{n}=I ,\](in fact we just need to check only one) it follows that $I-T$ is invertible and \[(I-T)^{-1}=I+T+\cdots+T^{n-1}.\] (b) From the…
1. Solution: Assume $V$ is 5-dimensional vector space with a basis $e_1$, $\cdots$, $e_5$. Define $T\in\ca L(V,V)$ by \[Te_1=e_1,Te_2=e_2,Te_3=Te_4=Te_5=0.\]Then $\mathrm{null} T=\mathrm{span}(e_3,e_4,e_5)$, hence $\dim \mathrm{null} T=3$. Similarly, $\mathrm{range} T=\mathrm{span}(e_1,e_2)$, hence $\dim…
1. Solution: The only vector spaces is $\{0\}$. For if there is a nonzero vector $v$ in a basis, then we can get a new basis by changing $v$ to…
1. Solution: By definition, we have\[(-v)+(-(-v))=0\quad\text{and}\quad v+(-v)=0.\]This implies both $v$ and $-(-v)$ are additive inverses of $-v$, by the uniqueness of additive inverse, it follows that $-(-v)=v$. 2. Solution: If…