# Chapter 5 Exercise A

1. Solution: (a) For any $u\in U$, then $Tu=0\in U$ since $U\subset \m{null} T$, hence $U$ is invariant under $T$.

(b) For any $u\in U$, then $Tu\in\m{range} T \subset U$, hence $U$ is invariant under $T$.

2. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 4. Let $\lambda=0$.

3. Solution: For any $u\in \m{range} S$, there exists $v\in V$ such that $Sv=u$, hence $Tu=TSv=STv\in \m{range} S.$Therefore range $S$ is invariant under $T$.

4. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 1.

5. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 2.

6. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 3.

7. Solution: Let $(x,y)$ be an eigenvector of $T$ corresponding to eigenvalue $\lambda$, then we have $T(x,y)=\lambda(x,y),$i.e., $(\lambda x,\lambda y)=(-3y,x)$. Hence we have $\lambda x=-3y$ and $\lambda y=x$, it follows that $\lambda^2xy=-3xy$. If $xy\ne 0$, then $\lambda^2=-3$, this is impossible.

If $x=0$, then $y=0$ by $\lambda x=-3y$. However $(x,y)$ is an eigenvector, hence $(x,y)\ne (0,0)$. We get a contradiction.

If $y=0$, then $x=0$ by $\lambda y=x$. Similarly, we get a contradiction.

Hence no such eigenvectors exist, namely $T$ has no eigenvalues.

8. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 5.

9. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 6.

10. Solution: (a) Suppose $v=(v_1,\cdots,v_n)$ is a eigenvector of $T$ corresponding to eigenvalue $\lambda$. Then we have $Tv=\lambda v$, hence $$\label{5AP1}(v_1,2v_2,\cdots,v_n)=(\lambda v_1,\lambda v_2,\cdots,\lambda v_n).$$ As $v\ne 0$ by definition of eigenvectors, there is some $i\in \{1,2,\cdots,n\}$ such that $v_i\ne 0$. Note that we have $iv_i=\lambda v_i$ by $(\ref{5AP1})$, it implies $\lambda=i$. For $\lambda=i$, it is easy to solve $(\ref{5AP1})$. We can conclude the corresponding eigenvectors are of the form $(0,\cdots,0,a,0,\cdots,0),a\in\mb F$ with $a$ in $i$-th component. Similarly, all eigenvalues of $T$ are $1$, $2$, $\cdots$, $n$. All eigenvectors with respect to $i$ are of the form $(0,\cdots,0,a,0,\cdots,0),a\in\mb F$ with $a$ in $i$-th component.

(b) Suppose $W$ is an invariant subspace of $T$. Assume $e_i=(0,\cdots,0,1,0,\cdots,0)$ with $1$ in $i$-th component. Then $e_1$, $\cdots$, $e_n$ is a basis of $\mb F^n$ and $e_i$ is an eigenvector of $T$ corresponding to $i$. If $a_1e_1+\cdots+a_ke_k\in W$ with $a_1\cdots a_k\ne 0$, we will show $\m{span}(e_{1},e_{2},\cdots,e_{k})\subset W$. Note that $a_1e_1+\cdots+a_ke_k\in W$ and $W$ is invariant with respect to $T$, it follows that $T(a_1e_1+\cdots+a_ke_k)=a_1e_1+\cdots+ka_ke_k\in W.$Hence $k(a_1e_1+\cdots+a_ke_k)-(a_1e_1+\cdots+ka_ke_k)=(k-1)a_1e_1+\cdots+a_{k-1}e_{k-1}\in W,$and the coefficients are nonzero. Inductively, we will get some $\lambda_1e_1+\cdots+\lambda_ie_i\in W$for $\lambda_1\cdots\lambda_i\ne 0$ for any $i\leqslant k$($\lambda_1$, $\cdots$, $\lambda_i$ change as $i$ changes). In particular, $\mu_1e_1\in W$ and $\mu_1\ne 0$. Hence $e_1\in W$. Then, consider $\eta_1e_1+\eta_2e_2\in W$ where $\eta_1\eta_2\ne 0$, we will get $e_2\in W$. Inductively, we can show that $\{e_{1},e_{2},\cdots,e_{k}\}\subset W$. Hence $\m{span}(e_{1},e_{2},\cdots,e_{k})\subset W$. Similarly, if $a_{i_1}e_{i_1}+\cdots+a_{i_k}e_{i_k}\in W$ with $a_{i_1}\cdots a_{i_k}\ne 0$ and all $\{a_{i_j}\}$ distinct, then $\m{span}(e_{i_1},\cdots,e_{i_k})\subset W$. Now let us consider the general form of $W$. Suppose $W\cap \{e_1,\cdots,e_n\}=\{e_{i_1},\cdots,e_{i_k}\}$, then we will show $\m{span}(e_{i_1},\cdots,e_{i_k})= W$. It is obvious that $\m{span}(e_{i_1},\cdots,e_{i_k})\subset W$. If there some $w\in W$ but $w\notin \m{span}(e_{i_1},\cdots,e_{i_k})$. Then $w$ can be written as $w=b_1e_1+\cdots+b_ne_n,\quad b_1,\cdots,b_n\in \mb F$such that there is some $s\notin \{i_1,\cdots,i_k\}$ and $b_s\ne 0$. By previous argument, we have $e_s\in W$. This contradicts with $W\cap \{e_1,\cdots,e_n\}=\{e_{i_1},\cdots,e_{i_k}\}$. Hence we show that $\m{span}(e_{i_1},\cdots,e_{i_k})= W$. Moreover, all invariant subspaces of $T$ have this form.

I am not satisfied with this solution. ❗

11. Solution: Suppose $\lambda$ is an eigenvalue of $T$ with an eigenvector $q$, then $q’=Tq=\lambda q.$Note that in general $\deg p'<\deg p$(because we consider $\deg 0=-\infty$). If $\lambda\ne 0$, then $\deg \lambda q>\deg q’$. We get a contradiction. If $\lambda=0$, then $q=c$ for nonzero $c\in\R$. Hence the only eigenvalue of $T$ is zero with nonzero constant polynomials as eigenvectors.

12. Solution: Suppose $\lambda$ is an eigenvalue of $T$ with an eigenvector $q$. Let $q=a_nx^n+\cdots+a_1x+a_0$ such that $a_n\ne 0$, then $\lambda q=Tq=xq’,$namely $\lambda a_nx^n+\cdots+\lambda a_1x+\lambda a_0=na_nx^n+\cdots+2a_2x^2+a_1x.$Since $a_n\ne 0$, it follows that $\lambda =n$ by considering the leading coefficient. Then we have $a_0=a_1=\cdots=a_{n-1}=0$, hence $q=a_nx^n$. Hence all eigenvalues of $T$ are $0,1,2,\cdots$ and all eigenvectors correspond to $m$ is $\lambda x^m$ such that $m\in \mb{N}$, $\lambda\ne 0$ and $\lambda\in\R$.

13. Solution: Let $\alpha_i\in\mb F$ such that $\left|\alpha_i-\lambda\right| = \frac{1}{1000+i},\quad i=1,\cdots,\dim V+1.$These $\alpha_i$ exist and are different from each other since $F=\R$ or $\C$. Note that each operator on $V$ has at most $\dim V$ distinct eigenvalues by 5.13. Hence there exists some $i\in\{1,2,\cdots,\dim V+1\}$ such that $\alpha_i$ is not an eigenvalue of $T$. Then by 5.6, $T-\alpha_i I$ is invertible.

14. Solution: Note that any $v\in V$ can be written uniquely as $u+w$ for $u \in U$ and $w \in W$ since $V=U\oplus W$. It follow that this $P$ is well-defined. Maybe you also need to check$P\in\ca L(V)$. Now let us consider the eigenvalues of $P$. By consider $v\ne 0$, if there exists $\lambda\in\mb F$ such that $Pv=\lambda v$. Write $v=u+w$ for $u \in U$ and $w \in W$, then $u$ and $w$ can not be both zero. Hence by definition of $P$, we have $Pv=u,\quad \lambda v=\lambda u+\lambda w.$It follows that $u=\lambda u+\lambda w$, namely $(\lambda-1)u+\lambda w=0$. Note that $V=U\oplus W$, it follows that $(\lambda-1)u=\lambda w=0$. If $u\ne 0$, then $\lambda =1$. Hence $w=0$ and the corresponding eigenvectors are nonzero vectors $v\in U$. If $w\ne 0$, then $\lambda =0$. Hence $u=0$ and the corresponding eigenvectors are nonzero vectors $v\in W$.

15. Solution: (a) Suppose $\lambda$ is an eigenvalue of $T$, then there exists a nonzero vector $v\in V$ such that $Tv=\lambda v$. Hence $S^{-1}TS(S^{-1}v)=S^{-1}Tv=S^{-1}(\lambda v)=\lambda S^{-1}v.$Note that $S^{-1}v\ne 0$ as $S^{-1}$ is invertible, hence $\lambda$ is an eigenvalue of $S^{-1}TS$, namely every eigenvalue of $T$ is an eigenvalue of $S^{-1}TS$. Similarly, note that $S(S^{-1}TS)S^{-1}=T$, we have every eigenvalue of $S^{-1}TS$ is an eigenvalue of $T$. Hence $T$ and $S^{-1}TS$ have the same eigenvalues.

(b) From the process of (a), one can easily deduce that $v$ is an eigenvector of $T$ if and only if $S^{-1}v$ is an eigenvector of $S^{-1}TS$.

16. Solution: Although this problem is true for infinite-dimensional vector space, I will just consider finite-dimension case since we are considering the matrix of $T$(otherwise, it would be a infinite matrix). Suppose the matrix of $T$ with respect to basis $e_1$, $\cdots$, $e_n$ of $V$ contains only real entries. Then $Te_j=A_{1,j}e_1+\cdots+A_{n,j}e_n,$where $A_{i,j}\in\R$ for all $i,j=1,2,\cdots,n$. Let $v=k_1e_1+\cdots+k_ne_n$be a eigenvector with respect to $\lambda$, where $k_i\in\C$, $i=1,\cdots,n$. Then we have $Tv=\lambda v,$namely $$\label{5A161} \lambda\sum_{i=1}^nk_ie_i=\sum_{i=1}^nk_iTe_i=\sum_{i=1}^n\sum_{j=1}^nk_iA_{j,i}e_j.$$ Consider the complex conjugation of $(\ref{5A161})$, we have $$\label{5A162} \overline{\lambda}\sum_{i=1}^n\overline{k_i}e_i=\sum_{i=1}^nk_iTe_i=\sum_{i=1}^n\sum_{j=1}^n\overline{k_i}A_{j,i}e_j$$ since $A_{i,j}\in\R$ for all $i,j=1,2,\cdots,n$. (why? consider components)Note that $(\ref{5A162})$ implies $$\label{5A163} T(\overline{k_1}e_1+\cdots+\overline{k_n}e_n)=\overline{\lambda}\sum_{i=1}^n\overline{k_i}e_i.$$Since $v=k_1e_1+\cdots+k_ne_n\ne 0$, it follows that not all $k_i$ is zero, so is $\overline{k_i}$. Hence $\overline{k_1}e_1+\cdots+\overline{k_n}e_n\ne 0$, hence $(\ref{5A163})$ tell us $\bar{\lambda}$ is an eigenvalue of $T$.

17. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 23.

18. Solution: Suppose $\lambda$ is an eigenvalue of $T$ and one corresponding eigenvector is $(w_1, w_2,\cdots)$. Then not all of $w_i$ is zero. Moreover, we have $(0,w_1, w_2,\cdots)=T(w_1, w_2,\cdots)=\lambda(w_1, w_2,\cdots).$If $\lambda=0$, then $(0,w_1, w_2,\cdots)=0$implies $w_i\equiv 0$ for any $i\in\mb N^+$. We get a contradiction. If $\lambda\ne 0$. Consider the first component, we have $0=\lambda w_1$, hence $w_1=0$. Then consider the second component, we have $\lambda w_2=w_1=0$, hence $w_2=0$. By induction, one can easily deduce that $w_i\equiv 0$ for any $i\in\mb N^+$. We get a contradiction as well. Hence $T$ has no eigenvalues.

19. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 7.

20. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 8.

21. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 10. (b) is almost proved there.

22. Solution: Note that we have $T(v+w)=Tv+Tw=3w+3v=3(v+w),$and$T(v-w)=Tv-Tw=3w-3v=-3(v-w).$If $v-w$ or $v+w$ is nonzero, then $3$ or $-3$ is an eigenvalue of $T$. In fact if $v-w=0$ and $v+w=0$, it is easy to see $v=w=0$. It contradicts with $v\ne 0$ and $w\ne 0$.

23. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 11.

24. Solution: (a) (a) If the sum of the entries in each row of $A$ equals $1$, then one can easily deduce that $T\left( \begin{array}{c} 1 \\ \vdots \\ 1 \\ \end{array} \right) =\left( \begin{array}{c} 1 \\ \vdots \\ 1 \\ \end{array} \right).$Hence $1$ is an eigenvalue of $T$ with $\left( \begin{array}{c} 1 \\ \vdots \\ 1 \\ \end{array} \right)$ as a corresponding eigenvector.

(b) This problem is interesting. It is simple by considering determinant. However it is complicated here. We just need to show that $T-I$ is not invertible. It suffices to show $T-I$ is not surjective by 5.6. Note that we have $$\label{5AP241} (T-I)\left( \begin{array}{c} x_1 \\ \vdots \\ x_n \\ \end{array} \right)=\left( \begin{array}{c} \sum_{i=1}^n A_{1,i}x_i-x_1 \\ \vdots \\ \sum_{i=1}^n A_{n,i}x_i-x_n \\ \end{array} \right)=\left( \begin{array}{c} y_1 \\ \vdots \\ y_n \\ \end{array} \right),$$where $A_{i,j}$ is the $(i,j)$-component of $A$. Moreover, we have $1=\sum_{i=1}^n A_{i,j}\quad j=1,\cdots,n.$Hence \begin{align*} y_1+\cdots+y_n=&\sum_{j=1}^n\sum_{i=1}^n A_{j,i}x_i-\sum_{j=1}^n x_j\nonumber\\ =&\sum_{i=1}^nx_i \sum_{j=1}^n A_{j,i}-\sum_{j=1}^n x_j\\ =&\sum_{i=1}^nx_i-\sum_{j=1}^n x_j=0\nonumber.\end{align*} By $(\ref{5AP241})$ and previous equation, it follows that $\m{range}(T-I)\subset \{(x_1,\cdots,x_n)^T\in\mb F^{n}:x_1+\cdots+x_n=0\},$where $(x_1,\cdots,x_n)^T$ means $\left( \begin{array}{c} x_1 \\ \vdots \\ x_n \\ \end{array} \right)$. It follows that $T-I$ is not surjective, hence completing the proof.

25. Solution: Let the eigenvalues corresponding to $u,v$ are $\lambda_1,\lambda_2$ respectively, then we have $Tu=\lambda_1u,\quad Tv=\lambda_2 v.$If the eigenvalue corresponding to $u+v$ is $\lambda$, we have$\lambda(u+v)=T(u+v)=Tu+Tv=\lambda_1u+\lambda_2v.$It follows that $(\lambda-\lambda_1)u+(\lambda-\lambda_2)v=0$. If $\lambda_1\ne\lambda_2$, then $\lambda-\lambda_1$ and $\lambda-\lambda_2$ can not be both zero. Hence $u$, $v$ is not linearly independent. By 5.10, it follows that $u$ and $v$ correspond to the same eigenvalue. Hence $\lambda_1=\lambda_2$.

26. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 12.

27. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 13.

28. Solution: For any nonzero vector $v\in V$, extend it to a basis of $V$ as $v=v_1$, $v_2$, $\cdots$, $v_n$. Then $Tv_1=\sum_{k=1}^n\lambda_kv_k.$Consider $U=\m{span}(v_1,v_2)$, as $U$ is invariant under $T$ by assumption. It follows that $Tv_1\in U$. Hence $\lambda_3=\cdots=\lambda_n=0$. Similarly, consider $U=\m{span}(v_1,v_3)$ (note that $\dim V \ge 3$), we will conclude $\lambda_2=\lambda_4=\cdots=\lambda_n=0$. Hence $\lambda_2=\cdots=\lambda_n=0$. This means $v_1$ is an eigenvector of $T$. That is every nonzero vector in $V$ is an eigenvector of $T$ since $v$ is chosen arbitrarily. By Problem 26, we deduce that $T$ is a scalar multiple of the identity operator.

29. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 9.

30. Solution: Note that $T$ has at most $\dim(\R^3)=3$ eigenvalues (by 5.13) and $4$, $5$, and $\sqrt{7}$ are eigenvalues of $T$, it follows that $9$ is not an eigenvalues of $T$. Hence $(T-9I)$ is surjective (by 5.6). Thus there exists $x\in\R^3$ such that $(T-9I)x=(4,5,\sqrt{7})$, namely $Tx-9x=(4,5,\sqrt{7})$.

31. Solution: If there exists $T\in \ca L(V)$ such that $v_1$, $\cdots$, $v_m$ are eigenvectors of $T$ corresponding to distinct eigenvalues. Then $v_1$, $\cdots$, $v_m$ is linearly independent by 5.10.

Conversely, if $v_1$, $\cdots$, $v_m$ is linearly independent. Then we can extend it to a basis of $V$ as $v_1$, $\cdots$, $v_m$, $v_{m+1}$, $\cdots$, $v_n$. Define $T\in \ca L(V)$ by $Tv_i=iv_i,\quad i=1,\cdots,n.$Then $v_1$, $\cdots$, $v_m$ are eigenvectors of $T$ corresponding to eigenvalues $1$, $\cdots$, $m$, respectively.

32. Solution: Let $V=\m{span}($ $e^{\lambda_1 x}$, $\cdots$, $e^{\lambda_n x})$, and define an operator $T\in \ca L(V)$ by $Tf=f’$(You should check $T\in \ca L(V)$). Then consider $Te^{\lambda_i x}=\lambda_ie^{\lambda_i x}.$Hence $\lambda_i$ is an eigenvalue of $T$ with an corresponding eigenvector $e^{\lambda_i x}$. As $\lambda_1$, $\cdots$, $\lambda_n$ is a list of distinct real numbers, by 5.10, it follows that $e^{\lambda_1 x}$, $\cdots$, $e^{\lambda_n x}$ is linearly independent.

33. Solution: By definition, for any $x+\m{range} T\in V/(\m{range} T )$, we have $T/(\m{range} T )(x+\m{range} T)=Tx+\m{range} T.$Note that $Tx\in \m{range} T$, it follows that $T/(\m{range} T )(x+\m{range} T)=0$. Since $x+\m{range} T$ is choosed arbitrarily, we conclude that $T/(\m{range} T )=0$.

34. Solution: By definition, for any $x+\m{null} T\in V/(\m{null} T )$, we have $T/(\m{null} T )(x+\m{null} T)=Tx+\m{null} T.$Hence $T/(\m{null} T )$ is injective if and only if$Tx\in \m{null} T\iff x\in\m{null}T.$Note that$Tx\in \m{null} T\iff x\in\m{null}T$is equivalent to $\m{null} T\cap\m{range} T=\{0\}$. Because if we assume $Tx\in \m{null} T\iff x\in\m{null}T$, then for any $v\in\m{null} T\cap\m{range} T$, then there exists $u\in V$ such that $Tu=v$, hence $Tu\in \m{null} T$ implies $u\in\m{null}T$. That is $v=Tu=0$. The other direction is also true. Hence the proof is completing.

35. Solution: Suppose $\lambda\in\mb F$ is an eigenvalue of $T/U$, we need to show $\lambda$ is an eigenvalue of $T$. There exists a nonzero $x+U\in V/U$(i.e. $x\not\in U$) such that $(T/U)(x+U)=\lambda(x+U)\Longrightarrow Tx-\lambda x\in U.$If $\lambda$ is an eigenvalue of $T|_U$, then we are done. If $\lambda$ is not an eigenvalue of $T$, then $T|_U-\lambda I:U\to U$ is invertible by 5.6 (here use $\dim V<\infty$). Hence there exists a $y\in U$ such that $(T|_U-\lambda I)y=Tx-\lambda x\Longrightarrow Ty-\lambda y=Tx-\lambda x$since $Tx-\lambda x\in U$. Hence we have $T(x-y)=\lambda(x-y),$and $x-y\ne 0$ since $x\not\in U$ and $y\in U$. It follows that $\lambda$ is an eigenvalue of $T$.

36. Solution: In Problem 32, we showed $1=e^{0x}$, $e^x$, $e^{2x}$, $\cdots$ is linearly independent in the vector space of real-valued functions on $R$. Consider $V=\m{span}(1,e^x,e^{2x},\cdots)$ and $U=\m{span}(e^x,e^{2x},\cdots)$, then $U$ and $V$ are subspaces of the vector space of real-valued functions on $R$. Define $T\in\ca L(V)$ by $T(f)=e^xf$. Please check $T\in \ca L(V)$ and $U$ is invariant under $T$. Consider $T/U$, we have $(T/U)(1+U)=e^x+U=0.$Since $1\not\in U$, it follows that $0$ is an eigenvalue of $T/U$. However $0$ is not an eigenvalue of $T$. Otherwise suppose there exists a nonzero $f\in V$ such that $Tf=0$, then we have $e^xf=0$. Hence $f=0$ since $e^x\ne 0$ for any $x\in \R$. We get a contradiction.