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 \begin{equation}\label{5AP1}(v_1,2v_2,\cdots,v_n)=(\lambda v_1,\lambda v_2,\cdots,\lambda v_n).\end{equation} 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 \begin{equation}\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. \end{equation} Consider the complex conjugation of $(\ref{5A161})$, we have \begin{equation}\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 \end{equation} since $A_{i,j}\in\R$ for all $i,j=1,2,\cdots,n$. (why? consider components)Note that $(\ref{5A162})$ implies \begin{equation}\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. \end{equation}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 \begin{equation}\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), \end{equation}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.

## Marie

14 Aug 2020Can someone explain what the general method is to find the corresponding eigenvectors from eigenvalues like in Q.8. I understand how to solve for eigenvalue, but from there how to I solve for corresponding eigenvector? I only get the answer by doing some guess and check. In the case of lamba = 1, I know that for arbitrary (w,z) in F^2 we want to find (w,z) such that T(w,z)=(z,w)=1*(w,z), but how would we proceed to do so?

## Jonathan Sharir-Smith

12 Aug 2020In 16, the solution presented takes the complex conjugate of equation 2. I'm wondering how or why this can be done, seeing as we have not defined taking the complex conjugate of a vector (the basis vectors are present in equation 2). Thus, I'm wondering why this procedure is valid?

## Yuheng

14 Aug 2020Hi, I understand your doubt as the solution jumps a bit too fast. What the solution does is to focus on the coefficients of each ei in the equation Tv=lambda*v, and take conjugate of the coefficients. (I use parentheses to denote conjugate.) Specifically, given v=a1e1+...+anen, the coefficient of ei in lambda*v is lambda*ai, and coefficient of ei in Tv is a1Ai,1+...+anAi,n. Now we take conjugate of both sides to get (lambda)*(ai)=(a1)Ai,1+...+(an)Ai,n.(*) Now we let v'=(a1)e1+...+(an)en. The LHS of (*) is just coefficient of ei in (lambda)*v', and the RHS of (*) is just coefficient of ei in Tv'. Thus Tv'=(lambda)*v'.

## Jonathan Sharir-Smith

14 Aug 2020@Yuheng. Yup, I got you, and to be really deliberate we might even argue at each point (when we're jumping back and forth between the C^n and abstract vector space representations) that we're using the matrix "function" isomorphism. Mostly just suggesting this as a small tweak to the solution. Thank you again to the writer of these solutions, they are an invaluable tool!

## Mustafa Kemal Turak

24 Jun 2020Hi I want to give alternative proof of 24 b. Let Tx= Ax and considering standart basis thus M(T) = A then transpoze of M(T) equal to dual map matrix (3.114 ) also A^t ensure 24.a condition therefore 1 is eigenvalue of dual map of T. By 5.6 implies that null(T' - I)= null(T-I)' is nonempty . By 3.107 implies that dimnull(T-I)'=dimF^n - dim F^n + dimnull(T-I) = dimnull(T-I) hence dimnull(T-I) is not equal to 0 thus 5.6 implies 1 is eigenvalue of T

## Tom C

28 May 2020For number 11, shouldn't the Taylor series expansion of e^{\lambda x} also be an eigenvector, with corresponding eigenvalue of \lambda? It seems to me your argument is sound for finite-dimensional polynomials, but the exercise does not make the assumption that the deg p is finite. In the case of \lambda = 0, the series expansion reduces to a constant polynomial, so that part is still the same. But I think any real number is an eigenvalue.

## Linearity

28 May 2020The Taylor series of $e^x$ is not a polynomial. Check the definition of polynomials.

## Sam Tay

19 Jan 2019Just a note on 35, a simpler proof: given an eigenvalue c of T/U, note that T/U - cI is not surjective by 5.6 so there exists w + U in V/U such that (T/U - cI)(v + U) is not equal to w + U for all v in V. Equivalently (Tv - cv) + U is not equal to w + U for all v in V. Hence Tv - cv is not equal to w for all v in V and it follows quickly that T - cI is not surjective in V, so that c is an eigenvalue of T.

## Ling Min Hao

29 Jan 2018For Question 16, there is a simpler method. Suppose lambda is the eigenvalue of T. Then Tv=lambda.v . By taking conjugate at both side, this is solved.

## Ling Min Hao

28 Jan 2018For Question 10, since each eigenvector can form an invariant subspaces, we have n invariant subspaces. By question 4, we know the sum of invariant subspaces is still invariant. There are 2^n such possibilities (including zero subspace). Until here, we know it AT LEAST have 2^n invariant subspace. But we know from (a), the eigenvectors actually form basis of F^n.This means F^n=U_1+...+U_m (each U_j represents an eigenspace) and hence it can maximally have 2^n subspace. Therefore, there are 2^n invariant subspace in T where each invariant subspace, say W is actually an element of span(u_1,...,u_n).

To make it clear, one of the possibility of invariant subspace could be the invariant subspace U_1+U_2, where it can be written as a_1u_1+b_1u_2+0u_3+...+0u_n which is an element of span(u_1,...,u_n).

(I might skip a few logical steps here, hope you can understand what I meant as it's hard to type things without Latex.)

## Yuheng

31 Jul 2020The number of subspace is not 2^n. We can't determine the number of subspaces by decomposing V into direct sums. For example {(a1,a2,...,an) in Fn : a1+4a2=0} is a subspace (and obviously there are infinitely many subspaces).

## Yuheng

1 Aug 2020For 16, I don't understand how we take complex conjugate of something in V? In the proof we used the fact that conjugate of a1e1 = (conjugate of a1)*e1, where e1 is an element from V. But didn't we deduce the property of complex numbers from numbers of the form a+bi? Does Vectors in V also behave like complex numbers? But definition of complex vector space only says a*v is in V for a in C and v in V.