Chapter 5 Exercise C

1. Solution: It is not said $V$ is finite-dimensional, but I will do it by assuming $\dim V<\infty$.

If $T$ is invertible, then $\m{null}{T}=0$ and $\m{range} T=V$ since $T$ is bijective and surjective. Hence $V=\m{null} T \oplus\m{range} T$.

If $T$ is not invertible, let $0$, $\lambda_1$, $\cdots$, $\lambda_m$ be all eigenvalues of $T$, where $\lambda_i\ne 0$ for $i=1,\cdots,m$. Then by 5.41(d), we have \begin{equation}\label{5CP11} V=E(0,T)\oplus E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T). \end{equation} By definition, it follows $E(0,T)=\m{null} T $. Moreover, for any $v_i\in E(\lambda_i,T)$, \[T\left(\frac{1}{\lambda_i}v_i\right)=\frac{1}{\lambda_i}Tv_i=v_i.\]This implies $E(\lambda_i,T)\subset \m{range} T$. Therefore \begin{equation}\label{5CP12} E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T)\subset \m{range} T.\end{equation} On the other hand, any $v\in V$ can be written as \[v=v_0+v_1+\cdots+v_m,\]where $v_0\in E(0,T)$ and $v_i\in E(\lambda_i,T)$ for $i=1,\cdots,m$. Hence \[T(v)=T(v_0+v_1+\cdots+v_m)=\lambda_1v_1+\cdots+\lambda_mv_m\in E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T).\]This implies \begin{equation}\label{5CP13} \m{range} T\subset E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T). \end{equation} By $(\ref{5CP12})$ and $(\ref{5CP13})$, we have \begin{equation}\label{5CP14} E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T)=\m{range} T.\end{equation}Combining $(\ref{5CP11})$ and $(\ref{5CP14})$, it follows that $V=\m{null} T \oplus\m{range} T$.

If we can show something like $(\ref{5CP11})$ for infinite-dimensional vector spaces, then we can deduce this problem for infinite-dimensional case by using similar arguments.

3. Solution: (a) $\Longrightarrow$ (b): It is obvious.

(b) $\Longrightarrow$ (c): By 3.22, we have \begin{equation}\label{5CP3.1}\dim V=\dim\m{null}T+ \dim\m{range}T.\end{equation} Note that $V=\m{null}T+ \m{range}T$ and 2.43, we have \begin{equation}\label{5CP3.2} \dim V=\dim\m{null}T+ \dim\m{range}T-\dim(\m{null}T\cap \m{range}T). \end{equation} By $(\ref{5CP3.1})$ and $(\ref{5CP3.2})$, we have $\dim(\m{null}T\cap \m{range}T)=0$. Hence $\m{null}T\cap \m{range}T=\{0\}$.

(c) $\Longrightarrow$ (a): Again by 2.43 and 3.22, we have \[\dim (\m{null}T+ \m{range}T)=\dim\m{null}T+ \dim\m{range}T-\dim(\m{null}T\cap \m{range}T). \] and \[\dim V=\dim\m{null}T+ \dim\m{range}T.\] As $\dim(\m{null}T\cap \m{range}T)=0$, it follows that \[ \dim (\m{null}T+ \m{range}T)=\dim V. \]Hence $\m{null}T+ \m{range}T=V$, thus $\m{null}T\oplus \m{range}T=V$ since $\m{null}T\cap \m{range}T=\{0\}$.

5.Solution: If $T$ is diagonalizable, so is $T-\lambda I$. Hence by problem 1, we have\[V=\m{null}(T-\lambda I)\oplus \m{range}(T-\lambda I)\] for every $\lambda\in\C$.

Lemma 1: Let $U,W,S$ be subspaces of $V$, if $V=U\oplus W$ and $U\subset S$, then $S=U\oplus (W\cap S)$.

Proof of Lemma 1: For any $s\in S$, $s$ can be written as $u+w$, where $u\in U$ and $w\in W$, since $V=U\oplus W$. As $U\subset S$, it follows that $u\in S$. Hence $w=s-u\in S$, namely $w\in s\cap W$. This implies $S=U+ (W\cap S)$. Note that $U\cap W=\{0\}$, we have $S=U\oplus (W\cap S)$.

Lemma 2: For $a,b\in\C$ such that $a\ne b$, we have $\m{null}(T-a I)\subset\m{range}(T-b I)$.

Proof of Lemma 2: For any $v\in \m{null}(T-a I)$, we have $Tv=av$. Hence \[ (T-bI)\left(\frac{1}{a-b}v\right)=v, \]namely $v\in \m{range}(T-b I)$. Thus $\m{null}(T-a I)\subset\m{range}(T-b I)$.

Proof of Problem: Conversely, since $V$ is finite-dimensional, $T$ has only finitely many eigenvalues. Suppose $\lambda_1$, $\cdots$, $\lambda_m$ are all distinct eigenvalues of $T$. Note that we have \[ V=\m{null}(T-\lambda_1 I)\oplus \m{range}(T-\lambda_1 I) \]and $\m{null}(T-\lambda_2 I)\subset \m{range}(T-\lambda_1 I)$ (by Lemma 2), we have \[ \m{range}(T-\lambda_1 I)=\m{null}(T-\lambda_2 I)\oplus \m{range}(T-\lambda_1 I)\cap\m{range}(T-\lambda_2 I) \]by Lemma 1. Similarly, we also have \[ \m{null}(T-\lambda_3 I)\subset \m{range}(T-\lambda_1 I)\cap\m{range}(T-\lambda_2 I). \]By using Lemma 1 and Lemma 2 inductively, we have \begin{align*} V=&\m{null}(T-\lambda_1 I)\oplus\cdots\oplus \m{null}(T-\lambda_m I)\oplus \\ &(\m{range}(T-\lambda_1 I)\cap \cdots\cap \m{range}(T-\lambda_m I)). \end{align*} If $\m{range}(T-\lambda_1 I)\cap \cdots\cap \m{range}(T-\lambda_m I)=\{0\}$, we showed\[V=\m{null}(T-\lambda_1 I)\oplus\cdots\oplus \m{null}(T-\lambda_m I).\]Hence $T$ is diagonalizable. If $\Gamma=\m{range}(T-\lambda_1 I)\cap \cdots\cap \m{range}(T-\lambda_m I)\ne\{0\}$, then note that $(T-\lambda_i I)T=T(T-\lambda_i I)$, we have $\m{range}(T-\lambda_i I)$ is invariant under $T$ for all $i=1,\cdots,m$ by Problem 3 of Exercises 5A. Hence $\Gamma$ is invariant under $T$ by Problem 5 of Exercises 5A. Consider $T|_{\Gamma}$, it has an eigenvalue $\lambda\in\C$ with a corresponding eigenvector $\mu$ by 5.21. Hence $\lambda$ is also an eigenvalue of $T$. Suppose $\lambda=\lambda_i$ for some $i\in\{1,\cdots,m\}$. Then $\mu\in \m{null}(T-\lambda_i I)$, $\mu\in\Gamma$ and $\mu\ne 0$, this contradicts with\begin{align*} V=&\m{null}(T-\lambda_1 I)\oplus\cdots\oplus \m{null}(T-\lambda_m I)\oplus \\ &(\m{range}(T-\lambda_1 I)\cap \cdots\cap \m{range}(T-\lambda_m I)). \end{align*}Hence $\Gamma=0$ and therefore $T$ is diagonalizable.

6. Solution: Since $T\in\ca L(V)$ has $\dim V$ distinct eigenvalues, then $T$ is diagonalizable by 5.44. Let $v_1$, $\cdots$, $v_{\dim V}$ be the basis of $V$ defined in the proof of 5.44, then $v_1$, $\cdots$, $v_{\dim V}$ are eigenvectors of $T$. As $S\in\ca L(V)$ has the same eigenvectors as $T$, $v_1$, $\cdots$, $v_{\dim V}$ are eigenvectors of $S$. Hence there exists $\lambda_1$, $\cdots$, $\lambda_{\dim V}\in\mb F$ and $\theta_1$, $\cdots$, $\theta_{\dim V}\in\mb F$ such that \[Tv_i=\lambda_i v_i\text{ and }Sv_i=\theta_iv_i,\quad i=1,\cdots,\dim V.\]Hence we have \[STv_i=S(\lambda_iv_i)=\lambda_iSv_i=\lambda_i\theta_iv_i,\quad i=1,\cdots,\dim V\] and \[ TSv_i=T(\theta_iv_i)=\theta_iTv_i=\theta_i\lambda_iv_i,\quad i=1,\cdots,\dim V. \]It follows that $STv_i=TSv_i$ for $i=1,\cdots,\dim V$. Note that $v_1$, $\cdots$, $v_{\dim V}$ is a basis of $V$, we deduce that $ST=TS$.

8. Solution: Suppose $T-2I$ and $T-6I$ are not invertible, then $2$ and $6$ are eigenvalues of $T$. Note that $\lambda$ is an eigenvalue of $T$ if and only if $E(\lambda, T)\ne \{0\}$. Hence $\dim E(2,T)\ge 1$ and $\dim E(6,T)\ge 1$. By 5.38, we have \[ 4+1+1\le\dim E(8,T)+\dim E(2,T)+\dim E(6,T)\le \dim (\mb F^5)=5. \]This is impossible. Hence $T-2I$ or $T-6I$ is invertible.

9. Solution: For every $\lambda\in\mb F$ with $\lambda\ne 0$, let $v\in E(\lambda,T)$. Then we have $Tv=\lambda v$. Note that $T$ is invertible and $\lambda\ne 0$, it follows that $\frac{1}{\lambda}v=T^{-1}v$. Hence $v\in E(1/\lambda,T^{-1})$, we conclude $E(\lambda,T)\subset E(1/\lambda,T^{-1})$. By symmetry, we also have $E(1/\lambda,T^{-1})\subset E(\lambda,T)$. To sum up, we deduce $E(\lambda,T)=E(1/\lambda,T^{-1})$ for every $\lambda\in\mb F$ with $\lambda\ne 0$.

12. Solution: Note that $R$ and $T$ has three eigenvalues and $\dim(\mb F^3)=3$. By 5.44, we have $R$ and $T$ are diagonalizable. Hence there exist bases $e_1,e_2,e_3$ and $\xi_1,\xi_2,\xi_3$ such that \[Te_1=2e_1,Te_2=6e_2,Te_3=7e_3\]and \[R\xi_1=2\xi_1,R\xi_2=6\xi_2,R\xi_3=7\xi_3.\]Define $S\in\ca L(\mb F^3)$ by \[S\xi_i=e_i,\quad i=1,2,3.\]Then we have $S$ is invertible and $S^{-1}e_i=\xi_i$. Moreover, \[ S^{-1}TS\xi_1=S^{-1}Te_1=S^{-1}(2e_1)=2\xi_1=R\xi_1. \]Similarly $S^{-1}TS\xi_2=R\xi_2$ and $S^{-1}TS\xi_3=R\xi_3$. Hence $R=S^{-1}TS$ as they coincide in the basis $\xi_1,\xi_2,\xi_3$.

13. Solution: Let $e_1$, $e_2$, $e_3$, $e_4$ be a basis of $\mb F^4$, define $R,T\in\ca L(\mb F^4)$ by \[Re_1=2e_1,Re_2=2e_2,Re_3=6e_3,Re_4=7e_4\]and \[Te_1=2e_1,Te_2=2e_2+e_1,Te_3=6e_3,Te_4=7e_4.\]Then $R$ is diagonalizable. In fact $T$ is not diagonalizable since $\dim E(2,T)=1$, $\dim(6,T)=1$ and $\dim(7,T)=1$ imply \[ \dim(2,T)+\dim(6,T)+\dim(7,T)<\dim(\mb F^4). \]If there exist an invertible operator $S\in\ca L(\mb F^4)$ such that $R=S^{-1}TS$, $\iff SRS^{-1}=T$, then $Se_1$, $Se_2$, $Se_3$, $Se_4$ is a basis of $\mb F^4$. Moreover, \[ T(Se_1)=SRS^{-1}(Se_1)=SRe_1=S(2e_1)=2Se_1. \]Similarly, \[ T(Se_2)=2Se_2,T(Se_3)=6Se_3,T(Se_4)=7Se_4. \]This implies $T$ is diagonalizable. Thus we get a contradiction. Hence there does not exist an invertible operator $S\in\ca L(\mb F^4)$ such that $R=S^{-1}TS$.

14. Solution: Let $T\in \ca L(\C)$ defined by \[Te_1=6e_1,Te_2=6e_2+e_1,Te_3=7e_3,\]where $e_1$, $e_2$, $e_3$ is a basis of $\C^3$. Then for any nonzero $\alpha\in \C^3$, write $\alpha$ by $k_1e_1+k_2e_2+k_3e_3$, if there exists $\lambda\in\C$ such that $T\alpha=\lambda\alpha$, then we have \begin{equation}\label{5CP14.1} \lambda(k_1e_1+k_2e_2+k_3e_3)=T\alpha=(6k_1+k_2)e_1+6k_2e_2+7k_3e_3. \end{equation} If $k_3\ne 0$, then we have $\lambda k_3=7k_3$ by the previous equation. Hence $\lambda=7$. If $k_3=0$, we have \[ (6-\lambda)k_2=0\text{ and }(6-\lambda)k_1=-k_2. \]Note that $\alpha\ne 0$, it follows that $k_1$ or $k_2$ is not zero. If $k_2\ne 0$, then $\lambda=6$. If $k_1\ne 0$, then \[ (6-\lambda)^2k_1=(6-\lambda)(-k_2)=-(6-\lambda)k_2=0. \]Thus $\lambda =6$.

By above, all the eigenvalues of $T$ are $6$ and $7$. Moreover, $\dim E(6,T)=1$ and $\dim E(7,T)=1$ by solving $(\ref{5CP14.1})$. That imples \[2=\dim E(6,T)+\dim E(7,T)<\dim \C^3=3.\]Thus $T$ is not diagonalizable.

15. Solution: Since $T$ does not have a diagonal matrix with respect to any basis of $\C^3$, $T$ is not diagonalizable. Hence $8$ is not an eigenvalue of $T$, otherwise $T$ has $3$ eigenvalues hence diagonalizable by 5.44. This implies $T-8I$ is surjective by 5.6. Hence there exists $(x,y, z)\in\mb C^3$ such that \[(T-8I)(x,y, z)=(17,\sqrt{5},2\pi),\]namely\[ T(x,y,z)=(17+8x,\sqrt{5}+8y,2\pi+ 8z).\]

16. Solution: (a) We show this part by induction. Note that \[T(0,1)=(1,1)=(F_1,F_2).\]Hence it is true for $n=1$. Suppose we have $T^n(0,1)=(F_n,F_{n+1})$ then \begin{align*} T^{n+1}(0,1)=&T(T^n(0,1))=T(F_{n},F_{n+1})\\=&(F_{n+1},F_n+F_{n+1})=(F_{n+1},F_{n+2}). \end{align*}Hence if it is true for $n$, so is the case for $n+1$. Thus we get the conclusion by induction.

(b) and (c) We need solve this equation \[T(x,y)=\lambda(x,y),\]where $(x,y)\ne (0,0)$ and $\lambda\in \R$. By definition of $T$, it is equivalent to \[ \left\{ \begin{array}{ll} \lambda x=y,\\ \lambda y=x+y. \end{array} \right. \]If $y=0$, then $x=0$ by the second equation. Hence $y\ne 0$, it follows that $x\ne 0$ and $\lambda\ne 0$. By the first equation, we have $x/y=1/\lambda$. By the second one we have $\lambda=x/y+1$. Hence we have \[\lambda=\frac{1}{\lambda}+1,\]the solutions for this equation are \[\lambda=\frac{1\pm\sqrt{5}}{2}.\]By $x/y=1/\lambda$, we have the eigenvectors corresponding to $\dfrac{1\pm\sqrt{5}}{2}$ are $\left(1,\dfrac{1\pm\sqrt{5}}{2}\right)$ respectively.

(d) Denote $\left(1,\dfrac{1+\sqrt{5}}{2}\right)$ and $\left(1,\dfrac{1-\sqrt{5}}{2}\right)$ by $e_1$ and $e_2$ respectively. Then we have \[(0,1)=\frac{1}{\sqrt{5}}(e_1-e_2).\]It follows that \begin{align*} T^n(0,1)=&T^n\left(\frac{1}{\sqrt{5}}(e_1-e_2)\right)=\frac{1}{\sqrt{5}}(T^ne_1-T^ne_2)\\ =&\frac{1}{\sqrt{5}}\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^ne_1-\left(\dfrac{1-\sqrt{5}}{2}\right)^ne_2\right]. \end{align*} By (a) and comparing the first component, we deduce that \begin{equation}\label{5CP161}F_n= \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]. \end{equation}(e) Note that $\sqrt{5}\geqslant 2$, we have \begin{equation}\label{5CP162} \frac{1}{\sqrt{5}}\left|\frac{1-\sqrt{5}}{2}\right|^n= \frac{1}{\sqrt{5}}\left|\frac{2}{1+\sqrt{5}}\right|^n\le\frac{1}{2}\times\frac{2}{3}<\frac{1}{2}. \end{equation} Moreover, $F_n\in\mb Z$ is easily shown by induction. Combining $(\ref{5CP161})$ and $(\ref{5CP162})$, it follows that \[ \left|\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n-F_n\right|=\frac{1}{\sqrt{5}}\left|\frac{1-\sqrt{5}}{2}\right|^n<\frac{1}{2} \]By $F_n\in\mb Z$, we deduce that the Fibonacci number $F_n$ is the integer that is closest to \[\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n.\]

About Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This entry was posted in Chapter 5 and tagged .