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 \[\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n.\]
George
1 Jul 2021For number 15, isn't T surjective as well? Because if it isn't, then it cannot be injective (since it's an operator on a finite-dimensional space) , which means lambda = 0 is an eigenvalue, hence it would have 3 distinct eigenvalues, and thus 3 distinct eigenvectors, so it would be diagnolizable.
Allen
8 Feb 2021Seems there hasn't been any post on Q.4, so I just give it a try.
V is a vector space of infinite dimension, so any vector v in V has an expression of v = a1v1+a2v2+a3v3+...., where all a's belong to field F.
Define linear transformation T as T((a1,a2,a3, ...)) = (a2,a3,a4,...), i.e T shift the entry of v one entry forward.
If v belongs to null(T), then a2=a3=a4=...=0. Therefore, null(T) = (a1,0,0,0,...).
As for range(T), we can think of Tv as reindexing vector v, i.e think of a2 as new a1, and a3 as new a2, etc. Hence
V = range(T).
Combining the range and the null space of T, we get range(T) + null(T) = V (If U is a subspace of V, then U+V = V),
which satisfies the (b) of exercise 3. However, the intersection of range(T) and null(T) is (a1,0,0,...) not {0}.
The shifting of entries on finite-dimensional V does not satisfy range(T) + null(T) = V, because there is no vector that has
value at the last entry.
Carl
26 Jan 2021I'm not sure the solution to number 8 is correct. My line of reasoning followed yours; however, all we've shown is that at least one of (T-2I) and (T-6I) must be invertible. This does not rule out both are invertible so I don't think this satisfies the problem "Prove that T-2I or T-6I is invertible"
Rivers
31 Dec 2020I think i have a easier proof for Q1: you just have to prove that only 0 is in $$null T \cap range T$$, then by rank-nullity theorem they add to V. Since T is diagonalizable, there is a basis of V such that T is diagonal, for any $v \in V$ we have Tv = 0 implies v = 0 because the diagonal T just multiply a scalar to each coefficient of v represented by a linear combination of the basis.
(I'm not sure how you typeset equations here, my apologies)
Xinyu
30 Nov 2020Actually, a correct example should be T(x,y)=(-y,x)
Yu-Jen
7 Aug 2020My solution for exercise 10:
For any \(u \in E(\lambda_1,T) \oplus ... \oplus E(\lambda_m,T)\), we can write \(u = v_1 + ... + v_m\, v_i \in E(\lambda_i,T)). There exists \(v \in V, v = \frac{1}{\lambda_1}v_1 + ... + \frac{1}{\lambda_m}v_m\) such that \(Tv = u\). Thus, \(E(\lambda_1,T) \oplus ... \oplus E(\lambda_m,T) \subseteq range T\). Taking dimension on both side, we get \(dim E(\lambda_1,T) + ... + dim E(\lambda_m,T) \leq dim range T\)
zen
28 May 2020Warning! It may be incorrect, see here.
Thank you for your insightful website.
I am thinking about another way to prove the theorem in 5.C 5 - the 'if' direction. Here’s a draft.
1.Suppose T has eigenspace E (λ1,T), …,E(λi, T) in V (finite space over C) (at least 1 based on 5.21);
2. Let v1,..,vk be the basis of the direct sum of the eigenspaces. Extend it to a basis of V (dimV = n), v1,..vk,vk+1,...,vn, Suppose k<n, otherwise the prove is done.
Then any vector v in span(vk+1,...,vn) is not an eigenvector of T in V, in other words, for any λ，（T-λI）v ≠ 0, otherwise v is in span(v1,..vk).
What I want to prove is that this kind of ‘extra subspace’ cannot exist under the condition in the theorem.
3. span(vk+1,...,vn) is a complex vector space, denoted as U. thus operator T|U has an upper-triangular matrix with respect to some basis of U(5.27). let it be v’k+1,...,v’n. Then T|U(v’i) =
T(v’i)∈span(v’k+1,...,v’i) for i in k+1,..,n
4. v1,..,vk,v’k+1,..,v’n is a basis of V, and T has an upper-triangular matrix with regard to this new basis.
5. Suppose T(v’n) = ακ+1v’k+1 + …+αnv’n, let λ = αn, then null T-αnI + range T-αnI is a subset of span(v1,..,vk,v’k+1,..,v’n-1), which is contradictory. Thus, k = n, the basis of the direct sum of the eigenspaces of T is a basis of V. T is diagonalizable.
Zen
29 May 2020Opps, I found some errors.
The most significant one is that, I need to prove U is invariant under T so that T|U is indeed an operator. This is possible.
Let λ = 0, the direct sum of null T and range T is V. Let any vector u∈U, if u = 0, T(0) = 0∈U. For some u≠0, suppose T(u) = v0, and v0 ≠0（aforementioned reason） and v0∈span(v1,…,vk). Then v0,vk+1, …, vn are independent vectors. What I want to prove here is that this ‘extra vector’ must be absent under the condition in the theorem.
Null T is a subset of span (v1,..,vk), and span(v1,…,vk) is invariant under T. Thus, considering the restriction linear map T|U, for any nonzero vector u in U (also in V), u∈ range T|U, otherwise u∈ Null T or span (v1,…,vk) which is contradictory.
Thus span (v0, vk+1, …, vn) is a subset of range T|U, leading to dim null T|U + dim range T|U > dim U, which is impossible. Thus for any nonzero u in U, T(u)∈U. T|U is indeed an operator.
Zen
29 May 2020I realize I need to change some lines above (from ‘considering the restriction linear map’).
vk+1 = T(wk+1) + T(uk+1), since vk+1 is in range T. wi, ui are vectors in span(v1,..,vk), and span (vk+1,..,vn) respectively for i in k+1,..,n. Thus non-zero vector vi – T(wi)∈range T|U for i in k+1,..,n.Then, v0, vk+1 – T(wk+1),…,vn-T(wn) are independent, and dim range T|U > dim U.
Zen
29 May 2020I am sorry for the verbose comments above. The logic is not correct due to some inappropriate assumptions about the restriction linear map and v0. It can be deleted if possible, or it would be a funny wrong example.
Zen
5 Jul 2020Coming back from $ch.8 ~A-C$, here are some ideas about proving the 'if' direction (beyond the scope in this chaptor).
Since $V$ is a finite-dimensional complex vector space and $T\in\mathcal{L}(V).T$ has at least one eigenvalue.$(5.21)$.
Also,
$$
V= G(\lambda_1,T)\oplus\cdots\oplus G(\lambda_{m},T) \qquad (8.21~a)
$$
If $V= Null(T - \lambda I~)\oplus Range(T-\lambda I)$ for every $\lambda$, including all the distinct eigenvalue or eigenvalues $\lambda_1,\cdots,\lambda_m$ of $\mathit{T}$.
Then, for $\lambda_i ~in~ \lambda_1,\cdots,\lambda_m$ , $Null(T-\lambda_i I) \cap Range(T-\lambda_i I)=\{0\}$, which means $$Null(T-\lambda_i I)=Null(T-\lambda_i I)^2 =\cdots= Null(T-\lambda_i I)^n,$$ where $n=dim(V)$. Based on the definition of $G(\lambda,T)$ and $E(\lambda,T)$, thus $E(\lambda_i,T) = G(\lambda_i,T) $ , thus$$V = E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T) $$
$T$ is diagonalizable.$(5.41 e)$
Zen
5 Jul 2020Typos:
$1. Null(T-\lambda_iI) \cap Range(T-\lambda_iI)=\{0\}$
$2.Null(T-\lambda_iI)=Null(T-\lambda_iI)^2 =\cdots= Null(T-\lambda_iI)^n, n=dim(V) $
It could be better if the comment is editable.
Chi Yuan Lau
3 Aug 2019How about ex 7 ?$\lambda$ would only appears 1 time
Michael D. Nguyen
13 Feb 2019For question number for, an example would be the differentiate operator D over vector space P(F).
null D = {1} and range D = P(F).
So (b) holds but (a) and (c) don't.
Tim Nguyen
6 Jun 2018For question 1, since T is diagonalizable, doesn't that imply V is finite-dimensional?
Or is there a way to represent an operator in infinite-dimension as a matrix? The definition of a matrix of an operator (5.22) assumes basis of V as v_1,...v_n. Thank you!
Wu Jinyang
11 Aug 2017Hi. I wonder if you can post the answer to question number 2.
the converse of the statement seems to hold.
Currently, I am stuck on trying to prove all invertible T can be diagonalized. (which is very unlikely, or the book should have mentioned it).
Silly me, T(x,y) = (x,x+y) is probably the easiest example,
null T = (0,0) with range T = V and the only eigenvalue is 1 with eigenvector (0,1)
Still, I wonder if you can give me some advice on solving this type of question. It always took me forever searching for a counter example.
By the way, for question 1. equation 4. it should be an equal sign instead of a subset sign.
Mohammad Rashidi
11 Aug 2017Yes, your example is a counterexample. A diagonalizable operator (on finite-dimensional space) whoes eigenvalues are all 1 must be identity operator. Not all invertible operators are diagonalizable.
It is hard to give advice on that. But usually, it might be a counterexample.
Thank you for correcting me.
Michael D. Nguyen
12 Feb 2019Another example would be T(x,y)=(y,x). This is another invertible map and has no eigenvalues. And thus not diagonalizable.
Chi Yuan Lau
3 Aug 2019also,if $T^2=T$ can also be a counterexample, by ex4 in 5B, one can know that $T$ agrees the condition , however, not all the periodical operator can be diagonalized, because their eigenvalues could only be 0 or 1
Yu-Jen
7 Aug 2020T(x,y) = (y,x) actually has eigenvalues of 1 and -1, with corresponding eigenvector (1,1) and (1,-1).
Disintegrator
27 Nov 2020Let T(x,y) = (x, x+y). This has only one eigenvalue (namely, λ = 1) and one eigenvector, so T is not diagonalizable. But clearly, the null space of T is {0}, so Range T = V. In fact, any operator of the form T(x,y) = (x, bx+y) works as long as b != 0.
Masacroso
10 Jun 2017I think the exercise 14 can be shorter observing that if we choose a basis of C^3 x,y,z composed by two eigenvectors of T and other vector then it is enough to define T as Tx=6x, Ty=7y and Tz=w, where w is not a multiple of z.
By example T(1,0,0)=(6,0,0), T(0,1,0)=(0,7,0) and T(0,0,1)=(0,1,1) will work.
Batuhan Eroğlu
16 Apr 2017The solution for the 14th problem has an error. You state that there are two possiblities, namely k_1 =! 0 and k_2 =! 0. However (6−λ)k2=0 and (6−λ)k1=−k2 forces k_2 to be zero eitherway. So the only possibility is that k_1 =! 0 and k_2 =0
Mohammad Rashidi
22 Apr 2017I used "or", not "and".
Mickey Mike
16 Mar 2017Mild error:
For 5.C.1 equation (4) should be an equals sign, not a subset sign
Mohammad Rashidi
22 Apr 2017Thanks!