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 familiar formula \[1-x^n=(1-x)(1+x+\cdots+x^{n-1}).\]

2. Solution: Let $v$ be an eigenvector of $T$ corresponding to $\lambda$, then we have $Tv=\lambda v$. Similarly, we have \[T^2v=T(\lambda v)=\lambda^2 v\quad T^3v=T(\lambda^2 v)=\lambda^3 v,\]and $T^nv=\lambda^n v$ for $n\in \mb N^+$. This implies for any polynomial $p$, we have $p(T)v=p(\lambda)v$. Hence \[0=(T-2I)(T-3I)(T-4I)v=(\lambda-2)(\lambda-3)(\lambda-4)v.\]As $v\ne 0$, it follows that \[(\lambda-2)(\lambda-3)(\lambda-4)=0.\]Thus $\lambda=2$ or $\lambda=3$ or $\lambda=4$.

3. Solution: Note that for any $v\in V$, we have \begin{equation}\label{5BP31}v=\frac{1}{2}(v-Tv)+\frac{1}{2}(v+Tv).\end{equation} Since $T^2-I=0$, it follows that \[(T+I)\left(\frac{1}{2}(v-Tv)\right)=\frac{1}{2}(I-T^2)v=0.\]Hence $\frac{1}{2}(v-Tv)\in \m{null}(T+I)$.Similarly we have \[ \frac{1}{2}(v+Tv)\in \m{null}(T-I). \]By $(\ref{5BP31})$, it follows that $V=\m{null}(T-I)+\m{null}(T+I)$. However, $-1$ is not an eigenvalue of $T$. Hence $\m{null}(T+I)=\{0\}$. Thus $V=\m{null}(T-I)$. This implies $T=I$, since for any $v\in V$ we have $Tv-v=(T-I)v=0$.

4. Solution: Note that for any $v\in V$, we have \begin{equation}\label{5BP41} v=Pv+(v-Pv). \end{equation} It is clear that $Pv\in\m{range}P$. Since $P^2=P$, it follows that $P(v-Pv)=(P-P^2)v=0$. Thus $v-Pv\in\m{null}P$. By $(\ref{5BP41})$, we have $V=\m{null}P+ \m{range}P$. Suppose $v\in \m{null}P\cap \m{range}P$, then there exists $u\in V$ such that $Pu=v$. Moreover $Pv=0$. Hence\[0=Pv=P(Pu)=P^2u=Pu=v.\]This implies $\m{null}P\cap \m{range}P=\{0\}$. Therefore $V=\m{null}P\oplus \m{range}P$.

5. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 14.

6. Solution: Note that $TU\subset U$, one can easily deduce $T^n U\subset U$ for any $n\in\mb N^+$ by induction. Hence $\lambda T^n U\subset U$ for any $\lambda\in\mb F$ since $U$ is a vector space. If we assume for any $p\in\ca P(\mb F)$ with $\deg p\le n-1$, $U$ is invariant under $p(T)$. Then we will show $U$ is invariant under $q(T)$ for every polynomial $q\in\ca P(\mb F)$ with $\deg q=n$. Let $q=\sum_{k=0}^n a_kx^k$, then \begin{align*} q(T)U=&\left(\sum_{k=0}^n a_kT^k\right)U=\left(\sum_{k=0}^{n-1} a_kT^k\right)U+a_nT^nU\\ \subset&U+U=U. \end{align*} Here $U+U=U$ by Problem 15 of Exercise 1C. Hence $U$ is invariant under $q(T)$ for every polynomial $q\in\ca P(\mb F)$ with $\deg q=n$. By induction, we conclude $U$ is invariant under $p(T)$ for every polynomial $p\in\ca P(\mb F)$.

7. Solution: By problem 2, we have $T^2v=\lambda^2v$ if $v$ be an eigenvector of $T$ corresponding to $\lambda$. Hence $3$ or $-3$ is an eigenvalue of $T$, then $9$ is an eigenvalue of $T^2$ since $3^2=9$ and $(-3)^2=9$.

Conversely, if $9$ is an eigenvalue of $T^2$. It follows that $T^2-9I$ is not injective, namely $(T-3I)(T+3I)$ is not injective. By Problem 11 of Exercise 3B, we have $T-3I$ or $T+3I$ is not injective. Hence we conclude $3$ or $-3$ is an eigenvalue of $T$.

(My apologies for using 5.6, it is only true for finite-dimensional $V$.)

8. Solution: Denote $T\in \ca L(\mb \R^2)$ by \[T(x,y)=\left(\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}y,\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y\right).\]You can directly check that $T^4=-I$.

Here I use a fact that \[ \left( \begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{array} \right)^n=\left( \begin{array}{cc} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \\ \end{array} \right) \]

9. Solution: By definition, an eigenvalue of $T$ must be contained in $\mb F$, hence we should assume that every zero of $p$ is in $\mb F$. Let $\lambda$ be a zero of $p$, then by 4.11 we have $p(z)=(z-\lambda)q(z)$, where $q(z)\in \ca P(\mb F)$. Suppose $\lambda$ is not an eigenvalue of $T$, then $T-\lambda I$ is injective. Hence \[0=p(T)v=(T-\lambda I)q(T)v\]implies $q(T)v=0$. However $\deg q<\deg p$ and $q$ is nonzero(otherwise $p$ is zero). This contradicts with the choice of $p$. Thus every zero of $p$ is an eigenvalue of $T$.

10. Solution: By the proof of Problem 2, it follows that $T^nv=\lambda^n v$. Hence for $p\in P(\mb F)$, suppose \[p=\sum_{n=0}^ka_nx^n.\]Then \begin{align*} p(T)v=&\left(\sum_{n=0}^ka_nT^n\right)v =\sum_{n=0}^ka_nT^nv =\sum_{n=0}^ka_n\lambda^nv=p(\lambda)v. \end{align*}

11. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 15.

12. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 16.

13. Solution: Suppose a subspace $U$ of $W$ invariant under $T$ and $\dim U<\infty$. If $U\ne \{0\}$, then by 5.21, $T|_U$ has an eigenvalue with an eigenvector $v$($v\ne 0$). That is $T|_U(v)=\lambda v$, namely $Tv=\lambda v$. Note that $v\ne 0$, we conclude $T$ has an eigenvalue $\lambda$. We get a contradiction since $T\in\ca L(W)$ has no eigenvalues. Hence every subspace of $W$ invariant under $T$ is either $\{0\}$ or infinite-dimensional.

14. Solution: See Linear Algebra Done Right Solution Manual Chapter 5 Problem 18.

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

16. Solution: Define $\vp:\ca P_n(\C)\to V$ by $\vp(p)=p(T)v$, then $\vp$ is a linear map(check it). Note that $\dim \ca P_n(\C)=n+1$ and $\dim V=n$, it follows that $\vp$ is not injective by 3.23. Hence there exists a nonzero $p\in \ca P_n(\C)$ such that $p(T)v=0$. The remained is the same as 5.21.

17. Solution: Define $\vp:\ca P_{n^2}(\C)\to \ca L(V)$ by $\vp(p)=p(T)$, then $\vp$ is a linear map(check it). Note that $\dim \ca P_{n^2}(\C)=n^2+1$ and $\dim \ca L(V)=n^2$, it follows that $\vp$ is not injective by 3.23. Hence there exists a nonzero $p\in \ca P_n(\C)$ such that $p(T)=0$. This implies $p(T)v=0$. The remained is the same as 5.21.

18. Solution: Let $\lambda_0$ be an eigenvalue of $T$, then $T-\lambda_0I$ is not surjective by 5.6. Hence $\dim \m{range}(T-\lambda I)<\dim V$. Note that $T$ has only finitely many eigenvalues, there exist a sequence of number $\lambda_n$ such that \[ \lim_{n\to\infty}\lambda_n=\lambda_0 \]and $\lambda_n$ are not eigenvalues of $T$. Then by 5.6, $T-\lambda_nI$ is surjective. Hence\[\dim \m{range}(T-\lambda_nI)=\dim V.\]This implies $\lambda_n\to\lambda_0$, but \[f(\lambda_0)\ne\lim_{n\to\infty}f(\lambda_n).\]Thus $f$ is not a continuous function.

19. Solution: Note that $Tp(T)=p(T)T$, if $\{p(T): p\in\ca P(\mb F)\}=\ca L(V)$, then $ST = TS$ for every $S\in\ca L(V)$. By Problem 16 of Exercises 3D, it follows that $T$ is a scalar multiple of the identity. Suppose $T=\lambda I$, then \[ \{p(T): p\in\ca P(\mb F)\}=\{p(\lambda)I:p\in\ca P(\mb F)\}=\{\mu I:\mu\in\mb F\}. \]Since $\dim V>1$, we have \[\dim \ca L(V)=(\dim V)^2\]and\[\dim\{p(T): p\in\ca P(\mb F)\}=1.\]Hence\[\{p(T): p\in\ca P(\mb F)\}\ne\ca L(V).\]

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

## Sicheng Zhong

22 Jun 2021There's a slight error in the solution to Problem 10. It is not safe to write p(v)\lambda as a summation from 0 to n, as lambda could equal to 0 and 0^0 is not defined.

## Facundo Manuel Canale

5 Apr 2021Isn't it easier to prove that the map in exercise 4 is a bijection?

I think I found it easier to prove it that way, let me know if anyone is interested in this answer, as it is hard to type.

## S.H.Lee

9 Nov 2020I think 5.B.18 has a "tiny" error. If V={0}, then the value of f(\lambda) is always 0, the constant. So f is continuous...

## The Dark Knight

29 Apr 2020For Exercise 9, it seems that the hypothesis “ V is finite-dimensional ” is not needed.

It is not used in the proof, right ? Please check it . Thanks!

## Linearity

29 Apr 2020I think you are right.

## The Dark Knight

30 Apr 2020I do think so. Thanks for the confirmation!

## Brian Zhang

16 May 2019Something arguably simpler for 5b3: $$(T^2 - I) = 0 \Longrightarrow (T - I)(T + I) = 0$$ $(T + I)$ is invertible since $-1$ is not an eigenvalue, therefore $T-I = 0$, $T = I$.

Edited by Linearity: Maybe better argue it as follows. Because $$T^2 - I = 0 \Longrightarrow (T + I)(T - I) = 0$$Since $-1$ is not an eigenvalue, we conclude that $T + I$ is injective (not invertible because if $V$ is not finite-dimensional, this would be wrong). Therefore $T-I = 0$, $T = I$.

## Linearity

2 May 2020This requires $V$ to de finite-diemnsional.

## Keyang Zhou

3 May 2020Which step in this proof requires finite dimensionality?

## Linearity

3 May 2020That proof does not require but needs minor changes.See the remark with red color.

## Jonathan Sharir-Smith

14 Aug 2020The step where it is said that -1 not an eigenvalue implies $(T+I)$ invertible. That follows from 5.6 in the 3rd edition which requires the finite-dimensional hypothesis.

## Ying Kit Hui

4 Dec 2017For problem 19, I think in the following way. (I must admit that your solution is better.) For F=C, then by 5.27, there is a basis of V to which T has an upper triangular matrix. Denote this basis be v(1), ... , v(n). Note that n>1, since dimV>1. Thus span(v(1), ... , v(j)), is invariant under p(T) for any p. For any linear map S such that S maps v(1) to v(2), based on the fact that span(v(1)) is not invariant under S, we can conclude that S is not in the set containing all p(T). Thus the statement is proved.

For F=R, if T has eigenvalue, then proceeds with similar argument as above. However, if T has no eigenvalues, how can I proceed with this approach? Thx.

## Michael D. Nguyen

26 Jan 2019If T has no eigenvalues then it is trivially that the two sets are not equal. Because L(V) have some linear maps that have eigenvalues and the p(T) eigenvalues must always has the form p(lambda) for lambda being T's eigenvalue.

## Masacroso

8 Jun 2017For exercise 7: I think that you dont need to use 5.6, observe that if (T-L)v=0 for non-zero v then, regardless of the dimension of V, L is an eigenvalue of T. And is clear that T^2-9I=(T-3I)(T+3I)=(T+3I)(T-3I), hence or -3 or 3 is the eigenvalue of v, or -3 is the eigenvalue of (T-3I)v, or 3 is the eigenvalue of (T+3I)v.

And it seems that the exercise 11 is not correctly formulated. See here:

https://math.stackexchange.com/questions/2315342/incomplete-proof-concerning-eigenvalues

for a discussion about the matter.

## Mohammad Rashidi

28 Jul 2017Yes, thank you.

## Masacroso

8 Jun 2017The exercise 3.B.2 seems wrong, because it seems to says that if L is an eigenvalue of T then L must be 2,3 or 4, what is not necessarily true, T can have more eigenvalues than these described by the polynomial on T.

## Marcel Ackermann

12 Apr 2017In 5B3: "Since T²−I, it follows that" should be "Since T²−I=0, it follows that"

## ToeJam

12 Aug 2016In problem 19, the vector space under consideration is L(V), so we can't use the result in problem 3.D.

The result pertains to operators on the vector space, so we could use it for an operator W belonging to L(L(V))- not the elements of the vector space itself.

Or, put another way, this is similar to saying that u belonging to a vector space U is such that uv = vu for all v belonging to U. Correct me if I'm wrong, but I don't think you can use result 3.D in this context.

Also, T is an arbitrary operator on L(V), so it certainly isn't necessarily a scalar multiple of identity operator- and the result Tp(T) = p(T)T holds for any T belonging to L(V).

Edit: I'd also like to say- thank you for this website! It's been of huge help while working through Sheldon Axler's book.

## Mohammad Rashidi

28 Jul 2017I don't think you are correct. P(T) belongs to L(V), T also belongs to L(V).