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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.