Fibers are free modules of finite rank over SymbolicRing (actually vector spaces of finite dimension over the vector bundle field (K ), here (K= RR )): sage: Ep. Basering Symbolic Ring sage: Ep. Category Category of finite dimensional vector spaces over Symbolic Ring sage: Ep. Rank 2 sage: dim ( Ep ) 2. Let V be a finite-dimensional vector space, and let U and W be subspaces of V. True or False: If V = U+W, then dim V dimU + dim W. True False Get more help from Chegg Get 1:1 help now from expert Algebra tutors Solve it with our algebra problem solver and calculator.
1. Suppose $Tinca L(U, V)$ and $Sinca L(V, W)$ are both invertible linear maps. Prove that $STinca L(U, W)$ / is invertible and that $(ST)^{-1}=T^{-1}S^{-1}$.
Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 22. It is almost the same.
2. Suppose $V$ is finite-dimensional and $dim V > 1$. Prove that the set of noninvertible operators on $V$ is not a subspace of $ca L(V) $.
Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 25.
3. Suppose $V$ is finite-dimensional, $U$ is a subspace of $V$, and $Sinca L(U,V)$. Prove there exists an invertible operator $Tinca L(V)$ such that $Tu=Su$ for every $uin U$ if and only if $S$ is injective.
Solution: If there exists an invertible operator $Tinca L(V)$ such that $Tu=Su$ for every $uin U$, then $S$ is injective since $T$ is injective.
If $S$ is injective. Assume $u_1$, $cdots$, $u_m$ is a basis of $U$, we can extend it to a basis of $V$ as $u_1$, $cdots$, $u_m$, $v_{m+1}$, $cdots$, $v_{n}$. Since $S$ is injective, by Problem 9 of Exercises 3B, we have $Su_1$, $cdots$, $Su_m$ is linearly independent in $V$. Hence we can extend it to a basis of $V$ as $Su_1$, $cdots$, $Su_m$, $w_{m+1}$, $cdots$, $w_{n}$. Define $Tin ca L(V)$ as below [Tu_i=Su_i,quad Tv_{j}=w_j, quad 1le ile m, m+1le j le n.]The existence of $T$ is guaranteed by 3.5(unique). Then for any $u=a_1u_1+cdots+a_mu_m$, $a_iinmb F$, we have begin{align*} Tu=&T(a_1u_1+cdots+a_mu_m) =&a_1Tu_1+cdots+a_mTu_m =&a_1Su_1+cdots+a_mSu_m =&S(a_1u_1+cdots+a_mu_m)=Su. end{align*}Moreover, $T$ is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.
Compare this problem with Problem 11 of Exercises 3A.
Infinite Dimensional Vector Space
4. Suppose $W$ is finite-dimensional and $T_1, T_2inca L(V,W)$. Prove that null $T_1$= null $T_2$ if and only if there exists an invertible operator $Sinca L(W)$ such that $T_1=ST_2$.
Solution: If we assume $m{null} T_1 =m{null} T_2$. Since $W$ is finite-dimensional, so is range $T_2$. Let $w_1$, $cdots$, $w_n$ be a basis of $m{range} T_2$, then there exist $v_1$, $cdots$, $v_nin V$ such that [ T_2v_i=w_i,quad i=1,cdots,n. ]Now we will show that $V=m{null}T_2oplus m{span}(v_1,cdots,v_n)$. For any $vin V$, we [ T_2v=a_1w_1+cdots+a_nw_n ]for some $a_1$, $cdots$, $a_ninmb F$. Hence [ T_2(v-a_1v_1-cdots-a_nv_n)=0, ]namely [ v=(v-a_1v_1-cdots-a_nv_n)+(a_1v_1+cdots+a_nv_n), ]this implies $V=m{null}T_2 +m{span}(v_1,cdots,v_n)$. Moreover, if $a_1v_1+cdots+a_nv_nin m{null}T_2$, then we have [ T_2(a_1v_1+cdots+a_nv_n)=a_1w_1+cdots+a_nw_n=0. ]Note that $w_1$, $cdots$, $w_n$ is linearly independent, it follows $a_1=cdots=a_n=0$. Thus we have[V=m{null}T_2 oplusm{span}(v_1,cdots,v_n).] Similarly, $T_1v_1$, $cdots$, $T_1v_n$ is linearly independent. For if $a_1T_1v_1+cdots+a_nT_1v_n=0$, we have [ T_1(a_1v_1+cdots+a_nv_n)=0. ]Note that $m{null} T_1 =m{null} T_2$, it follows that [ 0=T_2(a_1v_1+cdots+a_nv_n)=a_1w_1+cdots+a_nw_n.]Thus $a_1=cdots=a_n=0$. Now extend $w_1$, $cdots$, $w_n$ to a basis of $W$ as $w_1$, $cdots$, $w_n$, $e_1$, $cdots$, $e_m$ and $T_1v_1$, $cdots$, $T_1v_n$ to a basis of $W$ as $T_1v_1$, $cdots$, $T_1v_n$, $f_1$, $cdots$, $f_m$. Define $Sinca L(W)$ by [Sw_i=T_1v_i,Se_j=f_j,i=1,cdots,n;j=1,cdots,m.]Note that [ V=m{null}T_2oplus m{span}(v_1,cdots,v_n), ]any $vin V$ can be expressed as [v=v_{m{null}}+a_1v_1+cdots+a_nv_n,]where $v_{m{null}}inm{null}T_2=m{null}T_1$ and $a_1$, $cdots$, $a_ninmb F$. Hence we have begin{align*} ST_2(v)=&ST_2(v_{m{null}}+a_1v_1+cdots+a_nv_n) =&ST_2(a_1v_1+cdots+a_nv_n) =&S(a_1w_1+cdots+a_nw_n) =&a_1T_1v_1+cdots+a_nT_1v_n =&T_1(a_1v_1+cdots+a_nv_n) =&T_1(v_{m{null}}+a_1v_1+cdots+a_nv_n)=T_1(v) end{align*}namely $ST_2=T_1$. Moreover, $S$ is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.
If there exists an invertible operator $Sinca L(W)$ such that $ST_2=T_1$, then for any $muinm{null} T_1$, we have [ST_2mu=T_1mu=0.]As $S$ is invertible, we have $T_2mu=0$. Hence $muin m{null} T_2$, it follows that $m{null} T_1 subset m{null} T_2$. Similarly, consider $T_2=S^{-1}T_1$, $m{null} T_2 subset m{null} T_1$. Thus we conclude null $T_1$= null $T_2$.
Compare this problem with Problem 24 of Exercises 3B.
5. Suppose $V$ is finite-dimensional and $T_1,T_2in ca L(V,W)$. Prove that range $T_1=$ range $T_2$ if and only if there exists an invertible operator $Sinca L(V) $ such that $T_1=T_2S$.
Solution: If we assume $m{range} T_1 = m{range} T_2$. Let $u_1$, $cdots$, $u_m$ be a basis of null $T_1$, then we can extend it to a basis of $V$ as $u_1$, $cdots$, $u_m$, $w_1$, $cdots$, $w_n$. Then range $T_1$ is $m{span}(T_1w_1,cdots,T_1w_n)$ and $T_1w_1,cdots,T_1w_n$ is linearly independent. There exist $v_1$, $cdots$, $v_nin V$ such that $T_1w_i=T_2v_i$ for $i=1,cdots,n$ since $m{range} T_1=m{range} T_2$. As $T_1w_1,cdots,T_1w_n$ is linearly independent, it follows that $v_1$, $cdots$, $v_n$ is linearly independent by Problem 4 of Exercises 3A. Note that range $T_1=$ range $T_2$ implies null $T_1$ and null $T_2$ have the same dimension. Let $zeta_1$, $cdots$, $zeta_m$ be a basis of null $T_2$, then $zeta_1$, $cdots$, $zeta_m$, $v_1$, $cdots$, $v_n$ is a basis of $V$ by the proof of 3.22. Define $Sinca L(V) $ by $Su_i=zeta_i$ and $Sw_j=v_j$, then we have [T_1w_j=T_2v_j=T_2Sw_j,quad j=1,cdots,n,]and [T_1u_i=0=T_2zeta_i=T_2Su_i,quad i =1,cdots, m,]hence $T_1=T_2S$ by uniqueness in 3.5. Moreover, $S$ is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.
If there exists an invertible operator $Sinca L(V) $ such that $T_1=T_2S$, then for any $muin V$, we have [T_1mu=T_2Smuin m{range} T_2.]Hence $m{range} T_1 subset m{range} T_2$. As $S$ is invertible, we have $T_2=T_1S^{-1}$. Similarly, we conclude $m{range} T_1 subset m{range} T_2$. Thus range $T_1=$ range $T_2$.
Compare this problem with Problem 25 of Exercises 3B.
6. Suppose $V$ and $W$ are finite-dimensional and $T_1,T_2inca L(V,W)$. Prove that there exist invertible operators $Rinca L(V)$ and $Sinca L(W)$ such that $T_1=ST_2R$ if and only if $dim$ null $T_1= dim$ null $T_2$.
Solution: If there exist invertible operators $Rinca L(V)$ and $Sinca L(W)$ such that $T_1=ST_2R$, then $S^{-1}T_1=T_2R$. Hence null $T_1$= null $T_2R$ by Problem 4. Note that we have range $T_2R$= range $T_2$ by Problem 5, it follows that [dimm{null} T_2R=dim V-dimm{range} T_2R=dim V-dimm{range} T_2=dimm{null} T_2.]Hence $dimm{null} T_1=dimm{null} T_2$.
Conversely, if $dim$ null $T_1= dim$ null $T_2$. Let $u_1$, $cdots$, $u_m$ be a basis of null $T_1$, then we can extend it to a basis of $V$ as $u_1$, $cdots$, $u_m$, $w_1$, $cdots$, $w_n$. Let $v_1$, $cdots$, $v_m$ be a basis of null $T_2$, then we can extend it to a basis of $V$ as $v_1$, $cdots$, $v_m$, $zeta_1$, $cdots$, $zeta_n$. Then $T_1w_1$, $cdots$, $T_1w_n$ is linearly independent in $W$, hence we can extend it to a basis of $V$ as $T_1w_1$, $cdots$, $T_1w_n$, $alpha_1$, $cdots$, $alpha_l$. Similarly, $T_2zeta_1$, $cdots$, $T_2zeta_n$ is linearly independent in $W$, hence we can extend it to a basis of $V$ as $T_2zeta_1$, $cdots$, $T_2zeta_n$, $beta_1$, $cdots$, $beta_l$. Define $Rinca L(V)$ by [Ru_i=v_i,Rw_j=zeta_j,i=1,cdots,m;j=1,cdots,n.]Define $Sinca L(W)$ by [ST_2zeta_j=T_1w_j,Sbeta_k=alpha_k,j=1,cdots,n;k=1,cdots,l.]Since $S$ and $T$ map basis to basis, hence $S$ and $T$ are invertible(surjective). Moreover, it is easy to check $T_1u_i=0=ST_2Ru_i$ and [ T_1w_j=ST_2zeta_j=ST_2Rw_j, ]hence $T_1=ST_2R$.
Compare this problem with Problem 4 and Problem 5.
7. Suppose $V$ and $W$ are finite-dimensional. Let $vin V$. Let [E={T inca L(V,W): T v= 0}.] (a) Show that $E$ is a subspace of $ca L(V,W)$. (b) Suppose $vne 0$. What is $dim E$?
Solution: a) Let $T,Sin E$, then [(T+S)v=Tv+Sv=0+0=0,]hence $T+Sin E$, namely $E$ is closed under addition. For every $lambdain mb F$, [(lambda T)v=lambda(Tv)=lambda 0=0,]hence $lambda Tin E$, namely $E$ is closed under scalar multiplication. Therefore $E$ is a subspace of $ca L(V,W)$.
(b) Since $vne 0$, we can extend it to a basis of $V$, namely $v$, $v_2$, $cdots$, $v_n$. Let $w_1$, $cdots$, $w_m$ be a basis of $W$. Under these bases, we have a isometric between $ca L(V,W)$ and $mb F^{m,n}$ by 3.60. Moreover, $Tv=0$ if and only if the first column vector of $ca M(T)$ is zero. Hence $dim E$ is exactly all matrices in $mb F^{m,n}$ such the first column vector is zero. Now , it is easily seen that[dim E=m(n-1)=dim W(dim V-1).]Or it is equivalent to the dimension all linear map from $m{span}(v_2,cdots,v_n)$ to $W$. Then we can use 3.61.
8. Suppose $V$ is finite-dimensional and $T: V to W$ is a surjective linear map of $V$ onto $W$. Prove that there is a subspace $U$ of $V$ such that $T|_U$ is an isomorphism of $U$ onto $W$. (Here $T|_U$ means the function $T$ restricted to $U$. In other words, $T|_U$ is the function whose domain is $U$, with $T|_U$ defined by $T|_U(u)=Tu$ for every $uin U$.)
Solution: Let $w_1$, $cdots$, $w_n$ be a basis of $W$. Since $T$ is surjective, there exist $v_1$, $cdots$, $v_n$ such that $Tv_i=w_i$. Moreover, by Problem 4 of Exercises 3A, it follows that $v_1$, $cdots$, $v_n$ is linearly independent. Consider $U=m{span}(v_1,cdots,v_n)$, then $T|_U$ maps a basis of $U$ to a basis of $W$. Hence $T|_U$ is an isomorphism of $U$ onto $W$.
9. Suppose $V$ is finite-dimensional and $S, T inca L(V)$. Prove that $ST$ is invertible if and only if both $S$ and $T$ are invertible.
Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 22.
10. Suppose $V$ is finite-dimensional and $S, T inca L(V)$. Prove that $ST=I$ if and only if $TS=I$.
Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 23.
11. Suppose $V$ is finite-dimensional and $S, T ,Uinca L(V)$ and $ST U = I$. Show that $T$ is invertible and that $T^{-1}=US$.
Solution: By Problem 9, we have $TU$ is invertible. Again by Problem 9, we have $T$ is invertible. Multiply both side of $STU=I$ by $S^{-1}$ on left, we get $TU=S^{-1}$. Multiply both side of $TU=S^{-1}$ by $S$ on right, we have $TUS=I$. Multiply both side of $TUS=I$ by $T^{-1}$ on left, we get $US=T^{-1}$.
12. Show that the result in the previous exercise can fail without the hypothesis that $V$ is finite-dimensional.
Solution: Consider $V=C^infty$. Define $T$, $S$, $U$ by [ T(z_1,z_2,z_3,cdots)=(0,z_1,z_2,z_3,cdots) ][ S(z_1,z_2,z_3,cdots)=(z_2,z_3,z_4,cdots) ]and $U=I$. Then $STU=I$. However $T$ is not surjective.
13. Suppose $V$ is a finite-dimensional vector space and $R,S, Tinca L(V)$ are such that $RST$ is surjective. Prove that $S$ is injective.
Solution: Since $V$ is finite-dimensional, the surjectivity of $RST$ implies $RST$ is invertible. Hence $S$ is invertible by Problem 11.
14. Suppose $v_1$, $cdots$, $v_n$ is a basis of $V$. Prove that the map $T: Vto mb F^{n,1}$ defined by [Tv=ca M(v)] is an isomorphism of $V$ onto $mb F^{n,1}$, here $ca M(v)$ is the matrix of $v in V$ with respect to the basis $v_1$, $cdots$, $v_n$.
Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 20.
Finite Dimensional Vector Spaces Solutions
15. Prove that every linear map from $mb F^{n,1}$ to $mb F^{m,1}$ is given by a matrix multiplication. In other words, prove that if $Tinca L(mb F^{n,1},mb F^{m,1})$, then there exists an $m$-by-$n$ matrix $A$ such that $Tx=Ax$ for every $xin mb F^{n,1}$.
Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 21.
16. Suppose $V$ is finite-dimensional and $Tin ca L(V)$. Prove that $T$ is a scalar multiple of the identity if and only if $ST=TS$ for every $Sinca L(V)$.
Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 24.
17. Suppose $V$ is finite-dimensional and $E$ is a subspace of $ca L(V)$ such that $ST in E$ and $TS in E$ for all $Sin ca L(V)$ and all $T in E$. Prove that $E={0}$ or $E=ca L(V)$.
Solution: Let $e_1$, $cdots$, $e_n$ be a basis of $V$. Suppose $Ene {0}$, then there is nonzero $Tin E$. This implies there is some $sin{1,cdots,n}$ such that $Te_sne 0$. Let [Te_s=a_1e_1+cdots+a_ne_n,quad a_jinmb F,j=1,cdots,n.]Since $Te_sne 0$, there is some $tin{1,cdots,n}$ such that $a_tne 0$. Define $E_{ij}in ca L(V)$ by $E_{ij}e_k=delta_{ik}e_j$, where $delta_{ik}=0$ if $ine k$ and $delta_{ii}=1$ for every $i$.
By definition, for every $iin{1,cdots,n}$, we have [ E_{ti}TE_{is}e_j=a_tdelta_{ij}e_i,quad j=1,cdots,n. ]By assumption, $E_{ti}TE_{is}in E$. Note that $E$ is a subspace of $ca L(V)$. It follows that [ (E_{t1}TE_{1s}+cdots+E_{tn}TE_{ns})e_j=a_te_j,quad j=1,cdots,n, ]and $(E_{t1}TE_{1s}+cdots+E_{tn}TE_{ns})in E$. This implies $a_jIin E$, hence $Iin E$. Therefore for any $Sin ca L(V)$, $S=SIin E$, namely $E=ca L(V)$.
18. Show that $V$ and $ca L(mb F, V)$ are isomorphic vector spaces.
Solution: For given $vin V$, define $vp_v:mb Fto V$ by $vp_v(lambda)=lambda v$. Then $vp_vin ca L(mb F, V)$ (check it). Hence we can define $vp: Vto ca L(mb F, V)$ by $vmapsto vp_v$. It suffices to show $vp$ is an isomorphic from $V$ to $ca L(mb F, V)$. First, we should check $vp$ is linear. For every $v_1,v_2in V$ and $lambda,ainmb F$, we have begin{align*} vp_{v_1+lambda v_2}(a)=&a(v_1+lambda v_2)=av_1+lambda av_2=&vp_{v_1}(a)+(lambdavp_{v_2})(a)=(vp_{v_1}+lambdavp_{v_2})(a). end{align*}Hence $vp$ is linear. Then $vp_vequiv 0$ imples $av=0$ for every $ainmb F$, thus $v=0$. We conclude injectivity. For any $fin ca L(mb F, V)$, if $f(1)=v$ then $f(lambda)=lambda v=vp_v(lambda)$. Hence every $fin ca L(mb F, V)$ can be express as $vp_{f(1)}$, namely $vp$ is surjective.
19. Suppose $Tinca L(ca P(R))$ is such that $T$ is injective and $deg Tple deg p$ for every nonzero polynomial $pinca P(R)$.
(a) Prove that $T$ is surjective.
(b) Prove that $deg Tp= deg p$ for every nonzero $pinca P(R)$.
Solution: (a) Since $deg Tple deg p$ for every nonzero polynomial $pinca P(R)$, we have $T|_{ca P_n(R)}:ca P_n(R)to ca P_n(R)$ for any $nin mb N^+$. Note that $T$ is injective, hence $T|_{ca P_n(R)}:ca P_n(R)to ca P_n(R)$ is injective. Note that $ca P_n(R)$ is finite-dimensional, we conclude $T|_{ca P_n(R)}:ca P_n(R)to ca P_n(R)$ is surjective by 3.56. Hence $T$ is surjective as any polynomial must be contained in some $ca P_n(R)$.
(b) We argue it by induction on $deg p$. It is true for $deg =0$. Suppose it is true for $n$. As $T|_{ca P_n(R)}:ca P_n(R)to ca P_n(R)$ is surjective, every polynomial $p$ with degree $le n$ can be attained by $Tq$ for some $qin ca P_n(R)$. Moreover, $T|_{ca P_{n+1}(R)}:ca P_{n+1}(R)to ca P_{n+1}(R)$ is surjective, every polynomial $p$ with degree $le n+1$ can be attained by $Tq$ for some $qin ca P_{n+1}(R)$. If there exist $pin ca P_{n+1}(R)$ with degree $n+1$ such that $deg Tp
20. Suppose $n$ is a positive integer and $A_{i,j}inmb F$ for $i, j = 1,cdots,n$. Prove that the following are equivalent (note that in both parts below, the number of equations equals the number of variables):
(a) The trivial solution $x_1=cdots=x_n=0$ is the only solution to the homogeneous system of equations [sum_{k=1}^nA_{1,k}x_k=0] [cdots] [sum_{k=1}^nA_{n,k}x_k=0.](b) For every $c_1,cdots,c_ninmb F$, there exists a solution to the system of equations[sum_{k=1}^nA_{1,k}x_k=c_1] [cdots] [sum_{k=1}^nA_{n,k}x_k=c_n.] Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 26.
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