# Chapter 3 Exercise E

1. Solution: Suppose $T$ is linear. Let $(v_1, Tv_1), (v_2, Tv_2) \in \operatorname{graph of} T$. We have

\begin{aligned} (v_1, Tv_1) + (v_2, Tv_2) &= (v_1 + v_2, Tv_1 + Tv_2)\\ &= (v_1 + v_2, T(v_1 + v_2))\\ &\in \operatorname{graph of} T, \end{aligned}

where the second line follows because $T$ is linear and the third by definition of $\operatorname{graph of} T$. Hence $\operatorname{graph of} T$ is closed under addition. Similarly, it also closed under scalar multiplication. Therefore, $\operatorname{graph of} T$ is a subspace of $V \times W$.

Conversely, suppose $\operatorname{graph of} T$ is a subspace of $V \times W$. Let $v_1, v_2 \in V$. Then

$$(v_1, Tv_1), (v_2, Tv_2), (v_1 + v_2, T(v_1 + v_2)) \in \operatorname{graph of} T. \tag{1}$$

Since $\operatorname{graph of} T$ is a subspace of $V \times W$, adding the first two vectors above shows that

$$(v_1 + v_2, Tv_1 + Tv_2) \in \operatorname{graph of} T. \tag{2}$$

Because $T$ is function, if $(v, w), (v, \hat{w}) \in \operatorname{graph of} T$ then $w = \hat{w}$. This, together with $(1)$ and $(2)$, implies that $T(v_1 + v_2) = Tv_1 + Tv_2$. Hence $T$ satiesfies the additivity property of linear maps. Similarly, $T$ satisfies the homogeneity property. Therefore, $T$ is indeed a linear map.

2. Solution: Define $T: V_1 \times \dots \times V_m \to V_j$ by

$$T(v_1, \dots, v_m) = v_j.$$

$T$ is clearly linear and surjective. The Fundamental Theorem of Linear Maps (3.22) now implies that $V_j$ is finite-dimensional.

3. Solution: Let $V = U_1 = U_2 = \mathbb{F}^\infty$. Note that $U_1 + U_2 = \mathbb{F}^\infty$. Define $T: \mathbb{F}^\infty \times \mathbb{F}^\infty \to \mathbb{F}^\infty$ by

$$T\big((x_1, x_2, x_3, \dots), (y_1, y_2, y_3, \dots)\big) = (x_1, y_1, x_2, y_2, x_3, y_3, \dots).$$

Thus $T$ is obviously surjective. To prove injectivity, let

$$\big((x_1, x_2, x_3, \dots), (y_1, y_2, y_3, \dots)\big), \big((z_1, z_2, z_3, \dots), (w_1, w_2, w_3, \dots)\big) \in \mathbb{F}^\infty \times \mathbb{F}^\infty$$

such that

$$T\big((x_1, x_2, x_3, \dots), (y_1, y_2, y_3, \dots)\big) = T\big((z_1, z_2, z_3, \dots), (w_1, w_2, w_3, \dots)\big).$$

This implies that

$$(x_1, y_1, x_2, y_2, x_3, y_3, \dots) = (z_1, w_1, z_2, w_2, z_3, w_3).$$

Therefore $x_j = z_j$ and $y_j = w_j$ for each $j$. Hence

$$\big((x_1, x_2, x_3, \dots), (y_1, y_2, y_3, \dots)\big) = \big((z_1, z_2, z_3, \dots), (w_1, w_2, w_3, \dots)\big)$$

and so $T$ is injective. Thus $T$ is an isomorphism, as desired.

4. Solution: For any $f\in \ca L(V_1\times \cdots\times V_m,W)$ and given $i\in \{1,\cdots,m\}$, define $f_i:V_i\to W$ by the rule:$f_i(v_i)=f(0,\cdots,0,v_i,0,\cdots,0),$where $v_i$ sits in the $i$-th slot. One can check that $f_i\in\ca L(V_i,W)$.

Define $\vp :\ca L(V_1\times \cdots\times V_m,W)\to \ca L(V_1,W)\times \cdots \times \ca L(V_m,W)$ by$\vp(f)=(f_1,\cdots,f_m),$ where $f_1$, $\cdots$, $f_m$ are defined in the previous paragraph. Now we are going to check that $\vp$ is an isomorphism between $\ca L(V_1\times \cdots\times V_m,W)$ and $\ca L(V_1,W)\times \cdots \times \ca L(V_m,W)$ by construct an inverse map.

Define $\psi :\ca L(V_1,W)\times \cdots \times \ca L(V_m,W)\to \ca L(V_1\times \cdots\times V_m,W)$ by$\psi(f_1,\cdots,f_m)(v_1,\cdots,v_m)=f_1(v_1)+\cdots+f_m(v_m).$

Please check that $\psi$ and $\vp$ are linear. It is also not hard to see that $\psi\circ \vp=Id$, and $\vp\circ \psi=Id$. Hence we are done.

7. Solution: Note that $v+U=x+W$, hence $v=x+w_1$, where $w_1\in W$. It follows that $v-x\in W$. Hence for any $u\in U$, we have $v+u=x+w_2$for some $w_2\in W$ since $v+U=x+W$. Hence we conclude that $u=(x-v)+w_2=-w_1+w_2\in W,$ it follows that $U\subset W$ for $u$ is chosen arbitrarily. Similarly, we deduce that $W\subset U$. Thus $U=W$.

8. Solution: If $A$ is an affine subset of $V$, then there exist a vector $a\in V$ and a subspace $U$ of $V$ such that $A=a+U$. Then any $v,w\in A$ can be written as $v=a+u_1$ and $w=a+u_2$ for some $u_1,u_2\in U$. Hence $\lambda v+(1-\lambda)w=a+[\lambda u_1+(1-\lambda)u_2]\in a+U=A.$ Conversely, since $A$ is nonempty, let $a\in A$. We will show that $A-a=\{x-a:x\in A\}$is a subspace of $V$. For $x-a\in A-a$ and $\lambda\in \mb F$ where $x\in A$, then $\lambda x+(1-\lambda)a\in A\Longrightarrow \lambda (x-a)=\lambda x+(1-\lambda)a-a\in A-a.$This implies $A-a$ is closed under scalar multiplication. For $x-a$ and $y-a\in A-a$, where $x,y\in A$. We have $\frac{x}{2}+\frac{y}{2}-a\in A-a.$Note that $A-a$ is closed under scalar multiplication, it follows that $(x-a)+(y-a)=2\left(\frac{x}{2}+\frac{y}{2}-a\right)\in A-a.$That is $A-a$ is closed under addition. Hence $A-a$ is a subspace of $V$. Note that $A=a+(A-a)$, it follows that $A$ is an affine subset of $V$.

9. Solution: Suppose $A_1\cap A_2\ne \varnothing$, then for any $x,y\in A_1\cap A_2$ and $\lambda\in \mb F$, we have $\lambda x+(1-\lambda) y\in A_1$and $\lambda x+(1-\lambda) y\in A_2$by Problem 8 since $A_1$ and $A_2$ are affine subsets of $V$. Hence $\lambda x+(1-\lambda) y\in A_1\cap A_2.$Again by Problem 8, we deduce that $A_1\cap A_2$ is an affine subset of $V$.

10. Solution: It is the same as Problem 9.

11. Solution: (a) For $v=\lambda_1 v_1+\cdots+\lambda_m v_m\in A$ and $w=\eta_1 v_1+\cdots+\eta_m v_m\in A$,where $\lambda_1,\cdots,\lambda_m\in\mb F$, $\lambda_1+\cdots+\lambda_m=1$ and $\eta_1,\cdots,\eta_m\in\mb F$, $\eta_1+\cdots+\eta_m=1$. For any $\lambda\in\mb F$, we have \begin{align*} \lambda v+(1-\lambda) w=\sum_{i=1}^m(\lambda\lambda_i+(1-\lambda)\eta_i)v_i. \end{align*} Note that $\sum_{i=1}^m(\lambda\lambda_i+(1-\lambda)\eta_i)=\lambda\sum_{i=1}^m\lambda_i+(1-\lambda)\sum_{i=1}^n\eta_i=\lambda+(1-\lambda)=1,$we deduce that $\lambda v+(1-\lambda) w\in A$. Hence $A$ is an affine subset of $V$ by Problem 8.

(b) We will use induction to show that for any affine subset $W$ of $V$ that contains $v_1$, $\cdots$, v_mk\le m, then if \lambda_1+\cdots+\lambda_k=1, we have $\sum_{j=1}^k\lambda_jv_j\in W.$It is obvious for k=1 and k=2 by Problem 8. Suppose this is true for k, then we will show it for k+1 (k+1\le m). Assume \lambda_1+\cdots+\lambda_{k+1}=1. If \lambda_{k+1}=1, then $\sum_{j=1}^{k+1}\lambda_jv_j=v_{k+1}\in W.$If \lambda_{k+1}\ne 1, then $\frac{1}{1-\lambda_{k+1}}(\lambda_1+\cdots+\lambda_k)=1.$Hence by assumption, we deduce that $\frac{1}{1-\lambda_{k+1}}(\lambda_1v_1+\cdots+\lambda_kv_k)\in W.$By Problem 8, we have $(1-a_{k+1})\left(\frac{1}{1-\lambda_{k+1}}(\lambda_1v_1+\cdots+\lambda_kv_k)\right)+a_{k+1}v_{k+1}\in W,$namely $\lambda_1v_1+\cdots+\lambda_{k+1}v_{k+1}\in W.$By mathematical induction, we conclude if \lambda_1,\cdots,\lambda_m\in\mb F, \lambda_1+\cdots+\lambda_m=1, then $\lambda_1 v_1+\cdots+\lambda_m v_m\in W,$thus A\subset W. (c) Note that if \lambda_1+\cdots+\lambda_m=1, we have $\lambda_1 v_1+\cdots+\lambda_m v_m=v_1+\lambda_2(v_2-v_1)+\cdots+\lambda_m(v_m-v_1).$Hence A\subset v_1+\m{span}(v_2-v_1,\cdots,v_m-v_1). Similarly, for any$v\in v_1+\m{span}(v_2-v_1,\cdots,v_m-v_1),$v can be written as $v_1+\sum_{i=2}^m \lambda_i(v_i-v_1)=(1-\lambda_2-\cdots-\lambda_m)v_1+\sum_{i=2}^m \lambda_iv_i$for some \lambda_2, \cdots, \lambda_m\in\mb F. Note that $(1-\lambda_2-\cdots-\lambda_m)+\sum_{i=2}^m \lambda_i=1,$we deduce that v_1+\m{span}(v_2-v_1,\cdots,v_m-v_1)\subset A. Hence$A=v_1+\m{span}(v_2-v_1,\cdots,v_m-v_1).$Let v=v_1 and U=\m{span}(v_2-v_1,\cdots,v_m-v_1), then \dim U\le m-1. 12. Solution: Since V/U is finite-dimensional, we can suppose v_1+U, \cdots, v_n+U is a basis of V/U. Then for any v\in V, there exist a unique list of \lambda_1, \cdots, \lambda_n\in\mb F such that $v+U=\sum_{i=1}^n\lambda_i(v_i+U).$Then$v-\sum_{i=1}^n\lambda_iv_i\in U.$Define \vp:V\to U\times V/U by $\vp(v)=\left(v-\sum_{i=1}^n\lambda_iv_i,\sum_{i=1}^n\lambda_i(v_i+U)\right).$We first check that \vp is linear. For any v,w\in V, we $v+U=\sum_{i=1}^n\lambda_i(v_i+U)$and $w+U=\sum_{i=1}^n\eta_i(v_i+U).$for some \lambda_1, \cdots, \lambda_n\in\mb F and \eta_1, \cdots, \eta_n\in\mb F. Then for any \lambda \in \mb F, we have $(v+\lambda w)+U=(v+U)+\lambda(w+U)=\sum_{i=1}^n(\lambda_i+\lambda\eta_i)(v_i+U).$Hence $\vp(v+\lambda w)=\left(v+\lambda w-\sum_{i=1}^n(\lambda_i+\lambda\eta_i)v_i,\sum_{i=1}^n(\lambda_i+\lambda\eta_i)(v_i+U)\right),$and $\vp(v)=\left(v-\sum_{i=1}^n\lambda_iv_i,\sum_{i=1}^n\lambda_i(v_i+U)\right),$and $\vp(w)=\left(w-\sum_{i=1}^n\eta_iv_i,\sum_{i=1}^n\eta_i(v_i+U)\right).$Therefore we have (check it) \begin{align*} \vp(v)+\lambda\vp( w)=\vp(v+\lambda w), \end{align*} namely \vp is linear. Injectivity: if \vp(v)=0, then \lambda_1=\cdots=\lambda_n=0(defined as above) and $0=v-\sum_{i=1}^n\lambda_iv_i=v.$ Surjectivity: For \left(u,\sum_{i=1}^n\xi(v_i+U)\right)\in U\times V/U, it is easy to see $\vp\left(u+\sum_{i=1}^n\xi v_i\right)=\left(u,\sum_{i=1}^n\xi(v_i+U)\right).$Hence \vp is an isomorphism, namely V is isomorphic to U\times V/U. 13. Solution: For any v\in V, since v_1+U, \cdots, v_m+U is a basis of V/U, there exist \lambda_1, \cdots, \lambda_m\in\mb F such that $v+U=\sum_{i=1}^m \lambda_i(v_i+U).$This implies $v-\sum_{i=1}^m\lambda_iv_i\in U.$Note that u_1, \cdots, u_n is a basis of U, it follows that $v-\sum_{i=1}^m\lambda_iv_i=\sum_{j=1}^n\eta_ju_j$for some \eta_1, \cdots, \eta_n\in \mb F. Hence $v=\sum_{i=1}^m\lambda_iv_i+\sum_{j=1}^n\eta_ju_j,$this implies V=\m{span}(v_1,\cdots,v_m,u_1,\cdots,u_n) since v is chosen arbitrarily. Hence it suffices to show that v_1, \cdots, v_m, u_1, \cdots, u_n is linearly independent. Suppose for some \lambda_1, \cdots, \lambda_m\in\mb F and \eta_1, \cdots, \eta_n\in \mb F, we have $\sum_{i=1}^m\lambda_iv_i+\sum_{j=1}^n\eta_ju_j=0.$Then $\sum_{i=1}^m\lambda_i(v_i+U)=0,$hence \lambda_1=\cdots=\lambda_m=0 since v_1+U, \cdots, v_m+U is a basis of V/U. It follows that $\sum_{j=1}^n\eta_ju_j=0.$Note that u_1, \cdots, u_n is a basis of U, we obtain that \eta_1=\cdots=\eta_n=0. Hence v_1, \cdots, v_m, u_1, \cdots, u_n is linearly independent. 15. Solution: Note that \varphi:V\to \mb F and \varphi\ne 0, there exists v\in V such that \varphi(v)\ne 0. Then for any a\in\mb F, we have$\vp\Big(\frac{av}{\varphi(v)}\Big)=\frac{a}{\varphi(v)}\vp(v)=a.$Therefore \mathrm{range}\,\vp=\mb F. By 3.91(d), we have V/\mathrm{null}\,\vp\cong \mb F. Hence\dim V/\mathrm{null}\,\vp=\dim \mb F=1.$$16. Solution: Since \dim V/U=1, there exists a vector v\notin U such that v+U is a basis of V/U. Define \theta: V/U\to \mb F by $\theta(kv+U)=k,$then \theta\in\ca L(V/U,\mb F). Let \pi be the quotient map \pi :V\to V/U. Let \vp=\theta\circ\pi, then \vp\in\ca L(V,\mb F). Now we check that \mathrm{null}\,\vp=U. Let w\in V such that \vp(w)=0, i.e. \theta(\pi(w))=0. By the definition of \theta, we have$$w+U=0v+U.$$Hence w\in U, which implies that \mathrm{null}\,\vp\subset U. On the other hand, let u\in U, then \pi(u)=u+U=0+U. Hence \theta\circ \pi(u)=0, which implies that U\subset \mathrm{null}\,\vp. Therefore \mathrm{null}\,\vp=U. (Please note that such \vp is not unique.) 17. Solution: Since V/U is finite-dimensional, let n=\dim (V/U), then there exist w_1,\cdots,w_n\in V such that w_1+U, \cdots, w_n+U form a basis of V/U. It is easy to see that w_1,\cdots,w_n is linearly independent otherwise w_1+U, \cdots, w_n+U would be linearly dependent. Consider the subspace W of V spanned by w_1,\cdots,w_n, then we have \dim W=n=\dim(V/U). Now we are going to show that V=U\oplus W. For any v\in V, since w_1+U, \cdots, w_n+U is a basis of V/U, there exist k_1,\cdots,k_n\in\mb F such that$$v+U=\sum_{i=1}^n k_i(w_i+U).$$This implies that$$v-\sum_{i=1}^n k_iw_i\in U.$$Therefore consider $v=\Big(v-\sum_{i=1}^n k_iw_i\Big)+\sum_{i=1}^n k_iw_i,$it tells us that v\in U+W. Since v is chosen arbitrarily, we conclude that V\subset U+W. To show that V=U\oplus W, it suffices to show that U\cap W=0. Suppose v\in U\cap W, then there exist k_1,\cdots,k_n\in\mb F such that$$v=\sum_{i=1}^n k_iw_i.$$Hence we have 0=\pi(v)=\sum_{i=1}^n k_i(w_i+U), where \pi is the canonical quotient map from V onto V/U. However, note that w_1+U, \cdots, w_n+U is a basis of V/U we have k_1=\cdots=k_n=0. Thus$$v=\sum_{i=1}^n k_iw_i=0.$$This shows that U\cap W=0. 18. Solution: Suppose there exists S\in\ca L(V/U,W) such that T=S\circ\pi. Since for any u\in U we have \pi u=0, hence$$Tu=(S\circ \pi )u=S(\pi u)=S0=0.$$Hence u\in \mathrm{null}\, T. Note that this is true for any u\in U, we conclude that U\subset \mathrm{null}\, T. Conversely, suppose U\subset \mathrm{null}\, T. Define a map S: V/U\to W by S(v+U)=Tv. We have to check that this map is well-defined (since there may exist different v_1,v_2\in V such that v_1+U=v_2+U and we need to check that Tv_1=Tv_2). For v_1,v_2\in V such that v_1+U=v_2+U, we have v_1-v_2\in U. Note that U\subset \mathrm{null}\, T, it follows that T(v_1-v_2)=0, hence we have Tv_1=Tv_2. This implies that S is well-defined. Note that T\in \ca L(V,W), it is easy to see that S\in \ca L(V/U,W). By definition, we have$$S\circ \pi (v)=S(v+U)=Tv$$for any v\in V. This implies that T=S\circ \pi. 19. Solution: For any given finite set A, denote by |A| the number of elements in A. Let S_i, i=1,\cdots,n, be finite sets. The union \bigcup_{i=1}^n S_i is a disjoint union if and only if$$|S_1|+\cdots+|S_n|=\Big|\bigcup_{i=1}^n S_i\Big|.$$20. Solution: (a) To show that \Gamma is linear, we need to show that for any S_1,S_2\in\ca L(V/U,W) and k_1,k_2\in\mb F we have$$\Gamma(k_1S_1+k_2S_2)=k_1\Gamma(S_1)+k_2\Gamma(S_2).This follows directly since \begin{align*}\Gamma(k_1S_1+k_2S_2)=&(k_1S_1+k_2S_2)\circ \pi\\=& k_1S_1\circ \pi+k_2S_2\circ \pi\\=& k_1\Gamma(S_1)+k_2\Gamma(S_2).\end{align*}(b) To show that\Gamma$is injective, one has to show that if$\Gamma(S)=0$then$S=0$. If$\Gamma(S)=0$, it implies that$S\circ \pi=0$. Hence$(S\circ \pi) v=0$for all$v\in V$. Thus$S(\pi(v))=S(v+U)=0$for all$v\in V$. We conclude that$S=0$, which shows$\Gamma\$ is injective.

(c) By Problem 18, we have\begin{align*}T\in \mathrm{range}\, \Gamma\iff & \text{there exists }S\in \ca L(V/U,W)\text{ s.t. }T=S\circ \pi\\ \iff & U\subset \mathrm{null}\, T\quad \text{ by Problem 18}\\ \iff &T\in\ca L(V,W): \, Tu=0 \text{ for every } u\in U.\end{align*}This completes the proof.

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