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_m$$k\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.