If you find any mistakes, please make a comment! Thank you.

## Chapter 3 Exercise E

Exercises 1,2 and 4. For Problem 2, please also see Carson Rogers’s comment.

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.

6. (By YiweiJiang2015) To prove that $V^n$ and $\mathcal{L}(\mathbf{F}^n, V)$ are isomorphic vector spaces, we need to establish a bijective linear map between them.

Let’s define the map $\Phi: V^n \rightarrow \mathcal{L}(\mathbf{F}^n, V)$ as follows:

For any vector $\mathbf{v} = (v_1, v_2, \ldots, v_n) \in V^n$, we define $\Phi(\mathbf{v}): \mathbf{F}^n \rightarrow V$ as
$\Phi(\mathbf{v})(\mathbf{f}) = v_1 \mathbf{f}_1 + v_2 \mathbf{f}_2 + \ldots + v_n \mathbf{f}_n,$
where $\mathbf{f} = (\mathbf{f}_1, \mathbf{f}_2, \ldots, \mathbf{f}_n) \in \mathbf{F}^n, \mathbf{f}_j \in \mathbf{F}$.

Now, let’s show that $\Phi$ is a well-defined linear map and establish its properties:

Linearity:

Addition preservation: For any $\mathbf{v}_1, \mathbf{v}_2 \in V^n$, and $\mathbf{f} \in \mathbf{F}^n$, we have:
\begin{align*}
\Phi(\mathbf{v}_1 + \mathbf{v}_2)(\mathbf{f}) &= (v_{1,1} + v_{2,1})\mathbf{f}_1 + (v_{1,2} + v_{2,2})\mathbf{f}_2 + \ldots + (v_{1,n} + v_{2,n})\mathbf{f}_n \\
&= (v_{1,1}\mathbf{f}_1 + v_{1,2}\mathbf{f}_2 + \ldots + v_{1,n}\mathbf{f}_n) + (v_{2,1}\mathbf{f}_1 + v_{2,2}\mathbf{f}_2 + \ldots + v_{2,n}\mathbf{f}_n) \\
&= \Phi(\mathbf{v}_1)(\mathbf{f}) + \Phi(\mathbf{v}_2)(\mathbf{f})
\end{align*}
Thus, addition is preserved under $\Phi$.

Scalar multiplication preservation: For any scalar $c \in \mathbf{F}$, vector $\mathbf{v} \in V^n$, and $\mathbf{f} \in \mathbf{F}^n$, we have:
\begin{align*}
\Phi(c\mathbf{v})(\mathbf{f}) &= (cv_{1,1})\mathbf{f}_1 + (cv_{1,2})\mathbf{f}_2 + \ldots + (cv_{1,n})\mathbf{f}_n \\
&= c(v_{1,1}\mathbf{f}_1 + v_{1,2}\mathbf{f}_2 + \ldots + v_{1,n}\mathbf{f}_n) \\
&= c\Phi(\mathbf{v})(\mathbf{f})
\end{align*}
Thus, scalar multiplication is preserved under $\Phi$.

Injectivity: To show that $\Phi$ is injective, we need to prove that if $\Phi(\mathbf{v}_1) = \Phi(\mathbf{v}_2)$, then $\mathbf{v}_1 = \mathbf{v}_2$.

Suppose $\Phi(\mathbf{v}_1) = \Phi(\mathbf{v}_2)$ for some $\mathbf{v}_1, \mathbf{v}_2 \in V^n$. Let $\mathbf{f} = (\mathbf{f}_1, \mathbf{f}_2, \ldots, \mathbf{f}_n) \in \mathbf{F}^n$. Then we have:
\begin{align*}
\Phi(\mathbf{v}_1)(\mathbf{f}) &= \Phi(\mathbf{v}_2)(\mathbf{f}) \\
v_{1,1}\mathbf{f}_1 + v_{1,2}\mathbf{f}_2 + \ldots + v_{1,n}\mathbf{f}_n &= v_{2,1}\mathbf{f}_1 + v_{2,2}\mathbf{f}_2 + \ldots + v_{2,n}\mathbf{f}_n
\end{align*}

Since this equation holds for any $\mathbf{f} \in \mathbf{F}^n$, it must hold for the standard basis vectors of $\mathbf{F}^n$. Thus, we have:
\begin{align*}
v_{1,1} &= v_{2,1} \\
v_{1,2} &= v_{2,2} \\
&\vdots \\
v_{1,n} &= v_{2,n}
\end{align*}

Therefore, we conclude that $\mathbf{v}_1 = \mathbf{v}_2$, and hence, $\Phi$ is injective.

Surjectivity: To show that $\Phi$ is surjective, we need to prove that for any $\mathcal{T} \in \mathcal{L}(\mathbf{F}^n, V)$, there exists $\mathbf{v} \in V^n$ such that $\Phi(\mathbf{v}) = \mathcal{T}$.

Let $\mathcal{T} \in \mathcal{L}(\mathbf{F}^n, V)$ be an arbitrary linear map. We define $\mathbf{v} = (\mathcal{T}(\mathbf{e}_1), \mathcal{T}(\mathbf{e}_2), \ldots, \mathcal{T}(\mathbf{e}_n)) \in V^n$, where $\mathbf{e}_i$ is the $i$-th standard basis vector of $\mathbf{F}^n$. Now, let’s show that $\Phi(\mathbf{v}) = \mathcal{T}$.

For any $\mathbf{f} = (\mathbf{f}_1, \mathbf{f}_2, \ldots, \mathbf{f}_n) \in \mathbf{F}^n$, we have:
\begin{align*}
\Phi(\mathbf{v})(\mathbf{f}) &= \Phi(\mathcal{T}(\mathbf{e}_1), \mathcal{T}(\mathbf{e}_2), \ldots, \mathcal{T}(\mathbf{e}_n))(\mathbf{f}) \\
&= \mathcal{T}(\mathbf{e}_1)\mathbf{f}_1 + \mathcal{T}(\mathbf{e}_2)\mathbf{f}_2 + \ldots + \mathcal{T}(\mathbf{e}_n)\mathbf{f}_n \\
&= \mathcal{T}(\mathbf{e}_1\mathbf{f}_1 + \mathbf{e}_2\mathbf{f}_2 + \ldots + \mathbf{e}_n\mathbf{f}_n) \\
&= \mathcal{T}(\mathbf{f})
\end{align*}
The third equality holds because $\mathcal{T}$ is a linear map.
Hence, we have $\Phi(\mathbf{v}) = \mathcal{T}$, which implies that $\Phi$ is surjective.

Since $\Phi$ is injective and surjective, it is a bijective linear map between $V^n$ and $\mathcal{L}(\mathbf{F}^n, V)$. Therefore, $V^n$ and $\mathcal{L}(\mathbf{F}^n, V)$ are isomorphic vector spaces.

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. 14. Solution: a) Simply note that multiplying a vector u \in U by a scalar doesn’t increase the number of it’s non-zero entries, and that given vectors u_1 \in U with M_1 non-zero entries and u_2 \in U with M_2 non-zero entries, the sum u_1 + u_2 has at most M_1 + M_2 non-zero entries. b) Throughout this exercise, given a vector v \in \boldsymbol{F}^{\infty}, we’ll denote by v(p) the number on the p-th entry of v. Consider the list of vectors e_m for m=1,2,\cdots in \boldsymbol{F}^{\infty} defined by the rule: e_m(p) = \cases{1, & if(p-1) \mod m = 0$\cr 0, & otherwise } $$For example,$$e_1 = (1,1,1,1,1,1,…)e_2 = (1,0,1,0,1,0,…)e_3 = (1,0,0,1,0,0,…)…$$We want to show that this list gives a corresponding linear independent list on \boldsymbol{F}^{\infty}/U. Let k_1,k_2,…k_m \in \boldsymbol{F} such that:$$k_1(e_1 + U) + k_2(e_2 + U) + … + k_m(e_m + U) = 0 + U$$Note that k_1(e_1 + U) + … + k_m(e_m + U) = (k_1 e_1+ … + k_m e_m) + U, so the above equality implies that k_1 e_1+ … + k_m e_m = u for some u \in U. Since u has a finite number of non-zero elements, it must contain only zeroes from some position onward. Let L be the last position of u that might contain a non-zero entry, that is, u(i) = 0 for any i > L. Now, let p = t \cdot m! + 1, where we choose t so p \geq L . Our choice of p imples that e_1(p) = e_2(p) = … = e_m(p) = 1 and that u(q) = 0 for any q > p. Now we proceed to show that k_1 = k_2 = … = k_m = 0, proving the desired linear independence. Note that e_1(p+1) = 1 while e_2(p+1) = e_3(p+1) = … = e_m(p+1) = 0, which means u(p+1) = k_1 e_1(p+1) + … + k_m e_m(p+1) = k_1. Since u(p+1) = 0, this implies k_1 = 0. Similarly, we have e_2(p+2) = 1 while e_3(p+2) = e_4(p+2) = … = e_m(p+2) = 0, and since k_1 = 0, u(p+2) = k_2, so k_2 = 0. Following the induction, it’s possible to show the desired result that k_1 = … = k_m = 0. We constructed a list of vectors in \boldsymbol{F}^{\infty}/U of arbitrary size that is linearly independent (namely, the list e_1 + U,…,e_m + U). This shows that \boldsymbol{F}^{\infty}/U is infinite dimensioned, since no finite l.i. list can span the whole vector space. 15. Solution: Note that \varphi:V\to \mb F and \varphi\ne 0, there exists v\in V such that \varphi(v)\ne 0. Then for any a\in\mb F, we have$\vp\Big(\frac{av}{\varphi(v)}\Big)=\frac{a}{\varphi(v)}\vp(v)=a.$Therefore \mathrm{range}\,\vp=\mb F. By 3.91(d), we have V/\mathrm{null}\,\vp\cong \mb F. Hence$$\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.