--- **Problem 1:** Suppose $b, c \in \mathbf{R}$. Define $T : \mathbf{R}^3 \to \mathbf{R}^2$ by $T(x, y, z) = (2x - 4y + 3z + b,\ 6x + cxyz).$ Show that $T$ is linear if and only if $b = c = 0$. **($\Rightarrow$):** Suppose $b=c=0$. Then $T(x,y,z)=(2x-4y+3z,6x)$. Let $u=(u_{1},u_{2},u_{3})\in \mathbb{R}^{3}$ and $v=(v_{1},v_{2},v_{3})\in \mathbb{R}^{3}$. Then $ \begin{align} T(u+v) & = (2(u_{1}+v_{1})-4(u_{2}+v_{2})+3(u_{3}+v_{3}),6(u_{1}+v_{1}))\\ & = (2u_{1}-4u_{2}+3u_{3},6u_{1})+(2v_{1}-4v_{2}+3v_{3},6v_{1}) \\ & = Tu + Tv \end{align} $ So it is additive. By inspection, homogeneity $T(\lambda u)=\lambda Tu$, holds. (**$\Leftarrow$):** Suppose $T$ is a linear map. Then $T(0)=0$. $T(0,0,0)=(b,0)$. Thus it must be that $b=0$. Consider $2T(1,1,1)=2(\ldots, 6+c)=(\ldots,12+2c)$ and $T(2,2,2)=(\ldots, 12 + 8c)$. Then $12+c=12+8c\implies c=0$. Thus $c=b=0$. --- **Problem 2:** Suppose $b, c \in \mathbf{R}$. Define $T : \mathcal{P}(\mathbf{R}) \to \mathbf{R}^2$ by $Tp = \left(3p(4) + 5p'(6) + bp(1)p(2),\ \int_{-1}^{2} x^3 p(x) dx + c \sin p(0)\right).$ Show that $T$ is linear if and only if $b = c = 0$. **($\Rightarrow$):** Suppose $b=c=0$. Then, $T(p+q)=(3(p+q)(4)+5(p'+q')(6), \int_{-1}^{2} x^{3}(q+p)(x) \, dx)$. by inspection $(p+q)$ can be separated and we get $Tp+Tq$. Again by inspection, homogeneity $T(\lambda u)=\lambda Tu$, holds. so $T$ is linear. **($\Leftarrow$):** Suppose $T$ is a linear map and let $p(x)=1$ and $\lambda=2$. Then $2Tp=2\left( 3+0+b, \int_{-1}^{2} x^{3} \, dx + c\sin(1)\right)=\left( 6+2b,\frac{15}{2} + c\sin(1)\right)$. And $T(2p)=\left( 6+4b,\frac{15}{2}+c\sin(2) \right)$. Then, $6+2b=6+4b\implies b=0$. And $\frac{15}{2}+c\sin(1)=\frac{15}{2}+c\sin(2)\implies c=0$. Thus $b=c=0$. --- **Problem 3:** Suppose $T \in \mathcal{L}(\mathbf{F}^n, \mathbf{F}^m)$. Show that there exist scalars $A_{j,k} \in \mathbf{F}$ for $j = 1, \ldots, m$ and $k = 1, \ldots, n$ such that $T(x_1, \ldots, x_n) = (A_{1,1}x_1 + \cdots + A_{1,n}x_n,\ \ldots,\ A_{m,1}x_1 + \cdots + A_{m,n}x_n)$ for every $(x_1, \ldots, x_n) \in \mathbf{F}^n$. Let $e_{1},\ldots,e_{n}$ be the standard basis for $\mathrm{F}^{N}$. Suppose $x = (x_{1},\ldots,x_{n})\in \mathrm{F}^{N}$. Then $x=x_{1}e_{1} + \ldots + x_{n}e_{n}$. So, $ \begin{align} T(x_{1},\ldots,x_{n}) & =T(x_{1}e_{1} + \ldots + x_{n}e_{n}) \\ & =T(x_{1}e_{1})+\ldots+T(x_{n}e_{n}) & \text{Additivity} \\ & =x_{1}T(e_{1})+\ldots+x_{n}T(e_{n}) & \text{Homogeneity \& Std Basis} \\ \end{align} $ Now define $T(e_{k})=(A_{1,k},\ldots,A_{m,k})$ the be the outputs for each $T(e_{k})$. > [!Note] > $A_{j,k}\in F$ are not chosen. _The exercise above shows that $T$ has the form promised in the last item of Example 3.4._ --- **Problem 4:** Suppose $T \in \mathcal{L}(V, W)$ and $v_1, \ldots, v_m$ is a list of vectors in $V$ such that $Tv_1, \ldots, Tv_m$ is a linearly independent list in $W$. Prove that $v_1, \ldots, v_m$ is linearly independent. Suppose $0 = \alpha_{1} v_{1} + \ldots + \alpha _{m}v_{m}$. Then $T(0)=0=\alpha_{1}T(v_{1})+\ldots+\alpha _{m}T(v_{m})$. Since $Tv_{1},\ldots,Tv_{m}$ are linearly independent $\alpha_{1}=\ldots=\alpha _{m}=0$. Thus, $v_{1}, \ldots, v_{m}$ is linearly independent. --- **Problem 5:** Prove the assertion in 3.7. With the [[3A1 - Definition of Linear Map|definitions of addition and scalar multiplication]] on $\mathcal{L(V,W)}$, then $\mathcal{L(V,W)}$ is a vector space. - Commutativity: Suppose $S,T\in \mathcal{L}(V,W)$. Then $ \begin{align} (S+T)(v) & =Sv+Tv \\ &= Tv+Sv & \text{Commutivity of W} & = (T+S)(v) \\ & = (T+S)v \end{align} $ - Additive Inverse: If $S\in \mathcal{L}(V,W)$, then the additive inverse of $S$ as a linear map $-S$ where $(-S)(v)=-(S(v))$ for all $v\in V$. Then $(S+ -S)v=Sv + -Sv = 0$. So $S + -S=0$. - Associativity: Suppose $S,T,U \in \mathcal{L}(V,W).$ Then $((S+T)+U)v= (S+T)v+Uv=(Sv+Tv+Uv)= Sv+(Tv+Uv)=(S+(T+U))v$ - Additive Identity: $(S+0)v=Sv+0v=Sv +0_{W}=Sv$. So $S+0=S$ - Multiplicative Identity: $1(S(v))=S(v)$. - Distributive: $a(S+T)v=a(Sv+Tv)=aSv+aTv=(aS+aT)v$. And $(a+b)Sv=aSv+bSv$, by distributive property in $W$ and $Sv\in W$. --- **Problem 6:** Prove the assertions in 3.9. Linear maps are associative $(T_{1}T_{2})T_{3}=T_{1}(T_{2}T_{3})$, identity $TI=IT=T$, and distributive $(S_{1}+S_{2})T=S_{1}T+S_{2}T$ and $S(T_{1}+T_{2})=S_{1}T_{1}+ST_{2}$. Product of linear maps: $(ST)v=S(Tv)$ Associative: Suppose $T_{1}\in \mathcal{L}(Y,Z)$, $T_{2}\in \mathcal{L}(X,Y), T_{3}\mathcal{L}(W,X)$ and $w\in W$. $ \begin{align} (T_{1}T_{2})T_{3}w & = (T_{1}T_{2})(T_{3}w) \\ & =T_{1}(T_{2}(T_{3}w)) & \text{Definition of product of LMs }\\ & =T_{1}(T_{2}T_{3})w & \text{Definition of product of LMs }\\ \end{align} $ Identity: For any $T\in \mathcal{L}(V,W)$, define $I\in \mathcal{L}(V,V)$ and $I\in \mathcal{L}(W,W)$ such that $I(v)=v$ and $I(w)=w$, respectively. Distributive: Suppose $v\in V$ then, by definition of addition on $\mathcal{L}(V,W)$, $(S_{1}+S_{2})(Tv)=S_{1}Tv+S_{2}Tv$, so $(S_{1}+S_{2})T=S_{1}T+S_{2}T$. Similarly, by definition of addition of LM, $S(T_{1}+T_{2})v=S(T_{1}v+T_{2}v)=ST_{1}v+ST_{2}v$. --- **Problem 7:** Show that every linear map from a 1-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\dim V = 1$ and $T \in \mathcal{L}(V, V)$, then there exists $\lambda \in \mathbf{F}$ such that $Tv = \lambda v$ for all $v \in V$. Let $b$ be the basis for $V$. Then for an arbitrary $v\in V$, $v=\alpha b$ with $\alpha \in F$. Then $Tv=T(\alpha b)=\alpha T(b)$. Since $T(b)\in V$, there exists some $\lambda \in V$ such that $\lambda b=T(b)$. So, $Tv=\alpha T(b)=\alpha \lambda b=\lambda\alpha b=\lambda v$. --- **Problem 8:** Give an example of a function $\varphi : \mathbf{R}^2 \to \mathbf{R}$ such that $\varphi(av) = a\varphi(v)$ for all $a \in \mathbf{R}$ and all $v \in \mathbf{R}^2$ but $\varphi$ is not linear. Choose $\varphi(x,y)=\sqrt[3]{x^{3}+y^{3}}$. _The exercise above and the next exercise show that neither homogeneity nor additivity alone is enough to imply that a function is a linear map._ --- **Problem 9:** Give an example of a function $\varphi : \mathbf{C} \to \mathbf{C}$ such that $\varphi(w + z) = \varphi(w) + \varphi(z)$ for all $w, z \in \mathbf{C}$ but $\varphi$ is not linear. (Here $\mathbf{C}$ is thought of as a complex vector space.) Choose $\varphi(x)=\bar{x}$. Then $\varphi (x+y)=\bar{x}+\bar{y}=\varphi(x)+\varphi(y)$. But $i\varphi(1)=i \text{ and }\varphi(1\cdot i)=-i$. Conjugation is not complex-linear. _There also exists a function $\varphi : \mathbf{R} \to \mathbf{R}$ such that $\varphi$ satisfies the additivity condition above but $\varphi$ is not linear. However, showing the existence of such a function involves considerably more advanced tools._ --- **Problem 10:** Suppose $U$ is a subspace of $V$ with $U \neq V$. Suppose $S \in \mathcal{L}(U, W)$ and $S \neq 0$ (which means that $Su \neq 0$ for some $u \in U$). Define $T : V \to W$ by $Tv = \begin{cases} Sv & \text{if } v \in U, \ 0 & \text{if } v \in V \text{ and } v \notin U. \end{cases}$ Prove that $T$ is not a linear map on $V$. Suppose $u\in U$ with $Su\neq{}0$ and $v\in V, v\not\in U$. Then $T(u+v)=0$ since $u+v\not\in U$ (ATC $u+v\in U\implies-u + (u+v)\in U$). but, $Tu+Tv=Su+0\neq{}0$. Thus $T$ is not a linear map on $V.$ --- **Problem 11:** Suppose $V$ is finite-dimensional. Prove that every linear map on a subspace of $V$ can be extended to a linear map on $V$. In other words, show that if $U$ is a subspace of $V$ and $S \in \mathcal{L}(U, W)$, then there exists $T \in \mathcal{L}(V, W)$ such that $Tu = Su$ for all $u \in U$. Let $u_{1},\ldots,u_{n}$ be a basis for $U$. Extend $u_{1},\ldots,u_{n}$ to be a basis for $V$ with $u_{1},\ldots,u_{n},v_{1},\ldots,v_{m}$. Then for any $w\in U$, it will be written as a linear combination of the $u's$ and $0$ coefficients on the $v's$. And at least one nonzero coefficient on the $v's$ for $w\in V,w\not\in U$. Let $w_{1},\ldots,w_{m}\in W$ be arbitrary elements of $W$. Then, define $T$ such that $Tv_{i}=w_{i}$ for $i=1,\ldots,m$. And $Tu_{i}=Su_{i}$ for $i=1,\ldots,n$. Then, for an arbitrary $u\in U$, $Tu=T(a_{1}u_{1} +\ldots+a_{n}u_{n}) =a_{1}Tu_{1}+\ldots+a_{n}Tu_{n}=a_{1}Su_{1}+\ldots+a_{n}Su_{n}=Su$. Now suppose $v\in V,v\not\in U$. $Tv=T(a_{1}u_{1} +\ldots+a_{n}u_{n}+b_{1}v_{1}+\ldots+b_{n}v_{m})$ $=Su + T(b_{1}v_{1}+\ldots+b_{n}v_{m})=Su+b_{1}w_{1}+\ldots+b_{m}w_{m}\in W$. --- **Problem 12:** Suppose $V$ is finite-dimensional with $\dim V > 0$, and suppose $W$ is infinite-dimensional. Prove that $\mathcal{L}(V, W)$ is infinite-dimensional. Suppose $\mathcal{L}(V,W)$ is finite-dimensional. let $l_{1},\ldots l_{m}$ be the basis. We want to show that every $w\in W$ can be mapped by a linear combination of the $l's$. Thus making $W$ finite dimensional. Suppose an arbitrary $w\in W$. Define $T\in \mathcal{L}(V,W)$ such that $Tv_{1}=w$. Then, for every $w\in W$ there exists some linear combination such that $w=Tv_{1}=a_{1}l_{1}v_{1}+\ldots+a_{m}l_{m}v_{1}$. So $a_{1}l_{1}v_{1}, \ldots, a_{m}l_{m}v_{1}$ spans $W$. This cannot be since $W$ is infinite-dimensional. Thus it must be that $\mathcal{L}(V,W)$ is infinite-dimensional. --- **Problem 13:** Suppose $v_1, \ldots, v_m$ is a linearly dependent list of vectors in $V$. Suppose also that $W \neq {0}$. Prove that there exist $w_1, \ldots, w_m \in W$ such that no $T \in \mathcal{L}(V, W)$ satisfies $Tv_k = w_k$ for each $k = 1, \ldots, m$. Consider $0=a_{1}v_{1}+\ldots+a_{m}v_{m}$, then since $v_{1},\ldots,v_{m}$ is linearly dependent, at least one coefficient $\neq{}0$. Suppose $a_{k}\neq{}0$. Let $w_{k}\in W$ with $w_{k}\neq{}0$. Then ATC that there exists a $T$ such that $Tv_{i}=w_{i}$ for $i=1,\ldots,m$ with $w_{k}=Tv_{k}$ and $w_{i\neq{}k}=0$. Now consider, $ \begin{align} 0 & = T(0) \\ & =T(a_{1}v_{1}+\ldots+a_{m}v_{m}) \\ & = a_{1}T(v_{1})+ \ldots+a_{m}T(v_{m}) \\ & = a_{1}w_{1}+ \ldots+a_{m}w_{m} \\ & = 0 + \ldots + a_{k}w_{k} + \ldots + 0\\ & = a_{k}w_{k} \end{align} $ However, this can't be the case since $a_{k}$ and $w_{k}$ are both defined to be nonzero. --- **Problem 14:** Suppose $V$ is finite-dimensional with $\dim V \geq 2$. Prove that there exist $S, T \in \mathcal{L}(V, V)$ such that $ST \neq TS$. Let $n=\dim V \geq{} 2$. Suppose $(v_{1},\ldots,v_{n})$ is the basis for $V.$ Then for an arbitrary $v\in V$, with $v\neq{}0,$ $v=a_{1}v_{1} + a_{n}v_{n}$. Define $ \begin{align} S(v) & =S(a_{1}v1_{1},\ldots,a_{n}v_{n})=(0,a_{1}v_{1},\ldots,a_{n-1}v_{n-1})\\ T(v) & =T(a_{1}v_{1},\ldots,a_{n}v_{n})=(a_{2}v_{2},\ldots,a_{1}v_{n}, 0) \end{align} $ Consider $ST(a_{1}v_{1},\ldots,v_{n})=S(a_{2}v_{2},\ldots, a_{n}v_{n},0)=(0,a_{2}v_{2},\ldots,a_{n}v_{n})$. $TS(a_{1}v_{1},\ldots,a_{n}v_{n})=T(0,a_{1}v_{1},\ldots,a_{n-1}v_{n-1})=(a_{1}v_{1},\ldots,a_{n-1}v_{n-1},0)$. So $ST\neq{}TS$. _Note: I'm abusing list notation where the list represents components of a linear combination of the basis._