#math/linear-algebra --- **Problem 1:** Give an example of a linear map $T$ such that $\dim \operatorname{null} T = 3$ and $\dim \operatorname{range} T = 2$. Suppose $T\in \mathcal{L}(\mathbb{R}^{5}, \mathbb{R}^{5})$ with $T(x_{1},x_{2},x_{3},x_{4},x_{5})=(0,0,0,x_{4},x_{5})$. --- **Problem 2:** Suppose $V$ is a vector space and $S, T \in \mathcal{L}(V, V)$ are such that $\operatorname{range} S \subset \operatorname{null} T.$ Prove that $(ST)^2 = 0$. Consider $STSTv$. **Case 1**: $Tv=0$, then $STSTv=STS(0)=0$. **Case 2:** $Tv\neq{}0$. **Case 2A**: $Tv\in \text{null } S$. Then $S(Tv)=0$ and $STSTv=ST(0)=0$. **Case 2B**: $Tv\not\in \text{null } S$. Then $STv\neq{}0$ and $STv\in \text{range }S\subset \text{null }T$. Thus $T(STv)\in \text{null }T$. Thus, $STSTv=S(TSTv)=S(0)=0.$ --- **Problem 3:** Suppose $v_1, \ldots, v_m$ is a list of vectors in $V$. Define $T \in \mathcal{L}(\mathbf{F}^m, V)$ by $T(z_1, \ldots, z_m) = z_1 v_1 + \cdots + z_m v_m.$ **(a)** What property of $T$ corresponds to $v_1, \ldots, v_m$ spanning $V$? **(b)** What property of $T$ corresponds to $v_1, \ldots, v_m$ being linearly independent? **(a)** If $T$ is surjective. **(b)** if $T$ is injective. Then only $z_{1}=\ldots=z_{m}=0$ maps to 0. --- **Problem 4:** Show that ${T \in \mathcal{L}(\mathbf{R}^5, \mathbf{R}^4) : \dim \operatorname{null} T > 2}$ is not a subspace of $\mathcal{L}(\mathbf{R}^5, \mathbf{R}^4)$. Suppose $T\in \mathcal{L}(\mathbb{R}^{5}, \mathbb{R}^{4})$ with $T(x_{1},x_{2},x_{3},x_{4},x_{5})=(0,0,x_{3},x_{4})$. Suppose $S\in \mathcal{L}(\mathbb{R}^{5}, \mathbb{R}^{4})$ with $S(x_{1},x_{2},x_{3},x_{4},x_{5})=(x_{1},x_{2},0,0)$. Consider $(S+T)v$ with $v=(x_{1},x_{2},x_{3},x_{4},x_{5})$. $(S+T)v=Sv+Tv=(x_{1},x_{2},x_{3},x_{4})$. Thus $\text{dim null }(S+T)=1$ and is not in the set $\{T \in \mathcal{L}(\mathbf{R}^5, \mathbf{R}^4) : \dim \operatorname{null} T > 2\}$. --- **Problem 5:** Give an example of a linear map $T : \mathbf{R}^4 \to \mathbf{R}^4$ such that $\operatorname{range} T = \operatorname{null} T.$ Suppose $T\in \mathcal{L}(\mathbb{R}^{4}, \mathbb{R}^{4})$ with $T(x_{1},x_{2},x_{3},x_{4})=(0,0,x_{1},x_{2})$. - Range $T = (0,0,x,y) : x,y\in \mathbb{R}$ - Null $T = (0,0,x,y) : x,y\in \mathbb{R}$ --- **Problem 6:** Prove that there does not exist a linear map $T : \mathbf{R}^5 \to \mathbf{R}^5$ such that $\operatorname{range} T = \operatorname{null} T.$ Suppose there exists a map $T : \mathbb{R}^{5}\to \mathbb{R}^{5}$ such that range $T$ = null $T$. Then, substituting into $\text{dim }\mathbb{R}^{5}=\text{range }T +\text{null }T$ we get $5=x+x\implies x=2.5\not\in \mathbb{Z}$. Thus is not possible. --- **Problem 7:** Suppose $V$ and $W$ are finite-dimensional with $2 \leq \dim V \leq \dim W$. Show that $U=\{T \in \mathcal{L}(V, W) : T \text{ is not injective}\}$ is not a subspace of $\mathcal{L}(V, W)$. Let $V=W=\mathbb{R}^{2}$ . Suppose $T\in U$ with $T(x,y)=(0,y)$ and $S\in U$ with $S(x,y)=(x,0)$. Then $(S+T)(x,y)=S(x,y)+T(x,y)=(x,0) +(0,y)=(x,y)$, which is injective. Thus $U$ is not a subspace. --- **Problem 8:** Suppose $V$ and $W$ are finite-dimensional with $\dim V \geq \dim W \geq 2$. Show that $U=\{T \in \mathcal{L}(V, W) : T \text{ is not surjective}\}$ is not a subspace of $\mathcal{L}(V, W)$. Let $V=W=\mathbb{R}^{2}$. Suppose $T\in U$ with $T(x,y)=(0,y)$ and $S\in U$ with $S(x,y)=(x,0)$. Then $(S+T)(x,y)=S(x,y)+T(x,y)=(x,0) +(0,y)=(x,y)$, which is surjective. Thus $U$ is not a subspace. --- **Problem 9:** Suppose $T \in \mathcal{L}(V, W)$ is injective and $v_1, \ldots, v_n$ is linearly independent in $V$. Prove that $Tv_1, \ldots, Tv_n$ is linearly independent in $W$. Consider $ \begin{align} 0 & = a_{1}v_{1}+\ldots+a_{n}v_{n} \\ T(0)& =T(a_{1}v_{1}+\ldots+a_{n}v_{n}) \\ 0& =a_{1}Tv_{1}+\ldots+a_{n}Tv_{n} \end{align} $ Then $a_{1}=\ldots=a_{n}=0$ and $Tv_{1},\ldots,Tv_{n}$ are linearly independent in $W.$ --- **Problem 10:** Suppose $v_1, \ldots, v_n$ spans $V$ and $T \in \mathcal{L}(V, W)$. Prove that the list $Tv_1, \ldots, Tv_n$ spans $\operatorname{range} T$. The range $T$ is $Tv : v\in V$. Suppose an arbitrary $v\in V$, then $v=a_{1}v_{1}+\ldots+a_{n}v_{n}$. Consider $ \begin{align} Tv & = T(a_{1}v_{1}+\ldots+a_{n}v_{n}) \\ & = a_{1}Tv_{1}+\ldots a_{n}Tv_{n} \end{align} $ So $Tv_{1},\ldots,Tv_{n}$ spans $\text{range }T$. --- **Problem 11:** Suppose $S_1, \ldots, S_n$ are injective linear maps such that $S_1 S_2 \cdots S_n$ makes sense. Prove that $S_1 S_2 \cdots S_n$ is injective. Step 1: Consider $S_{n}v$. Then since it is injective, the range $S_{n}$ is distinct and the domain of $S_{n-1}$ is distinct. Step i: Consider $S_{i}v$. Then, since $S_{i}$ is injective and the domain is distinct, the range $S_{i}$ is distinct. And $S_{i},\ldots,S_{n}$ is injective. Thus on step n, $S_{1},\ldots,S_{n}$ is injective. --- **Problem 12:** Suppose that $V$ is finite-dimensional and that $T \in \mathcal{L}(V, W)$. Prove that there exists a subspace $U$ of $V$ such that $U \cap \operatorname{null} T = {0}$ and $\operatorname{range} T = {Tu : u \in U}$. Suppose that $V$ is finite-dimensional and that $T \in \mathcal{L}(V, W)$. Since $V$ is finite dimensional, by [[2B1 - Bases]], since null $T$ is a subspace of $V$, there exists a subspace $U$ of $V$ such that $\text{null }T\oplus U=V.$ Then $U$ satisfies $U\cap \text{null }T=\{ 0 \}$. Thus since null V makes up the vectors which $T$ maps to 0, $U$ makes up the vectors which map to range $T$. ==TODO== Properly prove the range statement. --- **Problem 13:** Suppose $T$ is a linear map from $\mathbf{F}^4$ to $\mathbf{F}^2$ such that $\operatorname{null} T = {(x_1, x_2, x_3, x_4) \in \mathbf{F}^4 : x_1 = 5x_2 \text{ and } x_3 = 7x_4}.$ Prove that $T$ is surjective. $\text{dim null }T=2.$ Thus, range $T=2$. Then we can form a basis of range $T$ with $(x_{1},x_{2})\in F^{2}$. However, since $\text{dim }F^{2}=2$ and has a length two basis. It must be that $(x_{1},x_{2})$ is also a basis for $F^{2}$. --- **Problem 14:** Suppose $U$ is a 3-dimensional subspace of $\mathbf{R}^8$ and that $T$ is a linear map from $\mathbf{R}^8$ to $\mathbf{R}^5$ such that $\operatorname{null} T = U$. Prove that $T$ is surjective. Dimension of $U$ is 3. Thus by the fundamental theorem of linear maps, 8 = 3 + dim null $T$. So $\text{dim null }T =5$. Thus, by same logic as problem 13, spans $\mathbb{R}^{5}.$ --- **Problem 15:** Prove that there does not exist a linear map from $\mathbf{F}^5$ to $\mathbf{F}^2$ whose null space equals ${(x_1, x_2, x_3, x_4, x_5) \in \mathbf{F}^5 : x_1 = 3x_2 \text{ and } x_3 = x_4 = x_5}.$ The dimension of the null space is $2$, which forces the dim range $T$ to be 3. However, codomain has dim $F^{2}=2$. Thus, there is no linear map from $F^{5}\to F^{2}$ with that null space. --- **Problem 16:** Suppose there exists a linear map on $V$ whose null space and range are both finite-dimensional. Prove that $V$ is finite-dimensional. Suppose $T$ is a linear map on $V.$ Since the null space is finite dimensional, we know theres a $u_{1},\ldots,u_{n}$ which spans $\text{null }T.$ Since the range is finite dimensional, there is a $Tv_{1},\ldots,Tv_{m}$ which is a basis for range $T$. Then for any $v\in V$, $Tv=Tb_{1}v_{1}+\ldots+Tb_{m}v_{m}$. Rearranging we get, $T(v-b_{1}v_{1}+\ldots+\ldots b_{m}v_{m})=0$. Thus, $ \begin{align} v-b_{1}v_{1}+\ldots+\ldots b_{m}v_{m} & =0 \\ & =a_{1}u_{1}+\ldots+a_{n}u_{n} & \text{Null space basis} \\ v & =b_{1}v_{1}+\ldots+\ldots b_{m}v_{m}+a_{1}u_{1}+\ldots+a_{n}u_{n} \end{align} $ Thus any $v\in V$ in span$(v_{1},\ldots,v_{m},u_{1},\ldots,u_{n})$. --- **Problem 17:** Suppose $V$ and $W$ are both finite-dimensional. Prove that there exists an injective linear map from $V$ to $W$ if and only if $\dim V \leq \dim W$. $(\Rightarrow)$Suppose $T$ is injective. Then $\text{null }V=0$. $ \begin{align} \text{dim }V & =\text{null }V + \text{range }T \\ & =0 + \text{range }T \\ & \leq{} \text{dim }W \end{align} $ $(\Leftarrow)$ Suppose that dim $V=n$ $\leq{}$ dim $W=m$. Let $v_{1},\ldots,v_{n}$ and $w_{1},\ldots,w_{m}$ be a basis for $V$ and $W$, respectively. Consider $T\in \mathcal{L}(V,W)$ with $Tv_{i}=w_{i}$ for $i=1,\ldots,n$. Then $Tv_{1},\ldots,Tv_{n}$ is linearly independent in $W.$ Thus $ \begin{align} 0 & =Tv \\ &=T(a_{1}v_{1}+\ldots+a_{n}v_{n}) \\ &= a_{1}Tv_{1}+\ldots+a_{n}Tv_{n} \\ \end{align} $ Then $a_{1}=\ldots=a_{n}=0$ and $v=0$. Thus null $T=\{ 0 \}$. And $T$ is injective. --- **Problem 18:** Suppose $V$ and $W$ are both finite-dimensional. Prove that there exists a surjective linear map from $V$ onto $W$ if and only if $\dim V \geq \dim W$. $(\Rightarrow)$ Suppose $T\in \mathcal{L}(V,W)$ is surjective. Then, $ \begin{align} \dim V & = \dim \text{null } T + \text{range } T \\ & =\text{dim null }T + \dim W \\ & \geq{} \dim W \end{align} $ $(\Leftarrow)$ Suppose $\text{dim }V \geq{} \text{dim }W$. Let $v_{1},\ldots,v_{n}$ and $w_{1},\ldots,w_{m}$ be a basis for $V$ and $W$, respectively, with $n\geq{}m$. Then consider $T\in \mathcal{L}(V,W)$ with $Tv_{i}=w_{i}$ for $i=1,\ldots,m$ and $Tv_{i>m}=0$. Suppose an arbitrary $w\in W \text{ with }w=a_{1}w_{1}+\ldots+a_{m}w_{m}$. Then construct a $v\in V$ with $v=a_{1}v_{1}+\ldots+a_{m}v_{m}+0v_{m+1}+\ldots+0v_{n}$. $ \begin{align} Tv & =T(a_{1}v_{1}+\ldots+a_{m}v_{m} + \ldots) \\ & =a_{1}Tv_{1}+\ldots+a_{m}Tv_{m} + 0 + \ldots\\ & =a_{1}w_{1}+\ldots+a_{m}w_{m} \\ & =w \end{align} $ Thus every $w\in W$ can be mapped by a $v\in V$. --- **Problem 19:** Suppose $V$ and $W$ are finite-dimensional and that $U$ is a subspace of $V$. Prove that there exists $T \in \mathcal{L}(V, W)$ such that $\operatorname{null} T = U$ if and only if$\dim U \geq \dim V - \dim W$. _Problem is saying that the minimum dimension of the null space is equal to the dimension gap between V and W._ $(\Rightarrow)$ Suppose there exists $T \in \mathcal{L}(V, W)$ such that $\operatorname{null} T = U$. Then $ \begin{align} \dim V & = \dim U + \text{range } T \\ & \leq{} \dim U + \dim W \\ \dim V - \dim W & \leq{} \dim U \end{align} $ $(\Leftarrow)$ Suppose $\dim U \geq \dim V - \dim W$. Let $u_{1},\ldots,u_{n}$ be a basis for $U$ with $Tu_{i}=0$ for $i=1,\ldots,n$. Extend this basis to one of $V$, as $u_{1},\ldots,u_{n},v_{1},\ldots,v_{m}$. $m=\dim V - \dim U$. Let $w_{1},\ldots,w_{k}$ be a basis for $W$ where $\dim W \geq{} \dim V - \dim U=m$. Then map $Tv_{i}=w_{i}$ for $i=1,\ldots,m$. Thus $\text{null }T=U$. --- **Problem 20:** Suppose $W$ is finite-dimensional and $T \in \mathcal{L}(V, W)$. Prove that $T$ is injective if and only if there exists $S \in \mathcal{L}(W, V)$ such that $ST$ is the identity map on $V$. Let $W$ be finite-dimensional and $T\in \mathcal{L}(V,W)$. $(\Rightarrow)$ Suppose $T$ is injective. Then, by [[3B2 - Injective and Surjective]], $V$ must be finite dimensional with $\dim V \leq{} \dim W$. Suppose that $v_{1},\ldots,v_{n}$ is the basis for $V$. Then, since $T$ is injective, null $T=\{ 0 \}$. Then, $Tv=T(a_{1}v_{1}+\ldots+a_{n}v_{n})$ is the range for $T$. Thus, $(Tv_{1},\ldots, Tv_{n})=w_{1},\ldots,w_{n}$ is a basis for range $T$. Suppose $w_{1},\ldots,w_{n}, w_{n+1},\ldots,w_{m}$ is the extended basis for $W$. Suppose $S\in \mathcal{L}(W,V)$ such that $Sw_{i}=v_{i}$ for $i=1,\ldots,n$ and $Sw_{i}=0$ for $i=n+1,\ldots,m$. Consider an arbitrary $v\in V.$ $ \begin{align} STv & =ST(a_{1}v_{1}+\ldots a_{n}v_{n}) \\ & =S(a_{1}w_{1}+\ldots+a_{n}w_{n}) \\ & =a_{1}v_{1}+ \ldots+ a_{n}v_{n} \\ & = v \end{align} $ Thus $ST$ is the identity map on V. $(\Leftarrow)$ Suppose $ST$ is the identity map on $V$. Then, for any arbitrary $v\in V$, $STv=Sw=v$. $T$ cannot map two vectors $u,v$ to the same $w\in W$. ATC, it does. Then there are $u,v\in V$ with $u\neq{}v$ such that $Tu=Tv$. Then, $v=STv=STu=u$ but $v\neq{}u$, thus $T$ does not map two vectors $u,v$ to the same $w\in W$. Thus $T$ is injective. --- **Problem 21:** Suppose $V$ is finite-dimensional and $T \in \mathcal{L}(V, W)$. Prove that $T$ is surjective if and only if there exists $S \in \mathcal{L}(W, V)$ such that $TS$ is the identity map on $W$. $(\Rightarrow)$ Suppose $T$ is surjective. Let $u_{1},\ldots,u_{m}$ be the basis for null $T$. Extend it to a basis of $V$ with $v_{1},\ldots,v_{n}$. Since $T$ is surjective $Tv_{1},\ldots,Tv_{n}$ is the span of range $T=W$ . Construct $S\in \mathcal{L}(W,V)$ such that $Sw_{i}=v_{i}$ with $w_{i}=Tv_{i}$ for $i=1,\ldots,n$. Suppose an arbitrary $w\in W$. Then, $ \begin{align} TSw & =TS(a_{1}w_{1}+\ldots+a_{n}w_{n}) \\ & = T(a_{1}Sw_{1}) + \ldots T(a_{n}Sw_{n}) \\ & = a_{1}Tv_{1}+\ldots+a_{n}Tv_{n} \\ & = aw_{1} + \ldots + a_{n}w_{n} \end{align} $ Thus $TS$ is the identity map on $W$. $(\Leftarrow)$ Suppose $TS$ is the identity map on $W$. Let $u_{1},\ldots,u_{m}$ be the basis for null $T$. Extend it to a basis of $V$ with $v_{1},\ldots,v_{n}$. Since $T$ is surjective $Tv_{1},\ldots,Tv_{n}$ is the span of range $T=W$. Suppose an arbitrary $w\in W$. Then $TSw=a_{1}Tv_{1}+\ldots+a_{n}Tv_{n}$. Since $TS$ is the identity map we know $w=a_{1}Tv_{1}+\ldots+a_{n}Tv_{n}$. Thus $Tv_{1},\ldots Tv_{n}$ spans $W$. Thus $T$ is surjective. --- **Problem 22:** Suppose $U$ and $V$ are finite-dimensional vector spaces and $S \in \mathcal{L}(V, W)$ and $T \in \mathcal{L}(U, V)$. Prove that $ \dim \operatorname{null} ST \leq \dim \operatorname{null} S + \dim \operatorname{null} T. $ Let $u_{1},\ldots,u_{n}\in U$ be a basis for $\text{null }T$ and $\text{dim null }T=n$. Since $\text{null }T \subseteq \text{null }ST$, extend $u_{1},\ldots,u_{n}$ with $v_{1},\ldots,v_{m}\in U$ to be a basis of $\text{null }ST$. Suppose an $a_{1},\ldots,a_{m}\in F$ such that $ \begin{align} 0 & = a_{1}Tv_{1}+\ldots+a_{m}Tv_{m} \\ & = T(a_{1}v_{1} + \ldots + a_{m}v_{m}) \end{align} $ Note that by construction this linear combination lives in $\text{null }S.$ Then $a_{1}v_{1} + \ldots+a_{m}v_{m}$ is in $\text{null }T$ and can be expressed by its basis. Thus, $ b_{1}u_{1}+\ldots+b_{n}u_{n}=a_{1}v_{1}+\ldots+a_{m}v_{m} $ Rearranging we get $ b_{1}u_{1}+\ldots+b_{n}u_{n}-a_{1}v_{1}-\ldots-a_{m}v_{m}=0 $ . Since $u_{1},\ldots,u_{n},v_{1},\ldots,v_{m}$ are a basis, then $a_{i}=b_{i}=0$. Consequently, $Tv_{1}, \ldots, Tv_{m}$ is a linearly independent list. Note that by construction each $v_{i}\in \text{null }ST$, so $S(Tv_{i})=0$ and each $Tv_{i}\in \text{null }S$. Thus the linearly independent list $Tv_{1},\ldots,Tv_{m}$ of length $m$ is in $\text{null S}$ and $\text{dim null }S\geq{}m$. Therefore, $ \begin{align} \text{dim null }ST & = n+m\\ & =\text{dim null }T + m\\ & \leq{}\text{dim null }T + \text{dim null }S \\ \end{align} $ ```mermaid flowchart LR subgraph U[Space U] direction TB nullT(null T) nullST(null ST) end subgraph V[Space V] direction TB zeroV((0_v)) intersection(range T ∩ null S) end subgraph W[Space W] zeroW((0_w)) end nullT -- "T compresses to 0" --> zeroV nullST -- "T maps extension vectors (v_i) here" --> intersection zeroV -- "S maps to 0" --> zeroW intersection -- "S compresses to 0" --> zeroW style nullT fill:#acc2d9,stroke:#333,stroke-width:2px style nullST fill:#8da0cb,stroke:#333,stroke-width:2px style zeroV fill:#e6e6e6,stroke:#333,stroke-width:2px style intersection fill:#f4cae4,stroke:#333,stroke-width:2px style zeroW fill:#e6e6e6,stroke:#333,stroke-width:2px ``` --- **Problem 23:** Suppose $U$ and $V$ are finite-dimensional vector spaces and $S \in \mathcal{L}(V, W)$ and $T \in \mathcal{L}(U, V)$. Prove that $ \dim \operatorname{range} ST \leq \min\{\dim \operatorname{range} S, \dim \operatorname{range} T\}. $ Note $\text{dim }U=\text{dim range }ST + \text{dim null }ST$ and $\text{dim }U=\text{dim range }T + \text{dim null }T$. Since $\text{null }T \subseteq \text{null } ST$, then, $ \begin{align} \text{dim range }T + \text{dim null }T& =\text{dim range }ST + \text{dim null }ST \\ \text{dim range }T& =\text{dim range }ST + \text{dim null }ST -\text{dim null }T \\ \text{dim range }T& \geq{} \text{dim null }ST \\ \end{align} $ Note $\text{range }ST\subseteq \text{range }S$. Suppose $u\in U$. Then $STu=S(Tu)\in \text{range }S$. So $\text{dim range }ST \leq{}\text{dim range S}$. --- **Problem 24:** Suppose $W$ is finite-dimensional and $T_1, T_2 \in \mathcal{L}(V, W)$. Prove that $\operatorname{null} T_1 \subset \operatorname{null} T_2$ if and only if there exists $S \in \mathcal{L}(W, W)$ such that $T_2 = ST_1$. $(\Rightarrow)$ Suppose $\text{null }T_{1}\subset \text{null }T_{2}$. Then $\text{dim null }T_{1}\leq{}\text{dim null }T_{2}$ and $\text{dim range }T_{1}\geq{}\text{dim range }T_{2}$. Let $v_{1},\ldots,v_{n}$ be a basis for $\text{null }T_{1}$. Extend it to a basis of $\text{null }T_{2}$ with $q_{1},\ldots,q_{k}$. Again, extend it to a basis of $W$ with $u_{1},\ldots,u_{m}$. Then $\text{range }T_{2}=\text{span}(u_{1},\ldots,u_{m})$. Suppose an arbitrary $w\in W$ and $w=a_{1}v_{1}+\ldots+b_{1}q_{1}+\ldots +c_{1}u_{1}+\ldots$. Now construct $S$ such that $ST_{1}u_{i}=T_{2}u_{i}$ for $i=1,\ldots,m$ and $STv_{i}=STq_{i}=0$. $ \begin{align} ST_{1}w & =0+ST_{1}(b_{1}q_{1}+\ldots+b_{k}q_{k}) +ST_{1}(c_{1}u_{1}+ \ldots + c_{m}u_{k}) \\ & = c_{1}ST_{1}u_{1}+\ldots+c_{m}ST_{1}u_{k} \\ & = c_{1}T_{2}u_{1}+\ldots+c_{m}T_{2}u_{k} \\ & = T_{2}w \end{align} $ $(\Leftarrow)$ Suppose there exists $S\in \mathcal{L}(W,W)$ such that $T_{2}=ST_{1}$. Let $v_{1},\ldots,v_{n}$ be a basis for $\text{null }T_{1}$ and extend it to a basis $u_{1},\ldots,u_{m}$ of $W$. Suppose $w\in \text{null }T_{1}$. Then $w=a_{1}v_{1}+\ldots+a_{n}v_{n}$ and $ST_{1}w=S0=0\in \text{null }T_{2}$. Thus $\text{null }T_{1}\subset \text{null }T_{2}$. --- **Problem 25:** Suppose $V$ is finite-dimensional and $T_1, T_2 \in \mathcal{L}(V, W)$. Prove that $\operatorname{range} T_1 \subset \operatorname{range} T_2$ if and only if there exists $S \in \mathcal{L}(V, V)$ such that $T_1 = T_2 S$. $(\Rightarrow)$ Suppose $\text{range }T_{1}\subset \text{range }T_{2}$. Let $v_{1},\ldots,v_{n}$ be a basis for $V$. Then for each $T_{1}v_{i}$ there must exist a $w$ such that $T_{1}v_{i}=T_{2}w_{i}$. Then construct $S$ such that $Sv_{i}=w_{i}$. Then, $T_{2}(Sv_{i})=T_{2}w_{i}=T_{1}v_{i}$. The last line is shortcut for the following. Suppose an arbitrary $v\in V$ with $v=a_{1}v_{1}+\ldots+a_{n}v_{n}$. Then $T_{2}Sv=T_{2}S(a_{1}v_{1}+\ldots+a_{n}v_{n})=a_{1}T_{2}w_{1}+\ldots+a_{n}T_{2}w_{n}$=a_{1}T_{1}v_{1}+\ldots+a_{n}T_{1}v_{n}=T_{1}v$. > [!Simple mistake] > Remember $\operatorname{range} T_1 \subset \operatorname{range} T_2$ does **not** mean $T_1$ and $T_2$ do the same thing to the same vectors, they can each have different ordered pairs. $(\Leftarrow)$ Suppose there exists $S\in \mathcal{L}(V,V)$ such that $T_{1}=T_{2}S$. Suppose an arbitrary $v\in V$. Then $T_{2}Sv=T_{2}(Sv)=T_{2}(v')=T_{1}v$. Thus every $T_{1}v$ has an equivalent $T_{2}v'$. Thus --- **Problem 26:** Suppose $D \in \mathcal{L}(\mathcal{P}(\mathbf{R}), \mathcal{P}(\mathbf{R}))$ is such that $\deg Dp = (\deg p) - 1$ for every nonconstant polynomial $p \in \mathcal{P}(\mathbf{R})$. Prove that $D$ is surjective. Consider the restricted map, $T=D|_{\mathcal{P}_{n+1}} : \mathcal{P_{n+1}(R)\to \mathcal{P_{n}(R)}}$. Since $D$ reduces the total deg by 1, $\text{dim null }T=1$ and thus $\text{dim range }T=n+2-1=n+1=\text{dim }\mathcal{P_{n}}$. Thus $\text{range }T=\mathcal{P}_{n}(\mathcal{R})$. Now, suppose an arbitrary $q \in P(R)$ with $\text{deg }q=n\geq{}0$. Then, there must exists a $p \in \mathcal{P_{n+1}(R)}$ such that $D|_{P_{n+1}}(p)=q$. Thus $D$ is surjective. _The notation $D$ is used above to remind you of the differentiation map that sends a polynomial $p$ to $p'$. Without knowing the formula for the derivative of a polynomial (except that it reduces the degree by 1), you can use the exercise above to show that for every polynomial $q \in \mathcal{P}(\mathbf{R})$, there exists a polynomial $p \in \mathcal{P}(\mathbf{R})$ such that $p' = q$._ --- **Problem 27:** Suppose $p \in \mathcal{P}(\mathbf{R})$. Prove that there exists a polynomial $q \in \mathcal{P}(\mathbf{R})$ such that $5q'' + 3q' = p$. _This exercise can be done without linear algebra, but it's more fun to do it using linear algebra._ Suppose a linear map $T:\mathcal{P_{n+1}(R)\to P_{n}(R)}$ such that $T(q)=5q''+3q'$. Then $n+2=1+\text{dim range T}$. Thus $\text{dim range }T=n+1=\text{dim }\mathcal{P_{n}(R)}$ and $\text{range }T=\mathcal{P_{n}(R)}$. Then suppose a $p \in \mathcal{P}_{n}(R)$, since $T$ is surjective, there must be a $q \in \mathcal{P}_{n+1}\mathcal{(R)}$ such that $5q'' + 3q' = p$. --- **Problem 28:** Suppose $T \in \mathcal{L}(V, W)$, and $w_1, \ldots, w_m$ is a basis of $\operatorname{range} T$. Prove that there exist $\varphi_1, \ldots, \varphi_m \in \mathcal{L}(V, \mathbf{F})$ such that $Tv = \varphi_1(v) w_1 + \cdots + \varphi_m(v) w_m$ for every $v \in V$. Suppose an arbitrary $v\in V$ then $Tv$ can be written with unique scalars $a_{1},\ldots,a_{m}\in F$ such that $Tv=w=a_{1}w_{1}+\ldots+a_{m}w_{m}$ . Define $\varphi _{i}(v)=a_{i}$ for $i=1, \ldots, m$. Suppose $v,q\in V$. Since $T$ is a linear map, $T(u+v)=Tu + Tv= (\varphi_{1}(u)w_{1}+\ldots) + (\varphi_{1}(v)w_{1}+\ldots)$ and $T(u+v)=\varphi_{1}(u+v)w_{1}+\ldots$. And since the representation is unique, $\varphi _{i}u+\varphi _{i}v=\varphi _{i}(u+v)$. By inspection, $\varphi$ is homogenous. _I originally had some difficulty realizing this could be a function..., the ontology of a function is that its just ordered input-output pairs and by defining these pairs, however so, its a valid function. We only need to prove such a function exists, is well defined, and is linear._ --- _An attempt which proves something true about V but does not help in answering the question: $V$ must have at least $m$ independent vectors._ ATC $V$ has $n<m$ independent vectors. Then $V$ is finite-dimensional and $ \begin{align} \text{dim V } & = n \\ & =\text{dim null }T +m\\ & \geq{} m \end{align} $ Thus $V$ must have at least $n\geq{}m$ independent vectors. --- **Problem 29:** Suppose $\varphi \in \mathcal{L}(V, \mathbf{F})$. Suppose $u \in V$ is not in $\operatorname{null} \varphi$. Prove that $V = \operatorname{null} \varphi \oplus \{au : a \in \mathbf{F}\}.$ Suppose $v\in \text{null }\varphi \cap \{ au:a\in F \}$. Then $v=au$ for some $a\in F$ and $\varphi v=0$. Then $\varphi v=\varphi(au)=a\varphi u=0$. Since $u\not\in \text{null }\varphi$ and $u\neq{}0$, then $a=0$. Thus $v=0$. So $\text{null }\varphi \cap \{ au : a \in F \}=\{ 0 \}$ Now, suppose an arbitrary $v\in V$. We want to decompose $v$ into any two vectors from $\text{null } \varphi$ and $\{ au:a\in F \}$, respectively. Specifically we want some $q\in \text{null }\varphi$ such that $q=v-au$. Let $\frac{\varphi v}{\varphi u} u\in\{ au:a\in F \}$. Then Construct a $q\in V$ such that $q=v-\frac{\varphi v}{\varphi u}u$. Then $\varphi q=\varphi v-\varphi(\frac{\varphi v}{\varphi u}u)=\varphi v-\varphi v=0$. Thus, $v=q + \frac{\varphi v}{\varphi u}u$. ```tikz \usetikzlibrary{arrows.meta} \begin{document} \begin{tikzpicture}[scale=0.85,>=Stealth,font=\small,line width=0.6pt] \definecolor{cA}{RGB}{192,57,43} % red : line {au} \definecolor{cB}{RGB}{36,113,163} % blue : null phi \definecolor{cD}{RGB}{148,103,189} % purple : v \definecolor{ax}{RGB}{128,128,128} % gray : axes / guides %================ Panel 1 : the two subspaces meet only at 0 =============== \begin{scope} \draw[ax,->] (-3,0)--(3.2,0) node[right]{$x_1$}; \draw[ax,->] (0,-3)--(0,3.3) node[above]{$x_2$}; \node[ax] at (0.45,4.05) {$\mathrm{null}\,\varphi\;\cap\;\lbrace{}au\rbrace{}=\{0\}$}; \draw[cB,line width=1pt] (-2.8,-1.4)--(2.8,1.4) node[right]{$\mathrm{null}\,\varphi$}; \draw[cA,line width=1pt] (-1.4,-2.8)--(1.4,2.8); \node[cA,right] at (1.05,2.5) {$\{au\}$}; \fill[cB] (2,1) circle (2.6pt); \fill[cA] (1,2) circle (2.6pt); \fill[black] (0,0) circle (3pt) node[below right]{$0$}; \end{scope} %================ Panel 2 : unique decomposition of an arbitrary v ========== \begin{scope}[xshift=9.5cm] \draw[ax,->] (-2.4,0)--(3.4,0) node[right]{$x_1$}; \draw[ax,->] (0,-2.4)--(0,3.3) node[above]{$x_2$}; \node[ax] at (0.75,4.05) {$\displaystyle v=q+\frac{\varphi v}{\varphi u}\,u$}; \draw[cB,line width=1pt] (-2.8,-1.4)--(3.2,1.6) node[right]{$\mathrm{null}\,\varphi$}; \draw[cA,line width=1pt] (-1.5,-3.0)--(1.45,2.9); \node[cA,right] at (1.05,2.5) {$\{au\}$}; \draw[cA,dashed] (2,1)--(2.5,2); % q -> v (parallel to u) \draw[cB,dashed] (0.5,1)--(2.5,2); % cu -> v (parallel to null phi) \draw[cB,->,line width=1pt] (0,0)--(2,1); \draw[cA,->,line width=1pt] (0,0)--(0.5,1); \draw[cD,->,line width=1.1pt] (0,0)--(2.5,2); \fill[cB] (2,1) circle (3pt); \node[cB,below right] at (2,1) {$q$}; \fill[cA] (0.5,1) circle (3pt); \node[cA,above left] at (0.45,1.05) {$\frac{\varphi v}{\varphi u}u$}; \fill[cD] (2.5,2) circle (3pt); \node[cD,right] at (2.55,2.05) {$v$}; \fill[black] (0,0) circle (2.6pt) node[below left]{$0$}; \end{scope} \end{tikzpicture} \end{document} ``` --- **Problem 30:** Suppose $\varphi_1$ and $\varphi_2$ are linear maps from $V$ to $\mathbf{F}$ that have the same null space. Show that there exists a constant $c \in \mathbf{F}$ such that $\varphi_1 = c\varphi_2$. Suppose $v\in V$ and $v\not\in \text{null }\varphi_{1}$. Then $\varphi_{1}v=\alpha \neq{}0$ and $\varphi_{2}v=\beta \neq{}0$. Choose $c=\frac{\alpha}{\beta}$. Suppose an arbitrary $u\in V$. Construct a $w\in V$ such that $w=\varphi_{2}(v)u-\varphi_{2}(u)v$. Then $\varphi_{2}(w)=\varphi_{2}(\varphi_{2}(v)u-\varphi_{2}(u)v)=\varphi_{2}v\varphi_{2}u-\varphi_{2}u\varphi_{2}v=0$ and $w\in \text{null }\varphi_{2}=\text{null }\varphi_{1}$. Thus, $0=w=\varphi_{1}(\varphi_{2}(v)u-\varphi_{2}(u)v)=\varphi_{2}v\varphi_{1}u-\varphi_{2}u\varphi_{1}v=\beta \varphi_{1}u-\alpha \varphi_{2}u$ and $\varphi_{1}u=\frac{\alpha}{\beta}\varphi_{2}u$. Thus $\varphi_{1}=c\varphi_{2}$. ```tikz \usetikzlibrary{arrows.meta} \begin{document} \begin{tikzpicture}[>=Stealth,line cap=round,line join=round,scale=1.6] % ---------- palette ---------- \definecolor{flr}{RGB}{190,190,190} \definecolor{panel}{RGB}{190,190,190} \definecolor{txt}{RGB}{201,210,227} \definecolor{ax}{RGB}{139,151,179} \definecolor{cP2}{RGB}{36,113,163} % phi_2 (blue) \definecolor{cP1}{RGB}{192,57,43} % phi_1 (red) \definecolor{cNul}{RGB}{155,127,212} % null space (purple) \definecolor{cV}{RGB}{240,160,75} % the point v (orange) % ---------- 3D -> 2D projection ---------- \newcommand{\pt}[3]{({0.64*((#1)-(#2))},{0.36*((#3)-(#1)-(#2))})} % ---------- styles ---------- \tikzset{lbl/.style={inner sep=1pt,font=\footnotesize}} \tikzset{cont/.style={txt,opacity=0.65,line width=0.5pt}} \tikzset{contR/.style={txt,opacity=0.65,cP1,line width=0.5pt}} \tikzset{contB/.style={txt,opacity=0.65,cP2,line width=0.5pt}} % ---------- reserve drawing area (invisible) ---------- \path (-2.6,-2.75) rectangle (3.75,2.0); % ---------- floor z = 0 ---------- \fill[panel,opacity=0.3] \pt{-0.8}{-0.8}{0} -- \pt{2.2}{-0.8}{0} -- \pt{2.2}{2.2}{0} -- \pt{-0.8}{2.2}{0} -- cycle; \foreach \a in {0,1,2}{\draw[ax,opacity=0.18,line width=0.3pt] \pt{\a}{-0.8}{0} -- \pt{\a}{2.2}{0};} \foreach \b in {0,1,2}{\draw[ax,opacity=0.18,line width=0.3pt] \pt{-0.8}{\b}{0} -- \pt{2.2}{\b}{0};} \draw[ax,opacity=0.5,line width=0.5pt] \pt{-0.8}{-0.8}{0} -- \pt{2.2}{-0.8}{0} -- \pt{2.2}{2.2}{0} -- \pt{-0.8}{2.2}{0} -- cycle; % ---------- coordinate axes ---------- \draw[ax,->,line width=0.7pt] \pt{-0.9}{0}{0} -- \pt{2.5}{0}{0} node[anchor=north,font=\footnotesize,text=ax]{$x_1$}; \draw[ax,->,line width=0.7pt] \pt{0}{-0.9}{0} -- \pt{0}{2.5}{0} node[anchor=north,font=\footnotesize,text=ax]{$x_2$}; \draw[ax,->,line width=0.7pt] \pt{0}{0}{0} -- \pt{0}{0}{3.6} node[anchor=south,font=\footnotesize,text=ax]{$z$}; % ---------- phi_2 ramp (blue) ---------- \fill[cP2,opacity=0.30] \pt{-0.8}{-0.8}{0} -- \pt{2.2}{-0.8}{3} -- \pt{2.2}{2.2}{0} -- \pt{-0.8}{2.2}{-3} -- cycle; \draw[cP2,line width=1pt] \pt{-0.8}{-0.8}{0} -- \pt{2.2}{-0.8}{3} -- \pt{2.2}{2.2}{0} -- \pt{-0.8}{2.2}{-3} -- cycle; % ---------- phi_1 ramp (red, steeper) ---------- \fill[cP1,opacity=0.26] \pt{-0.8}{-0.8}{0} -- \pt{2.2}{-0.8}{6} -- \pt{2.2}{2.2}{0} -- \pt{-0.8}{2.2}{-6} -- cycle; \draw[cP1,line width=1pt] \pt{-0.8}{-0.8}{0} -- \pt{2.2}{-0.8}{6} -- \pt{2.2}{2.2}{0} -- \pt{-0.8}{2.2}{-6} -- cycle; % ---------- level (contour) lines on each plane ---------- \begin{scope} \clip \pt{-0.8}{-0.8}{0} -- \pt{2.2}{-0.8}{3} -- \pt{2.2}{2.2}{0} -- \pt{-0.8}{2.2}{-3} -- cycle; \foreach \k in {-4,-2,2,4}{\draw[contB] ({0.64*(\k)},-3.2) -- ({0.64*(\k)},2.0);} \end{scope} \begin{scope} \clip \pt{-0.8}{-0.8}{0} -- \pt{2.2}{-0.8}{6} -- \pt{2.2}{2.2}{0} -- \pt{-0.8}{2.2}{-6} -- cycle; \foreach \k in {-4,-2,2,4}{\draw[contR] ({0.32*(\k)},-3.2) -- ({0.32*(\k)},2.0);} \end{scope} % ---------- shared null line ---------- \draw[cNul,line width=2.2pt] \pt{-0.8}{-0.8}{0} -- \pt{2.2}{2.2}{0}; % ---------- point v = (2,1) and its two heights ---------- \draw[cV,dashed,line width=1pt, opacity=0.9] \pt{2}{1}{0} -- \pt{2}{1}{2}; \draw[cV,dashed,line width=1pt, opacity=0.9] \pt{0}{1}{0} -- \pt{2}{1}{0}; \draw[cV,dashed,line width=1pt, opacity=0.9] \pt{2}{0}{0} -- \pt{2}{1}{0}; \draw[cV,line width=1.1pt] \pt{2}{1}{0} circle (0.5pt); \fill[cP2] \pt{2}{1}{1} circle (2.2pt); \fill[cP1] \pt{2}{1}{2} circle (2.2pt); % ---------- leaders + labels for v ---------- \draw[cP1,dashed,line width=0.3pt] (1.30,-0.26) -- \pt{2}{1}{2}; \draw[cP2,dashed,line width=0.3pt] (1.30,-0.64) -- \pt{2}{1}{1}; \node[lbl,text=cP1,anchor=west] at (1.32,-0.22) {$\varphi_1 v=2$}; \node[lbl,text=cP2,anchor=west] at (1.32,-0.60) {$\varphi_2 v=1$}; \node[lbl,text=cV,anchor=north] at \pt{2.2}{0.85}{0} {$\mathbf{v=(2,1)}$}; % ---------- other labels ---------- \draw[cNul,line width=0.5pt] (-0.45,0.78) -- \pt{-0.2}{-0.2}{0}; \node[lbl,text=cNul,anchor=east] at (-0.45,0.82) {null space\,($\varphi=0$)}; \node[lbl,text=cP1,anchor=west] at (2.05,1.60) {$\varphi_1=2\varphi_2$}; \node[lbl,text=cP2,anchor=west] at (2.05,0.42) {$\varphi_2=x_1-x_2$}; \end{tikzpicture} \end{document} ``` --- **Problem 31:** Give an example of two linear maps $T_1$ and $T_2$ from $\mathbf{R}^5$ to $\mathbf{R}^2$ that have the same null space but are such that $T_1$ is not a scalar multiple of $T_2$. Choose $T_{1}(x_{1},x_{2},x_{3},x_{4},x_{5})=(x_{1},x_{2})$ and $T_{2}(x_{1},x_{2},x_{3},x_{4},x_{5})=(x_{2},x_{1})$. ```tikz \usetikzlibrary{arrows.meta} \begin{document} \begin{tikzpicture}[scale=0.85,>=Stealth,font=\small,line width=0.6pt] \definecolor{cA}{RGB}{192,57,43} \definecolor{cB}{RGB}{36,113,163} \definecolor{cC}{RGB}{44,160,44} \definecolor{cD}{RGB}{148,103,189} \definecolor{ax}{RGB}{128,128,128} % ---------- Panel 1: T1 = identity (equal map) ---------- \begin{scope} \draw[ax,->] (-3,0)--(3,0) node[right]{$x_1$}; \draw[ax,->] (0,-3)--(0,3) node[above]{$x_2$}; \node[ax] at (0,3.7) {$T_1:\ (x_1,x_2)\mapsto(x_1,x_2)$ \ (identity)}; \foreach \x/\y/\c/\n in {2/1/cA/A, -1/2/cB/B, -2/-1/cC/C, 1/-2/cD/D}{ \fill[\c] (\x,\y) circle (3pt) node[\c,xshift=8pt,yshift=8pt]{$\n$}; } \end{scope} % ---------- Panel 2: T2 = reflection across x2 = x1 ---------- \begin{scope}[xshift=8.5cm] \draw[ax,->] (-3,0)--(3,0) node[right]{$x_1$}; \draw[ax,->] (0,-3)--(0,3) node[above]{$x_2$}; \draw[ax,dashed] (-2.9,-2.9)--(2.9,2.9) node[above,xshift=-6pt]{$x_2=x_1$}; \node[ax] at (0,3.7) {$T_2:\ (x_1,x_2)\mapsto(x_2,x_1)$ \ (reflection)}; % originals hollow, images filled \draw[cA] (2,1) circle (3pt); \fill[cA] (1,2) circle (3pt) node[cA,xshift=-8pt,yshift=8pt]{$A'$}; \draw[cB] (-1,2) circle (3pt); \fill[cB] (2,-1) circle (3pt) node[cB,xshift=8pt,yshift=-8pt]{$B'$}; \draw[cC] (-2,-1) circle (3pt); \fill[cC] (-1,-2) circle (3pt) node[cC,xshift=-8pt,yshift=-8pt]{$C'$}; \draw[cD] (1,-2) circle (3pt); \fill[cD] (-2,1) circle (3pt) node[cD,xshift=-8pt,yshift=8pt]{$D'$}; \end{scope} \end{tikzpicture} \end{document} ```