Problem Suppose $v_1, \dots, v_n$ spans $V$ and $T \in \mathcal{L}(V,W)$. Prove that the list $Tv_1, \dots, Tv_n$ spans range $T$ Solution As $v_1, \dots, v_n$ spans $V$, any $v \in V$ can be written as a linear combination of $v_1, \dots, v_n$ as per definition 2.5. Therefore, for any $v \in V$ there exist $a_1, \dots, a_n \in \mathbb{F}$ such that $v = a_1v_1 + \dots + a_nv_n$ implying that: $$Tv = T(a_1v_1 + \dots + a_nv_n) \stackrel{D3....
Problem Suppose $T \in \mathcal{L}(V,W)$ is injective and $v_1, \dots, v_n$ is linearly independent in $V$. Prove that $Tv_1, \dots, Tv_n$ is linearly independent in $W$ Solution Let $a_1, \dots, a_n \in \mathbb{F}$ such that $$0 = a_1Tv_1 + \dots + a_nTv_n \stackrel{D3.2}{=} T(a_1v_1 + \dots + a_nv_n)$$ Then injectivity implies $a_1v_1 + \dots + a_nv_n = 0$ and linear independence of $v_1, \dots, v_n$ then implies $a_1 = \dots = a_n = 0$ which in turn implies that $Tv_1, \dots, Tv_n$ is a linearly independent list in $W$....
Problem Suppose $V$ and $W$ are finite-dimensional with $\dim V \geq \dim W \geq 2$. Show that $\lbrace T \in \mathcal{L}(V,W) | T$ is not surjective $ \rbrace$ is not a subspace of $\mathcal{L}(V,W)$ Solution We can copy the setup in the solution to P3B7: Let $v_1, v_2, \dots, v_n$ be a basis for $V$ and let $w_1, w_2, \dots, w_m$ be a basis for $W$, then define $T_1, T_2 \in \mathcal{L}(V,W)$ by $T_1v_{i} = 0$, $T_1v_{j} = w_{\textrm{min}(j,m)}$ and $T_2v_{i} = w_{i}$, $T_2v_{j} = 0$ and $T_2v_{i} = w_{\textrm{min}(i,m)}$....
Problem Suppose $V$ and $W$ are finite-dimensional with $2 \leq \dim V \leq \dim W$. Show that $\lbrace T \in \mathcal{L}(V,W) | T $ is not injective $\rbrace$ is not a subspace of $\mathcal{L}(V,W)$ Solution Let $v_1, v_2, \dots, v_n$ be a basis for $V$ and let $w_1, w_2, \dots, w_m$ be a basis for $W$, then define $T_1, T_2 \in \mathcal{L}(V,W)$ by $T_1v_{i} = 0$, $T_1v_{j} = w_{j}$ and $T_2v_{i} = w_{i}$, $T_2v_{j} = 0$ and $T_2v_{i} = w_{i}$....
Problem Prove that there does not exist a linear map $T: \mathbb{R}^5 \to \mathbb{R}^5$ such that $$\textrm{range} \thinspace T = \textrm{null} \thinspace T$$ Solution The condition $\textrm{range} \thinspace T = \textrm{null} \thinspace T$ implies $\dim \textrm{range} \thinspace T = \dim \textrm{null} \thinspace T$, which by the fundamental theorem of linear maps implies: $$5 = \dim \mathbb{R}^5 \stackrel{T3.22}{=} \dim \textrm{range} \thinspace T + \dim \textrm{null} \thinspace T = 2\dim \textrm{null} \thinspace T$$ Implying $\dim \textrm{range} \thinspace T = \dim \textrm{null} \thinspace T = 5/2$ a contradiction....