Problem Suppose $V$ is finite-dimensional with $\dim V \geq 2$. Prove that there exist $S,T \in \mathcal{L}(V,V)$ such that $ST \neq TS$ Solution Let $v_1, v_2$ be a linearly independent list of $V$ which exist as per the restriction $\dim V \geq 2$ and definition 2.36. Now we can uniquely the linear maps $S$ and $T$ as the extension described in P3A10 to the subspace $U = \textrm{span}(v_1, v_2)$ and where $T,S$ is defined uniquely on $U$ by theorem 3....

Problem Suppose $v_1, \dots, v_m$ is a linearly independent list of vectors in $V$. Suppose also that $W \neq \lbrace 0 \rbrace$. Prove that there exist $w_1, \dots, w_m \in W$ such that no $T \in \mathcal{L}(V,W)$ satisfies $Tv_k = w_k$ for each $k = 1,\dots,m$ Solution As $w \neq \lbrace 0 \rbrace$, then there is a non-zero $w \in W$. If we now let $w = w_1 = \dots = w_m$, then because $v_1, \dots, v_m$ is linearly dependent, there exists $a_1, \dots, a_m \in \mathbb{F}$, not all zero, such that $0 = a_1v_1 + \dots + a_mv_m$....

Problem Suppose $V$ is finite-dimensional with $\dim V > 0$, and suppose $W$ is infinite-dimensional. Prove that $\mathcal{L}(V,W)$ is infinite-dimensional Solution Let $w_1, w_2, \dots$ be an infinite sequence of linearly independent vectors in $W$ which exists as per P2A14. If we let $v_1, \dots, v_n$ be a basis of $V$, then by theorem 3.5 we can construct an infinite sequence of linear maps in $\mathcal{L}(V,W)$ by letting $T_i \in \mathcal{L}(V,W)$ be defined by $T_i v_j = w_{i+j-1}$....

Problem 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 $u \in U$ Solution Let $u_1, \dots, u_m$ be a basis for $U$ and extend it to the basis $u_1, \dots, u_m, w_1, \dots, w_n$ of $V$, then by theorem 3....

Problem 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: Prove that $T$ is not a linear map on $V$ Solution Let $u \in U$ and $v \in V$ such that $v$ $\cancel{\in}$ $V$, the existence of which follows from $U \neq V$. Then if we assume $v + u \in U$, the fact that $U$ is closed under addition then implies $v = (v + u) - u \in U$, which is a contradiction....