Problem Give an example of a linear map $T: \mathbb{R}^4 \to \mathbb{R}^4$ such that $$\textrm{range} \thinspace T = \textrm{null} \thinspace T$$ Solution Define $T$ by $T(x_1, x_2, x_3, x_4) = (0, x_1, 0, x_3)$ for all $x_1,x_2,x_3,x_4 \in \mathbb{R}$, then $\textrm{range} \thinspace T = \textrm{null} \thinspace T = \lbrace (0,x,0,y) | x,y \in \mathbb{R}\rbrace$.

Problem Show that $$\lbrace T \in \mathcal{L}(\mathbb{R}^5, \mathbb{R}^4) | \dim \textrm{null} \thinspace T > 2 \rbrace$$ is not a subspace of $\mathcal{L}(\mathbb{R}^5, \mathbb{R}^4)$ Solution Let $U = \lbrace T \in \mathcal{L}(\mathbb{R}^5, \mathbb{R}^4) | \dim \textrm{null} \thinspace T > 2 \rbrace$, then $T_1 \in U$ defined by $T_1(x_1, x_2, x_3, x_4, x_5) = (x_1,x_2,0,0)$ and $T_2 \in U$ defined by $T_2(x_1, x_2, x_3, x_4, x_5) = (0,0,x_3,x_4)$, where $x_1, \dots, x_5 \in \mathbb{R}$....

Problem Suppose $v_1, \dots, v_m$ is a list of vectors in $V$. Define $T \in \mathcal{L}(\mathbb{F}^m, V)$ by $$T(z_1, \dots, z_m) = z_1v_1 + \dots + z_mv_m$$ (a) What property of $T$ corresponds to $v_1, \dots, v_m$ spanning $V$? (b) What property of $T$ corresponds to $v_1, \dots, v_m$ being linearly independent? Solution (a) If $v_1, \dots, v_m$ spans $V$, then we can choose $z_1, \dots, z_m \in \mathbb{F}$ such that $(z_1, \dots, z_m) \in \mathbb{F}^m$ is mapped to any $v$ we want by the definition of span....

Problem Suppose $V$ is a vector space and $S,T \in \mathcal{L}(V,V)$ are such that $$\textrm{range} \thinspace S \subset \textrm{null} \thinspace T$$ Prove that $(ST)^2 = 0$ Solution As $Tv \in V$ we must have $S(Tv) \in \textrm{range} \thinspace S \subset \textrm{null} \thinspace T$ implying $T(S(Tv)) = 0$ for all $v \in V$. Therefore $(ST)^2v = STSTv = S(T(S(Tv))) = S(0) \stackrel{T3.11}{=} 0$ for all $v \in V$, implying $STST = 0$....

Problem Give an example of a linear map $T$ such that $\dim \textrm{null} \thinspace\thinspace T = 3$ and $\dim \textrm{range} \thinspace\thinspace T = 2$ Solution Let $T: \mathbb{R}^5 \to \mathbb{R}^2$ defined by $T(x_1, x_2, x_3, x_4, x_5) = (x_1,x_2)$ for all $x_1, x_2, x_3, x_4, x_5 \in \mathbb{R}$. Then $\textrm{null} \thinspace T = \lbrace (0,0,x_1,x_2,x_3) | x_1,x_2,x_3 \in \mathbb{R} \rbrace$ and also $\textrm{range} \thinspace T = \lbrace (x_1,x_2) | x_1,x_2 \in \mathbb{R} \rbrace = \mathbb{R}^2$....