Problem Suppose $v_1, v_2, v_3, v_4$ spans $V$. Prove that the list $$v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4$$ also spans $V$ Solution Let $u = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4$ where $a_1,a_2,a_3,a_4 \in \mathbb{F}$ and define $b_i$ as follows: $b_1 = a_1, b_2 - b_1 = a_2, b_3 - b_2 = a_3$ and $b_4 - b_3 = a_4$. Then we must have that: $$u = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4$$ $$= b_1v_1 + (b_2 - b_1)v_2 + (b_3 - b_2)v_3 + (b_4 - b_3)v_4$$ $$= b_1(v_1 - v_2) + b_2(v_2 - v_3) + b_3(v_3 - v_4) + b_4v_4$$ Thus $u \in \textrm{span}(v_1,v_2,v_3,v_4) \iff u \in \textrm{span}(v_1 - v_2,v_2 - v_3,v_3 - v_4,v_4)$

Problem A function $f: \mathbb{R} \to \mathbb{R}$ is called even if $$f(-x) = f(x)$$ for all $x \in \mathbb{R}$. A function $f: \mathbb{R} \to \mathbb{R}$ is called odd if $$f(-x) = -f(x)$$ for all $x \in \mathbb{R}$. Let $U_e$ denote the set of real-valued even functions on $\mathbb{R}$ and $U_o$ denote the set of real-valued odd functions on $\mathbb{R}$. Show that $\mathbb{R}^\mathbb{R} = U_e \oplus U_o$ Solution Any function $f: \mathbb{R} \to \mathbb{R}$ can be decomposed into an even and an odd part as follows: $$f_{e}(x) = \frac{f(x) + f(-x)}{2}$$ $$f_{o}(x) = \frac{f(x) - f(-x)}{2}$$ Where $f_{e}(x) = f_{e}(-x)$ and $f_{o}(-x) = -f_{o}(x)$ and also: $$f_e(x) + f_o(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2} = f(x)$$ Thus any function $f \in \mathbb{R}^\mathbb{R}$ can be written as $f_e + f_o$ where $f_e \in U_e$ and $f_o \in U_o$ and since we clearly have $U_e \cap U_o = \lbrace 0 \rbrace$, we must have that $\mathbb{R}^\mathbb{R} = U_e \oplus U_o$....

Problem Prove or give counterexample: if $U_1$, $U_2$, $W$ are subspaces of $V$ such that $$U_1 \oplus W = U_2 \oplus W$$ then $U_1 = U_2$ Solution We have that $U_1 \oplus W = V = U_2 \oplus W$, therefore $U_1 \cap W = U_2 \cap W = \lbrace 0 \rbrace$. Choosing any element $u \in U_1$ will therefore not be in $W$, but will be in $V$ as $U_1 \subseteq V$....

Problem Suppose $$U = \lbrace (x,y,x+y,x-y,2x) \in \mathbb{F}^5 | x,y \in \mathbb{F} \rbrace$$ Find three subspaces $W_1, W_2, W_3$ of $\mathbb{F}^5$, none of which equals $\lbrace 0 \rbrace$, such that $\mathbb{F}^5 = U \oplus W_1 \oplus W_2 \oplus W_3$. Solution We saw in P1C21 that $U \oplus W = \mathbb{F}^5$ if $W = \lbrace (0,0,x,y,z) \in \mathbb{F}^5 | x,y,z \in \mathbb{F} \rbrace$. But clearly $W = W_1 \oplus W_2 \oplus W_3$ if: $$W_1 = \lbrace (0,0,x,0,0) \in \mathbb{F}^5 | x \in \mathbb{F} \rbrace$$ $$W_2 = \lbrace (0,0,0,x,0) \in \mathbb{F}^5 | x \in \mathbb{F} \rbrace$$ $$W_3 = \lbrace (0,0,0,0,x) \in \mathbb{F}^5 | x \in \mathbb{F} \rbrace$$ And by P1C17 plus P1C21 we therefore see that: $$\mathbb{F}^5 \stackrel{P1C21}{=} U \oplus W = U \oplus (W_1 \oplus W_2 \oplus W_3) \stackrel{P1C17}{=} U \oplus W_1 \oplus W_2 \oplus W_3$$

Problem Suppose $$U = \lbrace (x,y,x+y,x-y,2x) \in \mathbb{F}^5 | x,y \in \mathbb{F} \rbrace$$ Find a subspace $W$ of $\mathbb{F}^5$ such that $\mathbb{F}^5 = U \oplus W$ Solution Let $W = \lbrace (0,0,x,y,z) \in \mathbb{F}^5 |x,y,z \in \mathbb{F} \rbrace$, then $U + W = \lbrace (x_1, x_2, x_3, x_4, x_5) \in \mathbb{F}^5 | x_1,x_2,x_3,x_4,x_5 \in \mathbb{F} \rbrace = \mathbb{F}^5$. Also $$U \cap W = \lbrace u \in \mathbb{F}^5 | u \in U \land u \in W \rbrace$$ But $(x_1,x_2,x_3,x_4,x_5) = u \in U$ implies that if $x_1 = x_2 = 0$ and $(x_1,x_2,x_3,x_4,x_5)$ $= u \in W$ implies that $x_3 = x_1 + x_2$, $x_4 = x_1 - x_2$ and $x_5 = 2x_1$, but $x_1 = x_2 = 0$ then implies $x_1 = x_2 = x_3 = x_4 = x_5 = 0$, thus $U \cap W = \lbrace 0 \rbrace$ implying that $U \oplus W = \mathbb{F}^5$....