#math/linear-algebra **Problem 1** For each of the following subsets of $F^3$, determine whether it is a subspace of $F^3$: 1. ${(x_{1},x_{2},x_{3}) \in F^3 : x_{1} + 2x_{2} + 3x_{3} = 0}$ 2. ${(x_{1},x_{2},x_{3}) \in F^3 : x_{1} + 2x_{2} + 3x_{3} = 4}$ 3. ${(x_{1},x_{2},x_{3}) \in F^3 : x_{1}x_{2}x_{3} = 0}$ 4. ${(x_{1},x_{2},x_{3}) \in F^3 : x_{1} = 5x_{3}}$ **Problem 1.1** $U = \{(x_{1},x_{2},x_{3}) \in F^3 : x_{1} + 2x_{2} + 3x_{3} = 0\}$ 1. $0 \in{} U$ 2. Let $x, y \in{} U$. Then if we have $x + y$ $ \begin{align} (x+y)&= (x_{1}+y_{1}) + 2(x_{2}+y_{2}) + 3(x_{3}+y_{3}) \\ &= (x_{1} + 2x_{2}+3x_{3}) + (y_{1} + 2y_{2} + 3y_{3}) \\ &= 0 + 0 \\ &=0 \end{align} $ 3. Let $u \in{} U$ and $c \in{} F$. Then if we have $cu$ $ \begin{align} c(u_{1}+2u_{2} + 3u_{3}) & = c(0) \\ & = 0 \end{align} $ $\therefore U$ is a subspace of $F^3$. **Problem 1.2** $U = \{(x_{1},x_{2},x_{3}) \in F^3 : x_{1} + 2x_{2} + 3x_{3} = 4\}$ 1. $0 \not\in U$. $(0,0,0)$ can not satisfy the constraint. $\therefore U$ is not a subspace of $F^3$. **Problem 1.3**$U = \{(x_{1},x_{2},x_{3}) \in F^3 : x_{1}x_{2}x_{3} = 0\}$ 1. $0 \in{} F^3$. 2. Let $u, v \in{} F^3$. $ \begin{align} u+v & = (u_{1}u_{2}u_{3}) + (v_{1}v_{2}v_{3}) \\ & = 0 + 0 \\ & = 0 \end{align} $ 3. Let $u \in{} F^3$ and $c\in{}F$. $ \begin{align} cu & = c(u_{1}u_{2}u_{3}) \\ & = cu \\ & = 0 \end{align} $ $\therefore U$ is a subspace of $F^3$. **Problem 1.4** $U = \{(x_{1},x_{2},x_{3}) \in F^3 : x_{1} = 5x_{3}\}$ Before even fully answering... we can see that all elements, lines or planes, go through the origin, so its probably the case that $U$ is a subspace of $F^3$. Wait this logic is wrong. I don't know the equations or planes or lines in this regard. Do you set one variable to the others? 1. $0 \in{} U$ 2. Let $u=(5u_{3},u_{2},u_{3}), v=(5v_{3},v_{2},v_{3}) \in{} U$ $ \begin{align} u+v & = (5u_{3},u_{2},u_{3}) + (5v_{3},v_{2},v_{3}) \\ & = (5(u_{3} + v_{3}), (u_{2}+v_{2}), (u_{3} + v_{3})) \end{align} $ 3. Let $c \in{} F, u \in{} U$. $ \begin{align} cu & = c(5u_{3}, u_{2}, u_{3}) \\ & = (5(cu_{3}),cu_{2}, cu_{3}) \end{align} $ $\therefore U$ is a subspace of $F^3$. --- **Problem 2** Verify all the assertions in Example 1.35. (a) If $b \in{} F$, then $ \{(x_{1},x_{2},x_{3},x_{4}) \in{} F^4 : x_{3} = 5x_{4}+b\} $ is a subspace of $F^4$ if and only if $b=0$. $(\Rightarrow)$Let $b=0$. $U = \{(x_{1},x_{2},5x_{4},x_{4}) \in{} F^4\}$ (1) $(0,0,0,0) \in{} U$ (2) Let $u = (a, b, 5d, d) : a,b,d \in{} F$ and $w=(x,y,5z, z) : x,y,z \in{} F$. $ \begin{align} u + w & = (a, b, 5d, d) + (x,y,5z,z) \\ & = (a + x, b + y, 5(d + z), d+z) \end{align} $ (3) Let $u = (a,b,5d,d)$ and $\alpha \in{} F$. $ \begin{align} \alpha u & = (\alpha a, \alpha b, 5(\alpha d), \alpha d) \end{align} $ $\therefore U$ is a subspace of $F^4$. $(\Leftarrow)$ Let $U$ be a subspace of $F^4$. $0 \in{} U$. Setting all the free variables to $0$, $(x_{1},x_{2},5x_{4}+b,x_{4}) = (0,0,b,0)$. $\therefore b = 0$ **** (b) The set of continuous real-valued functions on the interval $[0,1]$ is a subspace of $R^{[0,1]}$. Let $U = \{f \in{} R^{[0,1]}: f \text{ is continuous }\}$ . (1) $0 \in{} U$ since $f(x) = 0$ is continuous. (2) The sum of continuous functions, $(f + g)(x)$ is continuous. (3) Scaling a continuous function by $c \in{} \mathbb{R}$, $cf(x)$, is continuous. **** (c) The set of differentiable real-valued functions on $\mathbb{R}$ is a subspace of $\mathbb{R}^\mathbb{R}$. Let $U = \{f \in{} R^{R}: f \text{ is differentiable }\}$ . (1) $0 \in{} U$ since $f(x) = 0$ is differentiable. $f'(x) = 0$. (2) If $f \in{} U$ and $g \in{} U$, then $(f+g)(x) \in{} U$ is differentiable. $(f+g)'(x) = f'(x) + g'(x)$. (3) If $c\in{}\mathbb{R}$ and $f \in{} U$ then $cf \in{} U$ since $(cf)'(x) = cf'(x)$. **** (d) The set of differentiable real-valued functions $f$ on the interval (0, 3) such that $f'(2) = b$ is a subspace of $\mathbb{R}^{(0,3)}$ if and only if $b=0$. Let $U = \{f \in{} R^{(0,3)} : f\text{ is differentiable and }f'(2) = b \in{}\mathbb{R}\}$. Let $b = 0$ and $f'(2) = 0$. - (1) $0 \in{} U$, since $0'(x) = 0$. - (2) $f,g \in{} U$ then $(f + g)'(x) = f'(x) + g'(x)$. and $(f+g)'(2) = f'(2) + g'(2) = 0 + 0 = 0$. So $(f+g)(x) \in{} U$. - (3) $f \in{} U$ and $c \in{} \mathbb{\mathbb{R}}$. Then $(cf)(x) = cf(x)$ and $(cf)'(2) = cf'(2) = c(0) = 0$. So $cf(x) \in{} U$. Thus if $b=0$ then $U$ is a subspace of $R^{(0,3)}$ Let $U$ be a subspace of $\mathbb{R}^{(0,3)}$. Let $f,g \in{} U$ then $(f + g)'(x) = f'(x) + g'(x)$. and $(f+g)'(2) = f'(2) + g'(2) = b + b = 2b$. Thus $ \begin{align} b & =2b \\ \end{align} $ Is only true if $b=0$. Thus $b = 0$. $\therefore U \text{ is a subspace of } \mathbb{R}^{(0,3)} \iff b=0$. **** (e) The set of all sequences of complex numbers with limit $0$ is a subspace of $C^{\infty}$. Let $U = \{(x_{1}, x_{2}, \dots) : x_{i} \in{} C \text{ for } i = 1, 2, \dots \text{ and } \lim_{ i \to \infty } x_{i} = 0\}$. (1) $0 \in{} U$ since $\lim_{ i \to \infty } 0_{i} = 0$. (2) If $x, y \in{} U$, then $\lim_{ i \to \infty }(x_{i}+y_{i}) = \lim_{ i \to \infty }x_{i} + \lim_{ i \to \infty } y_{i} = 0 + 0 = 0$ (3) If $x \in{} U$ and $c\in{}C$, then $\lim_{ i \to \infty }(cx_{i}) = c\left(\lim_{ i \to \infty }x_{i}\right) = c0 = 0$. Thus $U$ is a subspace of $C^{\infty}$. **** --- **Problem 3** Show that the set of differentiable real-valued functions $f$ on the interval $(-4,4)$ such that $f'(-1) = 3f(2)$ is a subspace of $\mathbb{R}^{(-4,4)}$. Let $U = \{f\in{}\mathbb{R}^{(-4,4)} : f \text{ is differentiable and } f'(-1)=3f(2)\}$ 1. $0 \in U$. Since $0'(-1) = 0 = 3(0(2))$. 2. Let $u, v \in{} U$. Then . $ \begin{align} (u+v)'(-1) & = u'(-1) + v'(-1) & \text{ Definition for Sum of Functions}\\ & = 3u(2) + 3v(2) \\ & = 3(u+v)(2) & \text{ Definition for Sum of Functions} \end{align} $ 3. Let $c \in{} \mathbb{R}, u \in{} U$. Then . $ \begin{align} (cu)(-1) & = c(u(-1)) & \text{ Definition for Product with Function} \\ & = c(3u(2)) \\ & 3((cu)(2)) \end{align} $ $\therefore$ the set of differentiable real-valued functions $f$ on the interval $(-4,4)$ such that $f'(-1) = 3f(2)$ is a subspace of $\mathbb{R}^{(-4,4)}$. --- **Problem 4** Suppose $b \in \mathbb{R}$. Show that the set of continuous real-valued functions $f$ on the interval $[0,1]$ such that $\int_0^1 f = b$ is a subspace of $\mathbb{R}^{[0,1]}$ if and only if $b = 0$. Let $U = \{f \in{} \mathbb{R}^{[0,1]} : f \text{ is continuous and }\int_{0}^{1} f\, = b\}$. $\Rightarrow$ Let $U$ be a subspace of $F^{[0,1]}$ and $u \in U, c \in \mathbb{R}$. So, $\int_{0}^{1} \,cu = c\int_{0}^{1} \, u = cb$. For $cb = b$ to be true, it must be that $b = 0$. $\Leftarrow$ Let $b=0$. 1. $0 \in U$ 2. Let $u, v \in U$ $ \begin{align} \int_{0}^{1} \, (u+v) & = \int_{0}^{1} \, u + \int_{0}^{1} \, v & \text{Definition of Addition of Integrals}\\ & =b + b \\ & = 0 + 0 \\ & = 0 \end{align} $ 3. Let $c \in \mathbb{R}, u \in U$. $ \begin{align} \int_{0}^{1} \, cu & = c \int_{0}^{1} \, u & \text{ Definition of Mul with Integral}\\ & =c (b) \\ & = c(0) \\ & =0 \end{align} $ So, $U$ is a subspace of $\mathbb{R}^{[0,1]}$. $\therefore$ The set of continuous real-valued functions $f$ on the interval $[0,1]$ such that $\int_0^1 f = b$ where $b \in \mathbb{R}$ is a subspace of $\mathbb{R}^{[0,1]}$ if and only if $b = 0$. --- **Problem 5** Is $\mathbb{R}^2$ a subspace of the complex vector space $\mathbb{C}^2$? No. If $c=i \in \mathbb{C}, x=(1,1)\in \mathbb{R}^2$, then $cx = (i,i) \not\in \mathbb{R}^2$ --- **Problem 6** (a) Is ${(a,b,c) \in \mathbb{R}^3 : a^3 = b^3}$ a subspace of $\mathbb{R}^3$? (b) Is ${(a,b,c) \in \mathbb{C}^3 : a^3 = b^3}$ a subspace of $\mathbb{C}^3$? **Problem 6A** Is $U = \{(a,b,c) \in \mathbb{R}^3 : a^3 = b^3\}$ a subspace of $\mathbb{R}^3$? 1. $0 \in U$ 2. Let $x,y \in U$. Then $ \begin{align} (x+y) & = (x_{1}+y_{1}, x_{2}+y_{2},x_{3}+y_{3}) \\ & = (x_{1}+y_{1}, x_{1}+y_{1},x_{3}+y_{3}) & \text{Since } x \mapsto x^{3} \text{ is injective on R }\\ & \in U \end{align} $ 3. Let $x \in U, a\in \mathbb{R}$. Then $ \begin{align} ax & =(ax_{1},ax_{2},ax_{3}) \\ & =(ax_{1},ax_{1},ax_{3}) & \text{ Since } x \mapsto x^{3} \text{ is injective on R }\\ & \in U & \text{ Since } (ax_{1})^{3}=(ax_{2})^{3} \end{align} $ $\therefore U$ is a subspace of $\mathbb{R}^{3}$ **Problem 6B** Is $U=\{(a,b,c) \in \mathbb{C}^3 : a^3 = b^3\}$ a subspace of $\mathbb{C}^3$? We hinged the previous proof on the fact that $f(x)=x^{3}$ is an [[Injective Function]] allowing us to conclude that if $a^3=b^3$ then $a=b$ . We want to find a value of $a$ and $b$ such that $a^{3}=b^{3}$ but $a\neq b$. Then when we add it with another term $u \in U$, we'll _probably_ end up with something $\not\in U$. So lets look at the [[Roots of Polynomials|roots]] of the function. $ \begin{align} a^{3} - b^{3} & =0 \\ (a-b)(a^2+ab+b^2) & =0 \\ \end{align} $ The first is $a=b$, which is not we're looking for so lets rule that out. The roots of the second term, treating $b$ as the constant, should be $ \begin{align} a & = \frac{-b\pm \sqrt{b^2-4b^2}}{2} \\ & = \frac{-b\pm \sqrt{-3b^2}}{2} \\ & = \frac{-b\pm (\sqrt{3})bi}{2}\\ & = b\frac{-1\pm (\sqrt{3})i}{2} \end{align} $ Lets choose $b=1$, then $ a = \frac{-1\pm (\sqrt{3})i}{2} $ Now, (choosing the positive radical), let $u = (\frac{-1 + (\sqrt{3})i}{2}, 1, 0) \in U \quad\text{ Since }u_{1}^3=u_{2}^3$ and $ v = \left( \frac{1}{2}, \frac{1}{2}, 0 \right) $ Now $u+v = \left( \frac{\sqrt{3}}{2}i, \frac{3}{2} , 0\right)$, but $\left( \frac{\sqrt{3}}{2}i \right)^{3}=-\frac{3\sqrt{3}i}{8} \neq \left(\frac{3}{2}\right)^3$ So $(u+v)\not\in U$. --- **Problem 7** Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \in U$ whenever $u \in U$), but $U$ is not a subspace of $\mathbb{R}^2$. Since the inverse implies $0\in U$, we'd want an example where $\lambda \in R, u\in U, \text{ but }\lambda u\not\in U$. Let $ U = \{(-1,1), (1, -1), (0,0)\} $ $(-1,1) + (1,-1) = (0,0)$ $(-1,1) + (0,0) = (-1, 1)$ $(1,-1) + (0,0) = (1, -1)$ But, $3(-1,1) = (-3, 3) \not\in U$. So $U$ is not a subspace of $\mathbb{R}^2$. --- **Problem 8** Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $\mathbb{R}^2$. Scalar multiplication probably means we need an infinitely large set. Let $ U = \{(a,b) \in \mathbb{R}^2 : a=0 \text{ or } b= 0\} $ Evidently $U$ is closed under scalar multiplication $(0, \lambda b) \text{ and } (\lambda a, 0) \in U.$ Then, $(0, 1) \in U \text{ and } (1,0) \in U$, but $(0,1) + (1,0) = (1,1) \not\in U$. So, $U$ is not a subspace. --- **Problem 9** A function $f : \mathbb{R} \to \mathbb{R}$ is called periodic if there exists a positive number $p$ such that $f(x) = f(x + p)$ for all $x \in \mathbb{R}$. Is the set of periodic functions from $\mathbb{R}$ to $\mathbb{R}$ a subspace of $\mathbb{R}^\mathbb{R}$? Explain. Intuition says yes because frequency waves when added together are still periodic. Let $ U = \{f \in \mathbb{R}^\mathbb{R} : f(x) = f(x+p) \} $ 1. $0 \in U$ 2. Let $f \in U, \lambda \in \mathbb{R}$. Then $(\lambda f)(x) = \lambda(f(x)) = \lambda(f)(x+p) = (\lambda f)(x+p)$ 3. Let $f, g \in U$. Then $f(x) = f(x+p) \text{ and } g(x) = g(x+p).$ $(f+g)(x) = f(x)+g(x).$ $= f(x+p) + g(x+p) = (f+g)(x+p).$ So $U$ is a subspace of $\mathbb{R}^\mathbb{R}$ --- **Problem 10** Suppose $U_{1}$ and $U_{2}$ are subspaces of $V$. Prove that the intersection $U_{1} \cap U_{2}$ is a subspace of $V$. Let $I = U_{1}\cap U_{2}$ 1. Since $0\in U_{1}$ and $0\in U_{2}$, $0 \in I$. 2. Let $i,j \in I$. Since $i,j\in U_{1}$ and $i,j\in U_{2}$ and $U_{1}, U_{2}$ are closed under addition, $(i+j) \in U_{1} \text{ and } (i+j) \in U_{2}$. $\therefore(i+j)\in I$. 3. Let $i\in I, \lambda \in F$. Since $i\in U_{1} \implies \lambda i \in U_{1}$ and $i\in U_{2}\implies \lambda i\in U_{2}$. $\therefore i\in I.$ So $I$ is a subspace of $V.$ --- **Problem 11** Prove that the intersection of every collection of subspaces of $V$ is a subspace of $V$. Let $ I = \bigcap_{\alpha \in A} U_{\alpha} \text{ where } A \text{ is an arbitrary indexing set}$ 1. Since $0 \in U_{\alpha}$ for every $\alpha \in A$, $0 \in I$. 2. Let $x, y \in I$. Then $x, y \in U_{\alpha}$ for every $\alpha \in A$. Since each $U_{\alpha}$ is a subspace, $x + y \in U_{\alpha}$ for every $\alpha \in A$. $\therefore (x+y) \in I$. 3. Let $x \in I, \lambda \in F$. Since $x \in U_{\alpha} \implies \lambda x \in U_{\alpha}$ for every $\alpha \in A$. $\therefore \lambda x \in I.$ So $I$ is a subspace of $V.$ --- **Problem 12** Prove that the union of two subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other. Let $U_{1},U_{2}$ be subspaces of $V$ and $U=U_{1}\cup U_{2}$ ($\Rightarrow$) Assume $U$ is a subspace. Assume towards a contradiction, that $U_{1} \not\subseteq U_{2}$ and $U_{2} \not\subseteq U_{1}$. Since $U_{1} \not\subseteq U_{2}$ there exists a $u\in U_{1}$ with $u\not\in U_{2}$. And since $U_{2} \not\subseteq U_{1}$ and there exists $w\in U_{2}$ with $w\not\in U_{1}$. Since $U$ is a subspace, $(u+w) \in U$. Then either $(u+w) \in U_{1}$ or $(u+w)\in U_{2}$. Without loss of generality, $(u+w) \in U_{1}$, then $-u + (u + w) \in U_{1} \implies w\in U_{1}$. Which is a contradiction. $\therefore U_{1}\subseteq U_{2} \text{ or } U_{2}\subseteq U_{1}$ $(\Leftarrow)$ Assume without loss of generality that $U_{1}\subseteq U_{2}$. Then $U=U_{2}$. $\therefore U$ is a subspace. >[!Example] >Consider $U_{1}=\{(x,0) \in \mathbb{R}^2\}$ and $U_{2}=\{(0,x)\in \mathbb{R}^2\}$. --- **Problem 13** Prove that the union of three subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces contains the other two. Let $U = U_{1}\cup U_{2}\cup U_{3}$ $(\Rightarrow)$ Assume $U$ is a subspace of $V$. WLOG, assume $U_{1}\cup U_{2}$ is not a subspace. Then there exists $u\in U_{1}-U_{2}$ and $w\in U_{2}-U_{1}$. Then $(u+w) \not\in U_{1}\cup U_{2}$. WLOG, Let $u'\in U_{1}$. $\text{If }(u'\in U_{2})$ then consider $v=u'+w+u$. - $v\not\in U_{1}$ : ATC $v\in U_{1}\implies(-u')+v=(-u')+u+w+u=w+u\in U_{1}$. But $w+u\not\in U_{1}\implies v\not\in U_{1}$ - $v\not\in U_{2}$ : ATC $v\in U_{2}\implies(-u')+v=(-u')+u+w+u=w+u\in U_{2}$. But $w+u\not\in U_{2}\implies v\not\in U_{2}$ - Since $u'+w+u$ exists in $U$, then it must be $u'+w+u \in U_{3}$. $-(w+u) + (u'+w+u) = u'\in U_{3}$ $\text{If }(u'\not\in U_{2})$ then consider $v= u'+\lambda w : \lambda \in F$. - $v\not\in U_{1}$ : ATC $v = u' + \lambda w\in U_{1}\implies-u'+u'+\lambda w\in U_{1}\implies w\in U_{1}$. But $w\not\in U_{1}\implies v\not\in U_{1}$ - $v \not\in U_{2}$ : ATC $v = u'+\lambda w\in U_{2}\implies-\lambda w+u'+\lambda w\in U_{2}\implies u'\in U_{2}$. But $u'\not\in U_{2}\implies v\not\in U_{2}$. - Since $v=u'+\lambda w \in U \implies u'+\lambda w\in U_{3}$. Choose $\lambda=1,2$. $u'+1w, u'+2w\in U_{3} \implies 2(u'+1w) + -(u'+2w)=u'\implies u'\in U_{3}$. Note we cannot choose $\lambda=0$ due to division by 0 in earlier step. Thus in any case $u'\in U_{3}$. Similarly, a $w'\in U_{2}$ has $w'\in U_{3}$. $\therefore U_{1}\subseteq U_{3}$ and $U_{2}\subseteq U_{3}$. WLOG, assume $U_{1}\cup U_{2}$ is a subspace. WLOG, by Problem 12, assume $U_{2}\subseteq U_{1} \implies U_{1}\cup U_{2} = U_{1}$. Since $U$ is a subspace, it must be $(U_{1}\cup U_{2})\subseteq U_{3}$ or $U_{3}\subseteq (U_{1}\cup U_{2})=U_{1}$. In either case, one subspace contains the other two. $(\Leftarrow)$ WLOG, assume that $U_{3}\subseteq U_{1}$ and $U_{2}\subseteq U_{1}$. Then $U=U_{1}\cup U_{2}\cup U_{3}=U_{1}$. $\therefore U$ is a subspace. The union of three subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces contains the other two. $\quad \blacksquare{}$ >[!Tip] >If $x + y\in U$ it does not mean $x,y\in U$. Take for example, $x=(1,0), y=(0,1)$ and $U=\{(a,a)\in \mathbb{R}^2\}$. However, if $x+y\in U$ and $x \in U \implies y\in U$ since $-x \in U$. --- **Problem 14** Verify the assertion in Example 1.38. Suppose that $U=\{(x,x,y,y)\in F^4:x,y\in F\}$ and $W=\{(x,x,x,y)\in F^4:x,y,z \in F\}$. Prove that $ U+W=\{(x,x,y,z)\in F^{4}:x,y,z\in F\} = S $ $(\Rightarrow)$ $U+W\subseteq S$ Suppose $u=(a,a,b,b)\in U:a,b\in F$ and $w=(c,c,c,d)\in W: c,d\in F$. Then $ \begin{align} u+w & =(a+c,a+c, b + c, b + d) \end{align} $ Since all $a,b,c,d\in F$, let $x=a+c, y=b+c, z=b+d$ then we can rewrite $ u+w=(x,x,y,z) : x,y,z\in F $ $U+W\subseteq S$ $(\Leftarrow)$ $S \subseteq U + W$ Let $s \in S$, then $s=(x,x,y,z) : x,y,z\in F$. Then we need a $u=(a,a,b,b)\in U:a,b\in F$ and $w=(c,c,c,d)\in W: c,d\in F$ such that $ \begin{align} a+c & =x \\ b+c & =y \\ b+d & =z \\ \end{align} $ Conveniently set $b=0$ since we have 4 variables and 3 equations. $ \begin{align} a+c & =x \\ c & =y \\ d & =z \\ \end{align} $ Then we get $u=(x-y,x-y,0,0)\in U$ and $w=(y,y,y,z)\in W$. Checking $ \begin{align} u+w & =(x-y,x-y,0,0) + (y,y,y,z) \\ & =(x,x,y,z) \\ & = s \end{align} $ $\therefore S \subseteq U+W$ So, $S = U + W\quad \blacksquare{}$ --- **Problem 15** Suppose $U$ is a subspace of $V$. What is $U + U$? It should be the same as $U$. Let $S= \{u+w : u,w\in U\} = U + U$. Then every element of $S$ is simply the sum of two elements in $U$ which itself is an element in $U$. To show that every element of $U$ is in $S$, let $w=0$ then we get $\{u : u \in U\}=U$. To show that $S\subseteq U$ every $s \in S$ can be decomposed into $s=u+w : u,w \in U$by definition of $S$. Since $U$ is a subspace, $(u+w)\in U$. So $S=U\quad \blacksquare{}$ --- **Problem 16** Is the operation of addition on the subspaces of $V$ commutative? In other words, if $U$ and $W$ are subspaces of $V$, is $U + W = W + U$? Yes since $V$ requires commutativity over all its elements $u + v = v + u \quad\forall u,v\in V$. $ \begin{align} U+W & =\{(u+w) : u\in U , w\in W \} \\ & = \{(w+u) : u\in U , w\in W \} & \text{Commutativity of elements in }V \\ & =W+U \end{align} $ So, if $U$ and $W$ are subspaces of $V$, then $U + W = W + U$. $\quad \blacksquare{}$ --- **Problem 17** Is the operation of addition on the subspaces of $V$ associative? In other words, if $U_{1}, U_{2}, U_{3}$ are subspaces of $V$, is $(U_{1} + U_{2}) + U_{3} = U_{1} + (U_{2} + U_{3})?$ Yes since $V$ requires associativity over all its elements $(u + v) + w = u + (v+w) \quad\forall u,v,w\in V$. $ \begin{align} (U_{1}+U_{2}) + U_{3} & =\{(u_{1}+u_{2}) : u_{1}\in U_{1} , u_{2}\in U_{2} \} + U_{3}\\ & =\{(u_{1}+u_{2}) +u_{3} : u_{1}\in U_{1} , u_{2}\in U_{2}, u_{3}\in U_{3} \} \\ & =\{u_{1}+(u_{2} +u_{3}) : u_{1}\in U_{1} , u_{2}\in U_{2}, u_{3}\in U_{3} \} \\ & = U_{1}+ \{(u_{2} +u_{3}) : u_{2}\in U_{2}, u_{3}\in U_{3} \} \\ & = U_{1}+ (U_{2}+U_{3}) \end{align} $ $\therefore (U_1 + U_2) + U_3 = U_1 + (U_2 + U_3) \quad \blacksquare$ --- **Problem 18** Does the operation of addition on the subspaces of $V$ have an additive identity? Which subspaces have additive inverses? Yes take the subspace containing only 0 to be the additive identity. $0 = \{0\}$. Since all subspaces $U$ over $V$ have $0\in U$. Then $U + 0 = \{u + 0 : u \in U\} = \{u : u \in U\}= U$ Adding any two subspaces together cannot result in any loss of elements since $0\in \text{any subspace }U$. The only subspace with an additive inverse would be $\{0\}$. --- **Problem 19** Prove or give a counterexample: if $U_{1}, U_{2}, W$ are subspaces of $V$ such that $U_{1} + W = U_{2} + W,$ then $U_{1} = U_{2}$. False. Let $U_{1}=\{0\}, U_{2} = W = \{x : x \in \mathbb{R}\}$. Then $U_{1}+W=W$ and $U_{2}+W=W$. Clearly, $U_{1} \neq U_{2} \quad \blacksquare{}$. --- **Problem 20** Suppose $U = {(x, x, y, y) \in F^4 : x, y \in F}.$ Find a subspace $W$ of $F^4$ such that $F^4 = U \oplus W$. One of the difficulties is that we need to avoid when $x=y$, otherwise their intersection $\neq \{0\}$ $ W = \{(x+y,y,y,x) : x,y \in F\} $ $W$ is a subspace. - $0\in W$ - $u=(a+b,b,b,a)\in W,w=(c+d,c,c,d)\in W$ $\to (u+w) = ((a+d)+(c+b), c+b, c+b, a + d)\in W$ - $\lambda \in F, w =(a+b,b,b,a)\in W$ then $\lambda w=(\lambda(a+b), \lambda b, \lambda b, \lambda a)=(\lambda a+ \lambda b), \lambda b, \lambda b, \lambda a)\in W$. Let $u=(a,a,b,b)\in U \text{ and } w=(c+d,d,d,c)\in W$. In order for $u+w=0$ $ \begin{align} a+c+d & =0 \\ a+d & =0 \\ b+d & =0 \\ b+c & =0 \end{align} $ Setting $d=-b, c=-b$ we get $a+c+d = a-2b=0$ and $a+d=a-b=0$ which $\implies a=b$ and $a=0.$ So for $u+w=0$ it must be that $a=b=c=d=0$. $\therefore W \oplus U$ Let $u=(a,a,b,b)\in U \text{ and } w=(c+d,d,d,c)\in W$. $ \begin{align} a+c+d & = x_{1} \\ a+d & =x_{2} \\ b+d & =x_{3} \\ b+c & =x_{4} \end{align} $ $ \begin{align} a+c+d & = x_{1} \implies c = x_{1}-x_{2}\\ a+d & =x_{2} \implies a=x_{2}-x_{3}+x_{4}-x_{1}+x_{2}\\ b+d & =x_{3} \implies d=x_{3}-x_{4}+x_{1}-x_{2}\\ b+c & =x_{4} \implies b=x_{4}-x_{1}+x_{2}\\ \end{align} $ With this we get - $u=(x_{2}-x_{3}+x_{4}-x_{1}+x_{2},x_{2}-x_{3}+x_{4}-x_{1}+x_{2}, x_{4}-x_{1}+x_{2}, x_{4}-x_{1}+x_{2})$ - $w=(x_{1}-x_{2} + x_{3}-x_{4}+x_{1}-x_{2}, x_{3}-x_{4}+x_{1}-x_{2}, x_{3}-x_{4}+x_{1}-x_{2}, x_{1}-x_{2})$ And $u+w=(x_{1},x_{2},x_{3},x_{4}) : x_{1},x_{2},x_{3},x_{4}\in F$. $\therefore U + W = F$ $U\oplus W=F\quad \blacksquare{}$ --- **Problem 21** Suppose $U = {(x, y, x+y, x-y, 2x) \in F^5 : x, y \in F}.$ Find a subspace $W$ of $F^5$ such that $F^5 = U \oplus W$. Maybe the idea it --- **Problem 22** Suppose $U = {(x, y, x+y, x-y, 2x) \in F^5 : x, y \in F}.$ Find three subspaces $W_{1}, W_{2}, W_{3}$ of $F^5$, none of which equals ${0}$, such that $F^5 = U \oplus W_{1} \oplus W_{2} \oplus W_{3}$. --- **Problem 23** Prove or give a counterexample: if $U_{1}, U_{2}, W$ are subspaces of $V$ such that $V = U_{1} \oplus W \quad \text{and} \quad V = U_{2} \oplus W,$ then $U_{1} = U_{2}$. I thought that it should be $U_{1}=U_{2}$... --- **Problem 24** 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 let $U_{o}$ denote the set of real-valued odd functions on $\mathbb{R}$. Show that $\mathbb{R}^\mathbb{R} = U_{e} \oplus U_{o}$.