If you find any mistakes, please make a comment! Thank you.

## Chapter 1 Exercise C

1. Solution: (a) $\{(x_1,x_2,x_3)\in\mathbb F^3:x_1+2x_2+3x_3=0\}$ is a subspace of $\mathbb F^3$. By 1.34, to show a subset is a subspace, we just need to check Additive identity, Closed under addition and Closed under scalar multiplication.

Additive identity: it is clear that the additive identity $(0,0,0)$ of $\mathbb F^3$ is contained in $\{(x_1,x_2,x_3)\in\mathbb F^3:x_1+2x_2+3x_3=0\}$.

Closed under addition: if $(a_1,a_2,a_3),(b_1,b_2,b_3)\in \{(x_1,x_2,x_3)\in\mathbb F^3:x_1+2x_2+3x_3=0\}$, then$a_1+2a_2+3a_3=0\quad\text{and} \quad b_1+2b_2+3b_3=0.$Hence$(a_1+b_1)+2(a_2+b_2)+3(a_3+b_3)=(a_1+2a_2+3a_3)+(b_1+2b_2+3b_3)=0,$this means $(a_1+b_1,a_2+b_2,a_3+b_3)=(a_1,a_2,a_3)+(b_1,b_2,b_3)$ is also in $\{(x_1,x_2,x_3)\in\mathbb F^3:x_1+2x_2+3x_3=0\}$.

Closed under scalar multiplication: if $(a_1,a_2,a_3)\in \{(x_1,x_2,x_3)\in\mathbb F^3:x_1+2x_2+3x_3=0\}$, then $a_1+2a_2+3a_3=0$. For any $\lambda\in\mathbb F$, we have$\lambda a_1+2(\lambda a_2)+3(\lambda a_3)=\lambda(a_1+2a_2+3a_3)=0.$This means $\lambda(a_1,a_2,a_3)=(\lambda a_1,\lambda a_2,\lambda a_3)\in \{(x_1,x_2,x_3)\in\mathbb F^3:x_1+2x_2+3x_3=0\}.$

(b) $\{(x_1,x_2,x_3)\in\mathbb F^3:x_1+2x_2+3x_3=4\}$ is not a subspace of $\mathbb F^3$ since $(0,0,0)$ is not in it.

(c) $\{(x_1,x_2,x_3)\in\mathbb F^3:x_1 x_2 x_3=0\}$ is not a subspace of $\mathbb F^3$ since $(1,1,0)$ and $(0,1,1)$ are in it, but the sum $(1,2,1)=(1,1,0)+(0,1,1)$ is not in it.

(d) $\{(x_1,x_2,x_3)\in\mathbb F^3:x_1=5x_3\}$ is a subspace of $\mathbb F^3$.

Additive identity: it is clear that the additive identity $(0,0,0)$ of $\mathbb F^3$ is contained in $\{(x_1,x_2,x_3)\in\mathbb F^3:x_1=5x_3\}$.

Closed under addition: if $(a_1,a_2,a_3),(b_1,b_2,b_3)\in \{(x_1,x_2,x_3)\in\mathbb F^3:x_1=5x_3\}$, then $a_1=5a_3$ and $b_1=5b_3$. Hence$a_1+b_1=5a_3+5b_3=5(a_3+b_3),$this means $(a_1+b_1,a_2+b_2,a_3+b_3)=(a_1,a_2,a_3)+(b_1,b_2,b_3)$ is also in $\{(x_1,x_2,x_3)\in\mathbb F^3:x_1=5x_3\}$.

Closed under scalar multiplication: if $(a_1,a_2,a_3)\in \{(x_1,x_2,x_3)\in\mathbb F^3:x_1=5x_3\}$, then $a_1=5a_3$. For any $\lambda\in\mathbb F$, we have $\lambda a_1=\lambda(5 a_3)=5(\lambda a_3)$. This means $\lambda(a_1,a_2,a_3)=(\lambda a_1,\lambda a_2,\lambda a_3)\in \{(x_1,x_2,x_3)\in\mathbb F^3:x_1=5x_3\}.$

2. Solution: (a) If this set is a subspace of $\mathbb F^4$, then $(0,0,0,0)\in \mathbb F^4$, then $0=5\cdot 0+b$. Hence $b=0$. Follow the same steps of Problem 1 (d), we will see the set is a subspace of $\mathbb F^4$ if $b=0$.

(b) (c) and (d) is similar to Problem 3 and 4.

Now let us consider (e). Denote the set of all sequences of complex numbers with limit $0$ by $A$.

Additive identity: it is clear that $(0,0,\cdots)\in A$.

Closed under addition: if $(a_1,a_2,\cdots),(b_1,b_2,\cdots)\in A$, then$\lim_{n\to\infty}a_n=0\quad\text{and}\quad\lim_{n\to\infty}b_n=0.$It is easy to see$\lim_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}a_n+\lim_{n\to\infty}b_n=0+0=0.$This means $(a_1+b_1,a_2+b_2,\cdots)=(a_1,a_2,\cdots)+(b_1,b_2,\cdots)\in A$.

Closed under scalar multiplication: if $(a_1,a_2,\cdots)\in A$, then$\lim_{n\to\infty}a_n=0.$For any $\lambda\in\mathbb C$, it is easy to see$\lim_{n\to\infty}(\lambda a_n)=\lambda\lim_{n\to\infty} a_n=\lambda 0=0.$This means $\lambda(a_1,a_2,\cdots)=(\lambda a_1,\lambda a_2,\cdots)\in A$.

3. Solution: Denote the set of differentiable real-valued functions $f$ on the interval $(-4,4)$ such that $f'(-1)=3f(2)$ by $V$.

Additive identity: it is clear that the constant function $f\equiv 0$ is contained in $V$.

Closed under addition: if $f,g\in V$, then $f$ and $g$ are differentiable real-valued functions. So is $f+g$. Moreover,$(f+g)'(-1)=f'(-1)+g'(-1)=3f(2)+3g(2)=3(f(2)+g(2))=3(f+g)(2).$This concludes $V$ is closed under addition.

Closed under scalar multiplication: if $f\in V$, for any $\lambda\in \mathbb R$, then $f$ is differentiable real-valued functions. So is $\lambda f$. Moreover,$(\lambda f)'(-1)=\lambda f'(-1)=\lambda (3f)(2)=3(\lambda f)(2).$This deduces $V$ is closed under scalar multiplication.

4. Solution: Denote the set of continuous real-valued functions $f$ on the interval $[0,1]$ such that $\int_0^1f=b$ by $V_b$.

If $V_b$ is a subspace of $\mathbb R^{[0,1]}$, then for any $f\in V_b$, we have $\int_0^1f=b$. Because $V_b$ is a subspace of $\mathbb R^n$, it follows that $kf\in V_b$ for any $k\in\mathbb R$. Hence$b=\int_0^1(kf)=k\int_0^1f=kb,\quad \text{for all}~k\in\mathbb R,$this happens if and only if $b=0$.

Now if $b=0$, then for any $f,g\in V_0$ and $\lambda\in\mathbb R$. We have that$\int_0^1(f+g)=\int_0^1f+\int_0^1g=0+0=0$and $f+g$ is continuous real-valued functions since $f$ and $g$ are. This deduces $f+g\in V_0$, i.e. $V_0$ is closed under addition. Similarly,$\int_0^1(\lambda f)=\lambda\int_0^1f=k0=0$and $\lambda f$ is continuous real-valued functions since $f$ is. This implies $\lambda f\in V_0$, i.e. $V_0$ is closed under scalar multiplication. On the other hand, the constant function $f\equiv 0\in V_0$, which is also the additive identity in $\mathbb R^{[0,1]}$. Hence $V_0$ is a subspace of $\mathbb R^n$ by 1.34.

5. Solution: Note that we consider complex vector space, so if $\mathbb R^2$ is a subspace of the complex vector space $\mathbb C^2$, then$i(1,1)=(i,i)\in\mathbb R^2,$we get a contradiction. Hence $\mathbb R^2$ is not a subspace of the complex vector space $\mathbb C^2$.

6. Solution: (a) Because $a^3=b^3$ if and only if $a=b$ in $\mathbb R$, hence $\{(a,b,c)\in\mathbb R^3:a^3=b^3\}=\{(a,b,c)\in\mathbb R^3:a=b\}$is obviously a subspace of $\mathbb R^3$ by the similar arguments in Problem 1 and Problem 2.

(b) Note that$x=\left(1,\frac{-1+\sqrt{3}i}{2},0\right)\in\{(a,b,c)\in\mathbb C^3:a^3=b^3\}$and$y=\left(1,\frac{-1-\sqrt{3}i}{2},0\right)\in\{(a,b,c)\in\mathbb C^3:a^3=b^3\}.$However,$x+y=(2,-1,0)\notin \{(a,b,c)\in\mathbb C^3:a^3=b^3\}.$This implies $\{(a,b,c)\in\mathbb C^3:a^3=b^3\}$ is not closed under addition, hence not a subspace of $\mathbb C^3$.

7. Solution: Denote $\{(x,y)\in\mathbb R^2:x,y\in\mathbb Z\}$ by $U$, then $U$ is not empty. If $(x_1,y_1)\in A$ and $(x_2,y_2)\in A$, then $x_1,x_2,y_1,y_2\in\mathbb Z$. Hence $x_1+x_2$ and $y_1+y_2$ are integers. This means $(x_1+x_2,y_1+y_2)=(x_1,y_1)+(x_2,y_2)\in U$, i.e. $U$ is closed under addition. Similarly, since $(-x_1,-y_1)\in U$, it follows that $U$ is closed under additive inverses. However, $U$ is not closed under scalar multiplication since $(1,1)\in A$ while $\dfrac{1}{2}(1,1)\notin U$. Hence $U$ is not a subspace of $\mathbb R^2$.

8. Solution: Denote $\{(x,y)\in \mathbb R^2: x=0~\text{or}~y=0\}$ by $U$, then $U$ is not empty. If $(x,0)\in U$, then for any $\lambda\in\mathbb R$, we have$\lambda(x,0)=(\lambda x,0)\in U.$Similarly, $\lambda(0,y)\in U$, hence $U$ is closed under scalar multiplication. However, $(1,0),(0,1)\in U$ while $(1,1)=(1,0)+(0,1)\notin U$. This implies $U$ is not closed under addition, hence not a subspace of $\mathbb R^2$.

9. Solution: Denote the set of periodic functions from $\mathbb R$ to $\mathbb R$ by $S$. Then $S$ is not a subspace of $\mathbb R^{\mathbb R}$. Otherwise, we have $h(x)=\sin\sqrt{2}x+\cos x$ since both $f(x)=\sin\sqrt{2}x$ and $g(x)=\cos x$ are periodic functions from $\mathbb R$ to $\mathbb R$. Assume there exists a positive number $p$ such that $h(x)=h(x+p)$ for all $x\in\mathbb R$, then $1=h(0)=h(p)=h(-p)$. It is equivalent to $1=\cos p+\sin\sqrt{2}p=\cos p-\sin\sqrt{2}p,$this implies $\sin\sqrt{2}p=0$ and $\cos p=1$. $\cos p=1$ deduces that $p=2k\pi$ where $k\in \mathbb Z$. However $\sin\sqrt{2}p=0$ implies $\sqrt{2}p=l\pi$ where $l\in\mathbb Z$. Hence $\sqrt{2}=\frac{l\pi}{2k\pi}=\frac{l}{2k}\in\mathbb Q,$which is impossible. Therefore we get the conclusion.

10. Solution: Additive identity: by definition $0\in U_1$ and $0\in U_2$, hence $0\in U_1\cap U_2$.

Closed under addition: if $x\in U_1\cap U_2$ and $y\in U_1\cap U_2$, then $x\in U_1$ and $y\in U_1$, hence $x+y\in U_1$ for $U_1$ is closed under addition. Similarly, $x+y\in U_2$. Therefore $x+y\in U_1\cap U_2$.

Closed under scalar multiplication: if $x\in U_1\cap U_2$, then $x\in U_1$. Then for any $\lambda\in\mathbb F$, we have $\lambda x\in U_1$ since $U_1$ is closed under scalar multiplication. Similarly, $\lambda x\in U_2$. Therefore $\lambda x\in U_1\cap U_2$.

11. Solution: Assume $U_i$ are subspaces of $V$, where $i\in I$. Now we will show $\cap_{i\in I}U_i$ is a subspace of $V$.

Additive identity: by definition $0\in U_i$ for every $i\in I$, hence $0\in \cap_{i\in I}U_i$.

Closed under addition: if $x\in \cap_{i\in I}U_i$ and $y\in \cap_{i\in I}U_i$, then for any given $i\in I$, we have $x\in U_i$ and $y\in U_i$, hence $x+y\in U_i$ for $U_i$ is closed under addition. Therefore $x+y\in \cap_{i\in I}U_i$.

Closed under scalar multiplication: if $x\in \cap_{i\in I}U_i$, then $x\in U_i$ for any given $i\in I$. Then for any $\lambda\in\mathbb F$, we have $\lambda x\in U_i$ since $U_i$ is closed under scalar multiplication. Therefore $\lambda x\in \cap_{i\in I}U_i$.

12. Solution: Suppose $U$ and $W$ are two subspaces of $V$. We argue it by contradiction. If $U\cup W$ is a subspace of $V$, moreover $U\nsubseteq W$ and $W\nsubseteq U$. Consider $u\in U\setminus W$ and $w\in W\setminus U$, then $u+w\in U\cup W$ since $U\cup W$ is a subspace of $V$. Hence $u+w\in U$ or $W$. If $u+w\in U$, then $w=(u+w)-u\in U$. We get a contradiction. If $u+w\in W$, then $u=(u+w)-w\in W$. We also get a contradiction. Hence if $U\cup W$ is a subspace of $V$, we must have $U\subset W$ or $W\subset U$.

If $U\subset W$ or $W\subset U$. Without loss of generality, we can assume $U\subset W$. Then $U\cup W=W$ is obviously a subspace of $V$.

14. Solution: It is clear that $U$ and $W$ are subspaces of $\mathbb{F}^4$. Now assume that $(x_1,x_1,y_1,y_1)\in U$ and $(x_2,x_2,x_2,y_2)\in W$, then \begin{align*} &(x_1,x_1,y_1,y_1)+(x_2,x_2,x_2,y_2)\\=&(x_1+x_2,x_1+x_2,y_1+x_2,y_1+y_2) \in \{(x,x,y,z):x,y,z\in\mathbb F\}. \end{align*} Hence $U+W\subset \{(x,x,y,z):x,y,z\in\mathbb F^4\}$. For any $x,y,z\in\mathbb F$, we have $(0,0,y-x,y-x)\in U$ and $(x,x,x,z+x-y)\in W$. However, $(x,x,y,z)=(0,0,y-x,y-x)+(x,x,x,z+x-y)\in U+W,$hence $\{(x,x,y,z):x,y,z\in\mathbb F^4\}\subset U+W$. Combining this with previous argument, it follows that $U+W=\{(x,x,y,z):x,y,z\in\mathbb F\}$.

15. Solution: Because $U$ is a subspace of $V$, hence closed under addition.Therefore for any $x,y\in U$, we have $x+y\in U$, i.e. $U+U\subset U$. Note that if $x\in U$, then $x=x+0\in U+U$, hence $U\subset U+U$. Combining with these arguments, it follows that $U+U=U$.

16. Solution: For $x\in U$ and $y\in W$, because addition in V is commutative, we have $x+y=y+x\in W+U$. This implies $U+W\subset W+U$. Similarly, we have $W+U\subset U+W$. Hence $U+W=W+U$.

17. Solution: Note that in V, we have $(x+y)+z=x+(y+z)$. Hence this is similar to Problem 16. Let $x_i\in U_i$, $i=1,2,3$, then $(x_1+x_2)+x_3=x_1+(x_2+x_3)\in U_1+(U_2+U_3).$Since every element in $(U_1+U_2)+U_3$ can be expressed as the form $(x_1+x_2)+x_3$, it follows that $(U_1+U_2)+U_3\subset U_1+(U_2+U_3)$. Similarly, we also have $U_1+(U_2+U_3)\subset (U_1+U_2)+U_3$. Hence $(U_1+U_2)+U_3=U_1+(U_2+U_3)$.

18. Solution: If $U$ is a additive identity, then for any subspace $W$ of $V$, we have $U+W=W$. This means $U\subset W$ by the similar arguments in Problem 15 and 16. Hence the only possibility is $U=0$, in fact this is the case. Hence $0$ is the additive identity. Suppose subspace $W$ of $V$ has additive inverses, then there exists a subspace $S$ of $V$ such that $W+S=0$. This can only happen when $W=0$ since $W\subset W+S$.

19. Solution: Here is a counterexample. Let $V=U_1=\{(a,b)\in\R^2:a,b\in\R\}$, $U_2=\{(a,0)\in\R^2:a\in\R\}$ and $W=\{(0,b)\in\R^2:b\in\R\}$. Then it is easy to see that $U_1+W=U_2+W$, but $U_1\ne U_2$.

See comments for the example from Neven Sajko.

20. Solution: Take $W=\{(0,z,0,w)\in\mb F^4:z,w\in\mb F\}$. For any $(x,y,z,w)\in\mb F^4$, we have$(x,y,z,w)=(x,x,z,z)+(0,y-x,0,w-z)\in U+W$since $(x,x,z,z)\in U$ and $(0,y-x,0,w-z)\in W$. We have $\mb F^4=U+W$.

Moreover, if $(x,y,z,w)\in U\cap W$, then we must have $x=y$ and $z=w$ since $(x,y,z,w)\in U$.

Similarly, since $(x,y,z,w)\in W$, we have $x=0$ and $z=0$. Therefore, $x=y=0$ and $z=w=0$, hence $(x,y,z,w)=(0,0,0,0)$. It follows that $U\cap W=\{0\}$. Hence $F=U\oplus W$ by 1.45.

21. Solution: Since we can only choose the first two coordinates arbitrarily in $U$, we can take $W$ in which the first two coordinates are zero and the last three are variables.

Let $W=\{(0,0,z,w,s)\in\mb F^5:x,y\in\mb F\}$. If $(x,y,z,w,s)\in U\cap W$, then $x=y=0$ since $(x,y,z,w,s)\in W$. Moreover, we have $z=x+y,w=x-y,s=2x$ since $(x,y,z,w,s)\in U$. Therefore $z=w=s=0$, hence $U\cap W=\{0\}$.

Again, for any $(x,y,z,w,s)\in\mb F^5$, we have$(x,y,z,w,s)=(x,y,x+y,x-y,2x)+(0,0,z-x-y,w-x+y,s-2x)\in U+W$since $(x,y,x+y,x-y,2x)\in U$ and $(0,0,z-x-y,w-x+y,s-2x)\in W$. Therefore $\mb F^5=U+W$.

Since $U\cap W=0$, it follows from 1.45 that $\mb F^5 =U\oplus W$.

22. Solution: Let $W_1=\{(0,0,z,0,0)\in\mb F^5:x,y\in\mb F\}$, $W_2=\{(0,0,0,z,0)\in\mb F^5:x,y\in\mb F\}$ and $W_3=\{(0,0,0,0,z)\in\mb F^5:x,y\in\mb F\}$. By the same argument as in Problem 21, we have$\mb F^5 =U\oplus W_1\oplus W_2\oplus W_3.$

23. Solution: Here is a counter example. Let $V=\R^2$, $U_1=\{(x,0)\in\R^2:x\in\R\}$, $U_2=\{(0,y)\in\R^2:y\in\R\}$ and $W=\{(z,z)\in\R^2:z\in\R\}$. By the same argument as in Problem 21, one has$V=U_1\oplus W \quad\text{and}\quad V=U_2\oplus W,$however $U_1\ne U_2$.

This is exactly the example from Neven Sajko.

24. Solution: Given any $f\in \R^\R$, define $f_e(x)$ to be $\dfrac{f(x)+f(-x)}{2}$ and define $f_o(x)$ to be $\dfrac{f(x)-f(-x)}{2}$ for all $x\in\R$ . Then $f_e,f_o\in\R^\R$. Moreover, for all $x\in \R$, we have$f_e(-x)=\frac{f(-x)+f(x)}{2}=\frac{f(x)+f(-x)}{2}=f_e(x)$and$f_o(-x)=\frac{f(-x)-f(x)}{2}=-\frac{f(x)-f(-x)}{2}=-f_o(x).$Hence $f_e\in U_e$ and $f_o\in U_o$. Note that we also have$f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}=f_e(x)+f_o(x),$hence $f=f_e+f_o\in U_e+U_o$. Since we can choose $f$ arbitrarily, one has $\R^\R=U_e+U_o$.

By 1.45, to show $\R^\R=U_e\oplus U_o$, it suffices to prove that $U_e\cap U_o=\{0\}$. Let $f\in U_e\cap U_o$, then $f(x)=f(-x)$ since $f\in U_e$ and $f(x)=-f(-x)$ since $f\in U_o$ for all $x\in\R$ . Sum up $f(x)=f(-x)$ and $f(x)=-f(-x)$, we have $f(x)=0$ for all $x\in \R$. Hence $f=0$, which implies $U_e\cap U_o=\{0\}$.

### This Post Has 51 Comments

1. For number 7 is it really valid to say that (x,y) is an element of R^2 only to immediately say that x and y are integers? It seems like a cheap trick and I'm skeptical if that would be a valid answer. I unfortunately don't have a better solution, so I guess I'll have to accept it for now, but it would be nice if I could hear some more justification.

2. I may be missing some basics. but i dont understand how in q14 one can write that (x1+x2,x1+x2,x1+y1,y1+y2) belongs to {(x,x,y,z), where x,y,z belongs to F. specially x1+y1 belonging to y and y1+y2 belonging to z

3. A possibly easier (and less general) proof of #13 compared to the links above. Let the three subspaces be $U_1$, $U_2$ and $U_3$. We consider two cases of $U_1$ and $U_2$: 1) one contains the other, and 2) neither one contains the other.
1): Suppose $U_1 \subseteq U_2$, then $U_1 \cup U_2 \cup U_3 = U_2 \cup U_3$. We can use the result of #12, so one of $U_2$ and $U_3$ contains the other.
2): We can find $u_1 \in U_1 \setminus U_2$ and $u_2 \in U_2 \setminus U_1$. By the result of #12, $u_1+u_2 \notin U_1 \cup U_2$. Since $U_1 \cup U_2 \cup U_3$ is a subspace, $u_1 + u_2 \in U_1 \cup U_2 \cup U_3$, so $u_1 + u_2 \in U_3$. If the field is not $\{0,1\}$, we can find another $c(u_1 + u_2) \in U_3$.

In this case, we'd like to prove that $U_3$ contains both $U_1$ and $U_2$. For the sake of contradiction, suppose $u_0 \in U_1 \setminus U_3$. Because $U_1 \cup U_2 \cup U_3$ is a subspace, both $u_0 + u_1 + u_2$ and $u_0 + c(u_1 + u_2)$ should be in it. Neither of these two can be in $U_1$ since $u_0$ and $u_1$ are in $U_1$, so it would mean $u_2 \in U_1$. Neither of these can be in $U_3$ since $u_1 + u_2$ and $c(u_1 + u_2)$ are in $U_3$, so it would mean $u_0 \in U_3$. Thus, both of them are in $U_2$. Then, their difference $(c-1)(u_1+u_2)\in U_2$, which means $(c-1)u_1\in U_2$. If the field is not $\{0,1\}$, this means $u_1 \in U_2$. Contradiction. Therefore, $U_1 \subseteq U_3$. The reasoning for $U_2$ is similar.

The other direction is simple: suppose $U_3$ contains $U_1$ and $U_2$, then the union of three is just $U_3$, which is a subspace.

4. I'm confused about 22 being solved by the same line of reasoning as 21.

The book explicitly states that the union-checking proving method for direct sums doesn't generalize past 2 sumands:

Regarding proof:
"Suppose U and W are subspaces of V. Then U C W is a direct sum if and only if: U union W = {0}."

The book states:
"The result above deals only with the case of two subspaces. When asking about a possible direct sum with more than two subspaces, it is not enough to test that each pair of the subspaces intersect only at 0. To see this, consider Example 1.43. In that nonexample of a direct sum, we have:

U_1 union U_2 = U_1 union U_3 = U_2 union U_3 = {0}."

1. That's a good point. I think we need to use the 'only one way to write summation' argument here.

5. In question 14,why can we directly prove U+W=（x,x,y,z）…？？

1. Given two sets A and B, if A ⊆ B (A is a subset of B) and B ⊆ A (B is a subset of A), then A = B.
In question 14, the answer first proves (U+W) ⊆ {(x,x,y,z):x,y,z∈F4} by picking an arbitrary element from (U+W), and expresses it by an element in the latter.
Then the answer proves {(x,x,y,z):x,y,z∈F4} ⊆ (U+W) by picking an arbitrary element from the former (namely (x,x,y,z)), and then expresses it by an element in (U+W).
By doing these two things, it proves that (U+W) = {(x,x,y,z):x,y,z∈F4}

6. For 9. "sin(sqrt(2)*p) = 0 and cos(p) = 1" doesn't immediately follow from 1 = cos(p) + sin(sqrt(2)*p) = cos(p) - sin(sqrt(2)*p), since there are many ways to get 1 as a sum of sin and cos.
A simpler solution is to define a piecewise function f(x) = 1 if x is an integer and 0 otherwise; g(x) = 1 if x is an integer multiple of pi and 0 otherwise. Then f(x) + g(x) = 2 if and only if x = 0, hence not periodic.

7. For #9, I understand the solution but can't we just simple show that there are multiple additive identities which is not possible for a vector space?

8. 20.
Would taking W = {(a, 0, 0, 0) ∈ F⁴ | a ∈ F} be easier and do the job?

W is a subspace of F, bc it has 0, it's is closed under addition (∀y we have a+y ∈ F) and multiplication (∀λ we have λa ∈ F).

There is only one way to express 0 in terms of U + W. Namely, by taking 0 from U and 0 from W. To see why let's start by fixing y = 0 in (x, x, y, y) ∈ U. It can't be otherwise, bc any element from W is not able to change those. Likewise for the second coordinate, thus x has to be equal to 0. This forces a in (a, 0, 0, 0) ∈ W to be equal to 0.
This means, by 1.44 Condition for a direct sum, that U + W is a direct sum.

Cheers!

I believe you are right when you say you can express 0 in terms of U + W, that's fine.
But I think you are missing the second requirement for it to be a direct sum.
We need every vЄF4 to be represented in exactly one way in U + W, and this does not happen with the W you defined.
Take v=(1,2,3,4)ЄF4. It can't be represented in U + W.

A neat way to approach this type of problems is to think of what is it that you are missing from F4 in the given U, always keeping in mind you need to avoid repetition.

Cheers :)

9. I was wondering where the idea for solving 6b comes from, and I remember Axler drops a hint of one of the complex cube roots of 1 in exercise 1.A.2. Clever.

with the point f'(-1)=0 and f(2)=0, but others area of [-4, 4] are continuous and deritive ? so this condition are more loose than all the range equals 0!

11. 23.
U_1=V-W+{0}
U_2=V-W+{0}

12. I do not understand the work presented for Q14. Isn't it enough to show that the addition of any two arbitrary elements (one belonging to U and the other to W) is equal to the set {(x,x,y,z) 'element' F^4 : x,y,z 'element' F} ? In short, doesn't the first part of your argument suffice?

1. The first part shows the addition of any two arbitrary elements is of form $(x,x,y,z)$. The second part shows all elements of form $(x,x,y,z)$ can be written as the addition of any two arbitrary elements.

Recall that if you would like to show two sets $A$ and $B$ are equal, then we usually show $A\subset B$ and $B\subset A$. That is why we have two parts here.

1. thank you so much!

13. I seem to have found a mistake. In question 14, it should be written like this"$x,y,z \in \mathbb{F}$", but you wrote"$x,y,z \in \mathbb{F}^4$".Right?

14. I don't quite follow the logic in #12. You say $u + w \in U \cup W$ since $U \cup W$ is a subspace of $V$, but $u + w$ is in $U \cup W$ by construction, not because $U \cup W$ is a subspace of V. Then, I'm not sure how it's asserted that $u + w$ must be in $U$ or $W$.

1. Oops, I misread. Got it now!

15. For 19: The examples given by @nevensajko:disqus seem needlessly complicated. Rather than making W the identity function (x = y), make W the entire xy plane. Then U1 (x-axis) and U2 (y-axis) are obviously not equal, but as each is a subset of the plane, U1 + W = W and U2 + W = W which means U1 + W = U2 + W

16. Exercise 24 means any function in R^R can be decomposed into the sum of two other functions, one odd and one even. Amazing!

17. Another solution for 8:
$$\left\{(x, y) \in \mathbb{R}^{2} :|x|=|y|\right\}$$

18. Another possible solution: $\left\{(x, y, z) \in \mathbb{R}^{2} \backslash(0,0,0)\right\}$

1. However, since you have an additive inverse, you must have an addiitive identity

19. 7 Another solution: $\{(x,y,z) \in \mathbb{R}^2 \setminus(0,0,0) \}$

1. if $u,-u\in U$,then $0= u+(-u)$ must be in $U$, otherwise it is not closed under addition.

20. I don't agree #14). It appears as though the second part of your argument where (x,x,x,z+x-y) ∈ W. I do not think this is valid. The (0,0,y-x,y-x) ∈ U is true if you use the closure by addition.

1. Given W = {(x, x, x, y) ∈ F^4: x, y ∈ F}, for any x, y, z ∈ F^4, (x, x, x, z+x-y) ∈ W.

21. 6) b) doesn't seem right. The vectors you chosen for x and y does not satisfy the condition a^3 = b^3; that condition isn't correct, the proof appears wrong and the answer should be that {(a,b,c) C^3: a^3 = b^3} is indeed a subspace of C^3

From the book: Definition of Subspace
A subset U of V is a subspace of V if and only if U satisfies the following
three conditions:
1) 0 ∈ U
2) u,w ∈ U
3) a ∈ F, u ∈ U implies au ∈ U

where F can be real or complex vector space.

since you're trying to say that x,y ∈ {(a,b,c), C^3: a^3 = b^3} and the vectors you chose are not part of the set. So, it's not right.

22. In 9, it is written "1 = cos p + sin sqrt2 p = cos p - sin sqrt2 p
this implies $\sin \sqrt{2 p} = 0$ and $\cos p = 1$"
but it is not necessarily so.
Numerically, p=1.45459662, sin sqrt2 p = 0.884061615 (not 0), cos p = 0.115938388 (not 1), is essentally 1.

1. btw, solution 9 is a very clever construction of a counter example.

23. For #11, showing that the intersection is closed under scalar multiplication we have: $x \in \cap_{i \in I} U_i \Rightarrow x \in U_i$. This is fine, but how do we suppose that $\lambda x \in U_i \Rightarrow \lambda x \in \cap_{i \in I} U_i$? Isn't it entire plausible that $\lambda x \in U_i$, but $\lambda x \notin \cap_{i \in I} U_i$?

1. Hmm, LaTeX doesn't appear to be supported any longer. So, my question is essentially: How does k*x ∈ U_i imply that k*x ∈ intersection of U_i's? Isn't it entirely plausible that x ∈ intersection of U_i's but that k*x is not ∈ intersection of U_i's (understanding that k*x ∈ U_i by definition)?

2. if $x\in U_i$, then $kx\in U_i$ because $U_i$ is closed under scalar multiplication. similarily, if $x$ is the element of the union of $U_i$, then $x\in U_i$ ($1\leq i\leq n$), so $kx\in U_i$ ($1\leq i\leq n$), which means $kx$ is also the element of the union.

24. 24

Space of real functions: R^R.
Uₑ= {f∈R^R : f(-x)=f(x)} (The set of even functions.)
Uₒ= {f∈R^R : f(-x)=-f(x)} (The set of odd functions.)
Show that R^R = Uₑ⊕Uₒ.

Proof:
Firstly we prove Uₑ,Uₒ≤R^R.
0∈Uₑ; ∀x∈R 0(x)=0; 0(-x)=0=0(x).
f,g∈Uₑ → (f+g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f+g)(x).
g∈Uₑ, λ∈R → (λg)(-x) = λg(-x) = λg(x) = (λg)(x).
Similarly Uₒ≤R^R.

We need to prove Uₑ∩Uₒ = {0}.
f∈Uₑ∩Uₒ → f(x) = f(-x) = -f(x) → f=-f → f=0.
Now we only need to prove Uₑ+Uₒ=R^R.
Uₑ+Uₒ⊆R because Uₑ,Uₒ⊂R^R and R^R is closed under addition.
Uₑ+Uₒ⊇R because ∀f∈R^R ∀x∈R f(x) = ½f(x) + ½f(x) + ½f(-x) -½f(-x) = ½(f(x)+f(-x)) + ½(f(x)-f(-x)) where the first summand is from Uₑ and the second from Uₒ.
QED

25. 23

The counter example from 19 still works.

1. No it doesnt work, for this is direct sum and so the answer is wrong.
For a direct sum a 0 can be written in only one way.
Your wrong solution allows it to be written in two ways, for example for W+U2:
(0,0)+(0,0) and also (0,-4)+(0,4)=(0,0) (and infinite others).
I think this exercise indeed has no counterexample and it can be proven.

26. 22

Just decompose W from exercise 21 into 3 parts.

27. 21

W = {(0,0,a,b,c)∈F⁵ : a,b,c∈F}

28. 20

W = {(0,x,y,0) : x,y∈F}

1. Can I ask why this is the case? It would appear to me that, using the W you've provided, that U+W=(x,x,y,y), which, doesn't seem to me to be all of F^4.

1. I think it should be W={ (0,y,x,0) : x,y in F}.

2. consider U = {(x1,x1,y1,y1):x1,y1 ∈ F}
and W = {(0,x2,y2,0) : x2,y2∈F}

F^4 = U+W = {(x1, x1+x2, y1+y2, y1) : x1, y1, x2, y2 ∈ F}

2. Indeed, it works if W is one of the four subspaces:

{(0, x, 0, y) : x, y ∈ F}
{(0, x, y, 0) : x, y ∈ F}
{(x, 0, 0, y) : x, y ∈ F}
{(x, 0, y, 0) : x, y ∈ F}

and many more.

3. How is it so when U ∩ W is {(0,x,y,0) : x,y ∈ F} and not {0} (correct me if I'm wrong)?

29. 19

A counter example:

U₁= {λ(1,0) : λ∈F} (The x axis.)

U₂= {λ(0,1) : λ∈F} (The y axis.)

W = {λ(1,1) : λ∈F} (The x=y line/the identity function graph.)

Obviously U₁≠U₂, but:

U₁+ W = {a(1,0)+b(1,1) : a,b∈F}

= {(a(1,0)-b(1,0))+b(1,1) : a,b∈F} (Because U₁ is closed under addition.)

= {a(1,0)+b(0,1) : a,b∈F} (Note that this is the x,y plane.)

= {a(1,0)+(a(0,1)+b(0,1)) : a,b∈F}

= {a(1,1)+b(0,1) : a,b∈F}

= W + U₂ = U₂+ W

1. I don't see clear how you do the step on the second equality of U_1+W.
I understand that U_1 is closed under addition because it is a vectorspace but why that imply that doesn't mean that a(1,0) = a(1,0)-b(1,0)

1. Because $a$ can be any number in $\mathbb F$. You can think it as $a(1,0)=\tilde a(1,0)-b(1,0)$where $\tilde a=a+b$ and then replace $\tilde a$ with $a$.