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 $\sqrt{2}p=2l\pi$ where $l\in\mathbb Z$. Hence \[\sqrt{2}=\frac{2l\pi}{2k\pi}=\frac{l}{k}\in\mathbb Q,\]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 $V$.


13. See If a field $\mb F$ is such that $|\mb F|>n−1|$ why is $V$ a vector space over $\mb F$ not equal to the union of proper subspaces of $V$ or A question on vector space over an infinite field.


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^4\}. \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 $U+W=\{(x,x,y,z):x,y,z\in\mathbb F^4\}$.


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 $\mb F^=U+W$.

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 $U_e\cap U_o=\{0\}$.


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