1. Solution: We just need to show that $v_1$, $v_2$, $v_3$, $v_4$ can be expressed as linear combination of $v_1-v_2$, $v_2-v_3$, $v_3-v_4$, $v_4$. Note that \[v_1=(v_1-v_2)+(v_2-v_3)+(v_3-v_4)+v_4,\]\[v_2=(v_2-v_3)+(v_3-v_4)+v_4,\]\[v_3=(v_3-v_4)+v_4,\quad v_4=v_4.\]
2. Solution: (a) If $v\ne 0$, then $av=0$ means $a=0$ by problem 2 in exercise 1B, hence $v\in V$ is linearly independent. Conversely, if $v\in V$ is linearly independent, then $v\ne 0$. Otherwise, we have $1v=v=0$, i.e. $v$ is linearly dependent. This is a contradiction.
(b) If $v_1\in V$, $v_2\in V$ is linearly independent, then neither vector is a scalar multiple of the other. Otherwise, without loss of generality, we can assume $v_1=cv_2$, then $1v_1+(-c)v_2=0$. It follows that $v_1\in V$, $v_2\in V$ is linearly dependent. We get a contradiction. Conversely, if $v_1\in V$, $v_2\in V$ is linearly dependent, then there exist $a$ and $b$ such that $av_1+bv_2=0$ and $a$ and $b$ are not both zero. Without loss of generality, we can assume $a\ne 0$, then $av_1+bv_2=0$ means $v_1=-\dfrac{b}{a}v_2$. We also get a contradiction.
(c) If there exist $x,y,z,w\in\mathbb F$ such that \[ x(1,0,0,0)+y(0,1,0,0)+z(0,0,1,0)=0, \]then it means $(x,y,z,0)=(0,0,0,0)$. Hence $x=y=z=0$, it follows that the list is linearly independent in $\mathbb F^4$.
(d) We just need the sentence before definition 2.12, that is, “Conclusion: the coefficients of a polynomial are uniquely determined by the polynomial”. Then use definition 2.17 and the similar method as (c), we can prove this case.
3. Solution: Let us consider the following equations \begin{equation}\label{2A2}3x+2y=5,\quad x-3y=9.\end{equation} We can get a solution $x=3,y=-2$. Hence let\[t=3\cdot 4+(-2)\cdot 5=2.\]Then we have \begin{equation}\label{2A1}3(3,1,4)+(-2)(2,-3,5)=(5,9,2),\end{equation} this means $t=2$ is the desired solution.
Remark: From $(\ref{2A1})$, one can know why we get these equations in $(\ref{2A2})$.
4. Solution: We already see that if $c=8$, the list is linearly dependent. Now we show that if the list is linearly dependent, then $c=8$. Suppose there are exist $x$, $y$ and $z$ such that they are not all zero and \begin{equation}\label{2A3}x(2,3,1)+y(1,-1,2)+z(7,3,c)=0.\end{equation} Then we have \[2x+y+7z=0\quad\text{and}\quad 3x-y+3z=0.\]From these equations, by solving in $x$ and $y$, we get $x=-2z$ and $y=-3z$. Since $x$, $y$, $z$ are not all zero, $z$ is not zero. However, $(\ref{2A3})$ also means $x+2y+cz=0$, plugging $x=-2z$ and $y=-3z$, we get \[-2z+2(-3)z+cz=0\iff (c-8)z=0.\]Hence we deduce that $c=8$ since $z\ne0$.
Using this method, we can also solve Problem 3.
5. Solution:
(a) Let $x$ and $y$ be in $\mathbb R$, then if $x(1+i)+y(1-i)=0$, we have \[0=x(1+i)+y(1-i)=(x+y)+(x-y)i.\]Hence $x+y=0$ and $x-y=0$, it follows that $x=0$ and $y=0$. Thus the list $(1+i,1-i)$ is linearly independent over $\mathbb R$.
(b) For this case, note that\[i(1+i)+1(1-i)=(i-1)+(1-i)=0,\]we conclude the list $(1+i,1-i)$ is linearly dependent over $\mathbb C$.
6. Solution: Suppose there exist numbers $x$, $y$, $z$, $w$ in the field such that \[x(v_1-v_2)+y(v_2-v_3)+z(v_3-v_4)+wv_4=0,\]then \[xv_1+(y-x)v_2+(z-y)v_3+(w-z)v_4=0.\]Because $v_1$, $v_2$, $v_3$, $v_4$ is linearly independent in $V$, it follows that \[x=0,\quad y-x=0,\quad z-y=0,\quad w-z=0.\]Hence we get $x=y=z=w=0$. This means the list\[v_1-v_2,v_2-v_3,v_3-v_4,v_4\] is also linearly independent in $V$.
7. Solution: This is true. For if there exist $a_1$, $\cdots$, $a_m$ $\in\mathbb F$ such that \[a_1(5v_1-4v_2)+a_2v_2+\cdots+a_mv_m=0,\]we have \[5a_1v_1+(a_2-4a_1)v_2+a_3v_3\cdots+a_mv_m=0.\] Because $v_1$, $v_2$, $\cdots$, $v_m$ is a linearly independent, it follows that \[5a_1=0,a_2-4a_1=0,a_3=\cdots=a_m=0.\]We get $a_1=a_2=\cdots=a_m=0$, hence the list\[5v_1-4v_2, v_2, v_3,\cdots,v_m\] is linearly independent by Definition 2.17.
8. Solution: This is true. For if there exist $a_1$, $\cdots$, $a_m$ $\in\mathbb F$ such that \[a_1(\lambda v_1)+a_2(\lambda v_2)+\cdots+a_m(\lambda v_m)=0,\]we have \[(a_1\lambda) v_1+(a_2\lambda) v_2+\cdots+(a_m \lambda) v_m=0.\]Because $v_1$, $v_2$, $\cdots$, $v_m$ is a linearly independent, it follows that \[a_1\lambda=a_2\lambda=\cdots=a_m\lambda.\]Since $\lambda\ne 0$, we deduce that $a_1=a_2=\cdots=a_m=0$, hence $\lambda v_1$, $\lambda v_2$, $\cdots$, $\lambda v_m$ is linearly independent.
9. Solution: Counterexample: let $w_i=-v_i$, then if $v_1$, $v_2$, $\cdots$, $v_m$ is linearly independent, we have $w_1$, $w_2$, $\cdots$, $w_m$ is linearly independent by Problem 8. However, $v_1+w_1=0$, $v_2+w_2=0$, $\cdots$, $v_m+w_m=0$ is linearly dependent.
10. Solution: Since $v_1+w$, $\cdots$, $v_m+w$ is linearly dependent, there exist $a_1$, $\cdots$, $a_m$ $\in\mathbb F$, not all $0$, such that \[a_1(v_1+w)+a_2(v_2+w)+\cdots+a_m(v_m+w)=0.\]Hence we have \[a_1v_1+\cdots+a_mv_m+(a_1+\cdots+a_m)w=0.\]If $a_1+\cdots+a_m=0$, we get $a_1v_1+\cdots+a_mv_m=0$, which will deduce that $a_i\equiv 0$. Hence $a_1+\cdots+a_m\ne 0$, it follows that \[w=-\frac{1}{a_1+\cdots+a_m}(a_1v_1+\cdots+a_mv_m)\in\mathrm{span}(v_1,\cdots,v_m).\]
11. Solution: It is equivalent to show $v_1$, $v_2$, $\cdots$, $v_m$, $w$ is linearly dependent if and only if $w\in \mathrm{span}(v_1,\cdots,v_m)$.
If $v_1$, $v_2$, $\cdots$, $v_m$, $w$ is linearly dependent, there exist $a_1$, $\cdots$, $a_m$, $b$ $\in\mathbb F$, not all $0$, such that \[a_1v_1+a_2v_2+\cdots+a_mv_m+bw=0.\]If $b=0$, we get $a_1v_1+\cdots+a_mv_m=0$, which will deduce that $a_i\equiv 0$. Hence $b\ne 0$, now we can obtain \[w=-\frac{1}{b}(a_1v_1+a_2v_2+\cdots+a_mv_m)\in \mathrm{span}(v_1,\cdots,v_m).\] Conversely, if $w\in \mathrm{span}(v_1,\cdots,v_m)$, then there exist $a_1$, $\cdots$, $a_m$ $\in\mathbb F$ such that \[w=a_1v_1+a_2v_2+\cdots+a_mv_m\Longrightarrow a_1v_1+a_2v_2+\cdots+a_mv_m-w=0.\]This implies $v_1$, $v_2$, $\cdots$, $v_m$, $w$ is linearly dependent.
12. Solution: Note that $1$, $z$, $z^2$, $z^3$, $z^4$ spans $\mathcal{P}_4(\mathbb F)$, hence any linearly independent list has no more than 5 polynomials by 2.23.
13. Solution: By the similar process of Problem 2, we can show that $1$, $z$, $z^2$, $z^3$, $z^4$ is a linearly independent list of $\mathcal{P}_4(\mathbb F)$. Due to 2.23, no list of four polynomials spans $\mathcal{P}_4(\mathbb F)$. Otherwise, the length of every linearly independent list of vectors is 5 while the length of some spanning list of vectors is 4.
14. Solution: If there is a sequence $v_1$, $v_2$, $\cdots$ of vectors in $V$ such that $v_1$, $v_2$, $\cdots$, $v_m$ is linearly independent for every positive integer $m$, then $V$ is obviously infinite-dimensional.
If $V$ is infinite-dimensional, then $V$ cannot be spanned by finitely many vectors. Now we obtain such a sequence $v_1$, $v_2$, $\cdots$ of vectors in $V$ by induction. Let $v_1\ne 0$ is a vector in $V$. Since $V$ is infinite-dimensional, there must exist some $v_2\in V$ such that $v_2\notin \text{span}\{v_1\}$. Similarly, if $v_1$, $v_2$, $\cdots$, $v_m$ is linearly independent, then there must exist some $v_{m+1}\in V$ such that $v_{m+1}\notin\text{span}\{v_1,\cdots,v_m\}$. Since $V$ is infinite-dimensional, we can always do this process. Hence we will get a sequence $v_1$, $v_2$, $\cdots$ of vectors in $V$. By 2.21, we deduce that $v_1$, $v_2$, $\cdots$, $v_m$ is linearly independent for every positive integer $m$.
15. Solution: Let $e_{i}=(0,\cdots,0,1,0,\cdots)$ be the vector that has $1$ in the $i$th-component and $0$ in other components. Then we can easily check that $e_1$, $e_2$, $\cdots$, $e_m$ is linearly independent for every positive integer $m$. Now by Problem 14, we conclude that $\mathbb{F}^{\infty}$ is infinite-dimensional.
16. Solution: Define $f_n=\left\{ \begin{array}{ll} x-1/n, & \hbox{$x\geqslant 1/n$;} \\ 0, & \hbox{$x\in[0,1/n]$.} \end{array} \right. $, then $f_n$ is continuous on the interval $[0,1]$. We will show that $f_{n+1}\notin\text{span}\{f_1,\cdots,f_n\}$, then by the argument in Problem 14 we can conclude that the real vector space of all continuous real-valued functions on the interval $[0,1]$ is infinite-dimensional. For if $f_{n+1}\in\text{span}\{f_1,\cdots,f_n\}$, then \[f_{n+1}(x)=a_1f_1(x)+\cdots+a_nf_n(x).\]Note that $f_1(1/n)=\cdots=f_n(1/n)=0$, but $f_{n+1}(1/n)=1/(n(n+1))\ne 0$, we get a contradiction.
17. Solution: If $p_0$, $p_1$, $\cdots$, $p_m$ is linearly independent in $\mathcal{P}_m(\mathbb F)$, then consider $z$, $p_0$, $p_1$, $\cdots$, $p_m$. By 2.23, we know that length of linearly independent list $\leqslant$ length of spanning list. Because $1$, $z$, $\cdots$, $z^m$ is a spanning list of $\mathcal{P}_m(\mathbb F)$, it follows that every linearly independent list in $\mathcal{P}_m(\mathbb F)$ has length no more than $m+1$. Now we obtain that $z$, $p_0$, $p_1$, $\cdots$, $p_m$ is linearly dependent(of length $m+2$), then by Problem 11, there exist $a_i\in\mathbb F$such that \begin{equation}\label{2A31}z=a_0p_0(z)+a_1p_1(z)+\cdots+a_{m+1}p_{m+1}(z).\end{equation} Note that $p_j(2)=0$ for each $j$, we deduce that $2=0$ from $(\ref{2A31})$. Hence we get a contradiction.
