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.

## Adam

12 Sep 2021How is does the list z,pO,p1,...,pm being linearly independent, implies that p0,p1,...,pm is linearly dependent????? exercise 17

## Ahm

25 Jul 2021@No. 10:

how is a1+a2+...=0 => a1*v1+a2*v2+...=0 ?

## Ivan Ivanov

11 Jul 2020Not sure I follow the logic in 17.

We assume that if the list of p's were linearly independent, it would become linearly dependent when we added z to it.

But that's also true if the list was linearly dependent in the first place, right?

Suppose that p0, p1, ..., pm is linearly dependent. Then p0, p1, ..., pm, z is sure as hell also linearly dependent, with the same consequences.

So I'm not sure what the contradiction we get at the end proves!

## Ivan Ivanov

6 Oct 2020Ok I got it now. We used problem 11 to assume that z is in the span of $p_i, and that requires that the list of p's is linearly independent.

## Chen Changning

28 Oct 2020No.We assumed that p0,p1,...,pm is linearly independent.Then we try to get a contradiction.And if p0,p1,...,pm is linearly dependent itself, we will not be sure whether z belongs to span(p0,p1,...,pm). So we can’t describe z like that.

## jassie

24 Jun 2020For exercise 16, I believe there is a much simpler solution.

## Linearity

24 Jun 2020Yes, you are right. I edited your answer so that it would be clearer.

## Marie

22 Jun 2020In 15, how come the solution picks a particular example? Aren't we supposed to show that F^inf is infinite-dimensional generally? So how come we get to let v1= (1,0,...),...,v_n=(0,...,1,0...) ?

## Jonathan Sharir-Smith

15 May 2020Just to clarify a little bit (for myself). In 16, what you have effectively shown is that some subspace of continuous real-valued functions is infinite dimensional, right (the subspace in question being all functions of that f_n form)? And thus, by the contrapositive of Theorem 2.26 in the book, 3rd edition (ie. that every subspace of a finite-dimensional vector space is finite dimensional) we get the required result?

## Daniel Benvenutti

12 May 2020Hello friends, I have a question in exercise 5. Does the solution presented consider a list with two vectors? For it seems to me that the list in the statement is a list of only one vector, with two coordinates in \C. My consideration is based on page 28 of the text book, where the author quotes: "To avoid confusion, we will usually write lists of vectors without surrounding parentheses". If they were two vectors I believe the list would be arranged like this: (1 + i) , (1 - i).

Anyway, congratulations on the site!

## Linearity

12 May 2020It is a list of two vectors. A list of one nonzero vector is always linearly independent.

## Daniel Benvenutti

12 May 2020Oh, it's true. Thank you so much!

## Brian Lee

21 Dec 2019The solution to Question 16 is kind of long. Just consider the list 1,x,...,x^m.

## Linearity

22 Dec 2019I would like to avoid that because I am not sure if it is known that $1,x,\dots,x^m$ is linearly independent at that section. Of course you can check that it is a linearly independent list by using the the fundamental theorem of algebra. But this theorem is not known at this moment.

## Elena Orins

8 Oct 2018I do not understand the solution for 11. How does b go from equaling 0 to no equaling 0? How do they get -1/b? Also, why did they decide to make w to be in the span (v1, ....... , vm) ? Can someone please explain

## Lionel Ngouh

3 May 2018I don't get #9. Anyone with a better explanation

## Atharva Chingre

4 Dec 2018[Note: Please correct me if my thinking/mathematical logic is wrong]

Explanation 1: To find a counterexample for v1+w1, v2+w2, .... vm+wm is linearly independent, we have to find a1(v1+w1) + a2(v2+w2) + .... am(vm+w2) = 0 [eq.1], where not all a1, a2, .... am are 0 according to definition 2.17.

The easiest way to do this would be to take any one term, aj(vj+wj), from the left hand side of [eq.1] and make it equal to 0, such that aj is not equal to 0. Therefore take vj+wj=0, i.e. vj=(-wj). Now, aj can be any value not equal to zero, which still satisfies [eq.1], making this list linearly dependent. Since both lists v1, v2, v3.... vm and the list -v1, -v2, -v3.... -vm [i.e. the list w1, w2, w3.... wm] are independent (see problem 8, where lamda=-1), and their combination, [eq.1] is dependent, we have a counterexample.

Explanation 2: Since the list v1, v2, v3.... vm of length of m and the list w1, w2, w3.... wm of length m are linearly independent, then any list greater than the length m is linearly dependent [2.23] (all lists in this sentence are over V). Simplifying [eq.1], we get a1v1+a1w1 + a2v2+a2w2 + ....amvm+amwm. This could be seen as list v1, v2, ....vm, w1, w2, ....wm, with length 2m. Since 2m>m, this list is linearly dependent.

## Hanson Char

12 Feb 2018Apparently the solution to 2.c is incorrect. There isn't a vector (0,0,0,1) in the question, which provides a list of only 3 vectors in F^4.

https://uploads.disquscdn.com/images/9a6449dc2447a219a23b4c059bd6073ab0b48bf634a1c7445b5d548125e00603.png

## Hanson Char

12 Feb 2018Question 5 is cool!

## Eric

14 Dec 2016Does question 17 need all of that work? The degree 0 polynomial (i.e., the constant polynomial) being equal to 0 is proof enough of linear dependence. "Every list of vectors in V containing the 0 vector is linearly dependent."

## Mohammad Rashidi

15 Jan 2017But no one tells you that p_j has degree 0 for some j=0,...,m.

## Abhinav Patil

11 Mar 2019If some p_j in p_0,...,p_m does not have degree 0, then it follows from the pidgeonhole principle that at least two polynomials in p_0,...,p_m have the same degree. A list of polynomials where at least two polynomials have the same degree is necessarily linearly dependent.

A detailed proof:

Suppose at least one term in p_0,...,p_m has degree 0. Then, by the proof given by @disqus_UJbD3gIU7Q:disqus above, the list p_0,...,p_m must be linearly dependent.

Now suppose at least no terms in p_0,...,p_m have degree 0. Note that p_0,...,p_m has m+1 different terms. The vector space P_m(F) contains polynomials with coefficients in F and degree at most m. This means polynomials in P_m(F) can have degree 0, degree 1,..., degree m. This means there are m+1 different degrees possible for a polynomial in P_m(F). However, since we stated no terms in p_0,...,p_m have degree 0, then there are m different degrees possible (1,...m) for each term in p_0,...,p_m.

By the pidgeonhole principle if we place m+1 different objects (terms) into at most m different containers (degree values), then at least two terms must be placed into the same container. That is, two terms have the same degree. Clearly, if two polynomials with degree at most m in a list of m + 1 polynomials have the same degree, then the list is not linearly independent.

## George

9 Jun 2020I believe that's in the definition of Pj (notation definition). J is the degree.

## Linearity

10 Jun 2020No, I don't think so. J would be degree.

## Rich Yu

17 Sep 2016Another way to see problem 10)

Let us suppose the w is not in the span (v1,...,vm)

This implies that no linear combination a1v1+....amvm = w.

Now the condition for a linearly dependence means that for some numbers a1,..., am in the field

a1 (v1+w) + ... am (vm+w) = 0 such that not all ai = 0.

Consider v = a1v1+....amvm. Then any v+w cannot = 0 because no linear combination of vectors in (v1,...,vm) = w.

If v =/= w in every case, then all ai must equal 0 for a1 (v1+w) + ... am (vm+w) = 0, since (vi + w) =/=0.

Therefore, if the list of vectors (v1+w),..., (vm+w) is linearly dependent, w must in the span of (v1,...,vm).

If w is not in the span of (v1,...,vm) then (v1+w),..., (vm+w) is independent since all ai must =0.

## Nuno Alvares Pereira

6 Sep 2016for 16), much easier is to think about the space P_m (polinomials up to degree m). it's easy to find lin. indep. vectors which span P_m, and then see that, for all positive integers, P_{m+1} is not spanned by vectors of smaller degree... P_m is a strict subset for V for all m.

Hence by 14) it must infinite dimensional.

## Hanson Char

13 Feb 2018Nice!