Exercises 1,2 and 4. For Problem 2, please also see Carson Rogers’s comment.

4. Solution: For any $f\in \ca L(V_1\times \cdots\times V_m,W)$ and given $i\in \{1,\cdots,m\}$, define $f_i:V_i\to W$ by the rule:\[f_i(v_i)=f(0,\cdots,0,v_i,0,\cdots,0),\]where $v_i$ sits in the $i$-th slot. One can check that $f_i\in\ca L(V_i,W)$.

Define $\vp :\ca L(V_1\times \cdots\times V_m,W)\to \ca L(V_1,W)\times \cdots \times \ca L(V_m,W)$ by\[\vp(f)=(f_1,\cdots,f_m),\] where $f_1$, $\cdots$, $f_m$ are defined in the previous paragraph. Now we are going to check that $\vp$ is an isomorphism between $\ca L(V_1\times \cdots\times V_m,W)$ and $\ca L(V_1,W)\times \cdots \times \ca L(V_m,W)$ by construct an inverse map.

Define $\psi :\ca L(V_1,W)\times \cdots \times \ca L(V_m,W)\to \ca L(V_1\times \cdots\times V_m,W)$ by\[\psi(f_1,\cdots,f_m)(v_1,\cdots,v_m)=f_1(v_1)+\cdots+f_m(v_m).\]

Please check that $\psi$ and $\vp$ are linear. It is also not hard to see that $\psi\circ \vp=Id$, and $\vp\circ \psi=Id$. Hence we are done.

7. Solution: Note that $v+U=x+W$, hence $v=x+w_1$, where $w_1\in W$. It follows that $v-x\in W$. Hence for any $u\in U$, we have \[ v+u=x+w_2 \]for some $w_2\in W$ since $v+U=x+W$. Hence we conclude that \[u=(x-v)+w_2=-w_1+w_2\in W,\] it follows that $U\subset W$ for $u$ is chosen arbitrarily. Similarly, we deduce that $W\subset U$. Thus $U=W$.

8. Solution: If $A$ is an affine subset of $V$, then there exist a vector $a\in V$ and a subspace $U$ of $V$ such that $A=a+U$. Then any $v,w\in A$ can be written as $v=a+u_1$ and $w=a+u_2$ for some $u_1,u_2\in U$. Hence \[ \lambda v+(1-\lambda)w=a+[\lambda u_1+(1-\lambda)u_2]\in a+U=A. \] Conversely, since $A$ is nonempty, let $a\in A$. We will show that \[A-a=\{x-a:x\in A\}\]is a subspace of $V$. For $x-a\in A-a$ and $\lambda\in \mb F$ where $x\in A$, then \[ \lambda x+(1-\lambda)a\in A\Longrightarrow \lambda (x-a)=\lambda x+(1-\lambda)a-a\in A-a. \]This implies $A-a$ is closed under scalar multiplication. For $x-a$ and $y-a\in A-a$, where $x,y\in A$. We have \[ \frac{x}{2}+\frac{y}{2}-a\in A-a. \]Note that $A-a$ is closed under scalar multiplication, it follows that \[ (x-a)+(y-a)=2\left(\frac{x}{2}+\frac{y}{2}-a\right)\in A-a. \]That is $A-a$ is closed under addition. Hence $A-a$ is a subspace of $V$. Note that $A=a+(A-a)$, it follows that $A$ is an affine subset of $V$.

9. Solution: Suppose $A_1\cap A_2\ne \varnothing$, then for any $x,y\in A_1\cap A_2$ and $\lambda\in \mb F$, we have \[ \lambda x+(1-\lambda) y\in A_1 \]and \[ \lambda x+(1-\lambda) y\in A_2 \]by Problem 8 since $A_1$ and $A_2$ are affine subsets of $V$. Hence \[ \lambda x+(1-\lambda) y\in A_1\cap A_2. \]Again by Problem 8, we deduce that $A_1\cap A_2$ is an affine subset of $V$.

10. Solution: It is the same as Problem 9.

11. Solution: (a) For $v=\lambda_1 v_1+\cdots+\lambda_m v_m\in A$ and $w=\eta_1 v_1+\cdots+\eta_m v_m\in A$,where $\lambda_1,\cdots,\lambda_m\in\mb F$, $\lambda_1+\cdots+\lambda_m=1$ and $\eta_1,\cdots,\eta_m\in\mb F$, $\eta_1+\cdots+\eta_m=1$. For any $\lambda\in\mb F$, we have \begin{align*} \lambda v+(1-\lambda) w=\sum_{i=1}^m(\lambda\lambda_i+(1-\lambda)\eta_i)v_i. \end{align*} Note that \[ \sum_{i=1}^m(\lambda\lambda_i+(1-\lambda)\eta_i)=\lambda\sum_{i=1}^m\lambda_i+(1-\lambda)\sum_{i=1}^n\eta_i=\lambda+(1-\lambda)=1, \]we deduce that $\lambda v+(1-\lambda) w\in A$. Hence $A$ is an affine subset of $V$ by Problem 8.

(b) We will use induction to show that for any affine subset $W$ of $V$ that contains $v_1$, $\cdots$, $v_m$$k\le m$, then if $\lambda_1+\cdots+\lambda_k=1$, we have \[ \sum_{j=1}^k\lambda_jv_j\in W. \]It is obvious for $k=1$ and $k=2$ by Problem 8. Suppose this is true for $k$, then we will show it for $k+1$ ($k+1\le m$). Assume $\lambda_1+\cdots+\lambda_{k+1}=1$. If $\lambda_{k+1}=1$, then \[\sum_{j=1}^{k+1}\lambda_jv_j=v_{k+1}\in W.\]If $\lambda_{k+1}\ne 1$, then \[ \frac{1}{1-\lambda_{k+1}}(\lambda_1+\cdots+\lambda_k)=1. \]Hence by assumption, we deduce that \[ \frac{1}{1-\lambda_{k+1}}(\lambda_1v_1+\cdots+\lambda_kv_k)\in W. \]By Problem 8, we have \[ (1-a_{k+1})\left(\frac{1}{1-\lambda_{k+1}}(\lambda_1v_1+\cdots+\lambda_kv_k)\right)+a_{k+1}v_{k+1}\in W, \]namely \[ \lambda_1v_1+\cdots+\lambda_{k+1}v_{k+1}\in W. \]By mathematical induction, we conclude if $\lambda_1,\cdots,\lambda_m\in\mb F$, $\lambda_1+\cdots+\lambda_m=1$, then \[\lambda_1 v_1+\cdots+\lambda_m v_m\in W,\]thus $A\subset W$.

(c) Note that if $\lambda_1+\cdots+\lambda_m=1$, we have \[ \lambda_1 v_1+\cdots+\lambda_m v_m=v_1+\lambda_2(v_2-v_1)+\cdots+\lambda_m(v_m-v_1). \]Hence $A\subset v_1+\m{span}(v_2-v_1,\cdots,v_m-v_1)$. Similarly, for any\[v\in v_1+\m{span}(v_2-v_1,\cdots,v_m-v_1),\]$v$ can be written as \[ v_1+\sum_{i=2}^m \lambda_i(v_i-v_1)=(1-\lambda_2-\cdots-\lambda_m)v_1+\sum_{i=2}^m \lambda_iv_i \]for some $\lambda_2$, $\cdots$, $\lambda_m\in\mb F$. Note that \[ (1-\lambda_2-\cdots-\lambda_m)+\sum_{i=2}^m \lambda_i=1, \]we deduce that $v_1+\m{span}(v_2-v_1,\cdots,v_m-v_1)\subset A$. Hence\[A=v_1+\m{span}(v_2-v_1,\cdots,v_m-v_1).\]Let $v=v_1$ and $U=\m{span}(v_2-v_1,\cdots,v_m-v_1)$, then $\dim U\le m-1$.

12. Solution: Since $V/U$ is finite-dimensional, we can suppose $v_1+U$, $\cdots$, $v_n+U$ is a basis of $V/U$. Then for any $v\in V$, there exist a unique list of $\lambda_1$, $\cdots$, $\lambda_n\in\mb F$ such that \[ v+U=\sum_{i=1}^n\lambda_i(v_i+U). \]Then\[v-\sum_{i=1}^n\lambda_iv_i\in U.\]Define $\vp:V\to U\times V/U$ by \[\vp(v)=\left(v-\sum_{i=1}^n\lambda_iv_i,\sum_{i=1}^n\lambda_i(v_i+U)\right).\]We first check that $\vp$ is linear. For any $v,w\in V$, we \[ v+U=\sum_{i=1}^n\lambda_i(v_i+U) \]and \[ w+U=\sum_{i=1}^n\eta_i(v_i+U). \]for some $\lambda_1$, $\cdots$, $\lambda_n\in\mb F$ and $\eta_1$, $\cdots$, $\eta_n\in\mb F$. Then for any $\lambda \in \mb F$, we have \[ (v+\lambda w)+U=(v+U)+\lambda(w+U)=\sum_{i=1}^n(\lambda_i+\lambda\eta_i)(v_i+U). \]Hence \[ \vp(v+\lambda w)=\left(v+\lambda w-\sum_{i=1}^n(\lambda_i+\lambda\eta_i)v_i,\sum_{i=1}^n(\lambda_i+\lambda\eta_i)(v_i+U)\right), \]and \[ \vp(v)=\left(v-\sum_{i=1}^n\lambda_iv_i,\sum_{i=1}^n\lambda_i(v_i+U)\right), \]and \[ \vp(w)=\left(w-\sum_{i=1}^n\eta_iv_i,\sum_{i=1}^n\eta_i(v_i+U)\right). \]Therefore we have (check it) \begin{align*} \vp(v)+\lambda\vp( w)=\vp(v+\lambda w), \end{align*} namely $\vp$ is linear.

Injectivity: if $\vp(v)=0$, then $\lambda_1=\cdots=\lambda_n=0$(defined as above) and \[0=v-\sum_{i=1}^n\lambda_iv_i=v.\] Surjectivity: For $\left(u,\sum_{i=1}^n\xi(v_i+U)\right)\in U\times V/U$, it is easy to see \[ \vp\left(u+\sum_{i=1}^n\xi v_i\right)=\left(u,\sum_{i=1}^n\xi(v_i+U)\right). \]Hence $\vp$ is an isomorphism, namely $V$ is isomorphic to $U\times V/U$.

13. Solution: For any $v\in V$, since $v_1+U$, $\cdots$, $v_m+U$ is a basis of $V/U$, there exist $\lambda_1$, $\cdots$, $\lambda_m\in\mb F$ such that \[ v+U=\sum_{i=1}^m \lambda_i(v_i+U). \]This implies \[ v-\sum_{i=1}^m\lambda_iv_i\in U. \]Note that $u_1$, $\cdots$, $u_n$ is a basis of $U$, it follows that \[ v-\sum_{i=1}^m\lambda_iv_i=\sum_{j=1}^n\eta_ju_j \]for some $\eta_1$, $\cdots$, $\eta_n\in \mb F$. Hence \[ v=\sum_{i=1}^m\lambda_iv_i+\sum_{j=1}^n\eta_ju_j, \]this implies $V=\m{span}(v_1,\cdots,v_m,u_1,\cdots,u_n)$ since $v$ is chosen arbitrarily. Hence it suffices to show that $v_1$, $\cdots$, $v_m$, $u_1$, $\cdots$, $u_n$ is linearly independent. Suppose for some $\lambda_1$, $\cdots$, $\lambda_m\in\mb F$ and $\eta_1$, $\cdots$, $\eta_n\in \mb F$, we have \[ \sum_{i=1}^m\lambda_iv_i+\sum_{j=1}^n\eta_ju_j=0. \]Then \[ \sum_{i=1}^m\lambda_i(v_i+U)=0, \]hence $\lambda_1=\cdots=\lambda_m=0$ since $v_1+U$, $\cdots$, $v_m+U$ is a basis of $V/U$. It follows that \[\sum_{j=1}^n\eta_ju_j=0.\]Note that $u_1$, $\cdots$, $u_n$ is a basis of $U$, we obtain that $\eta_1=\cdots=\eta_n=0$. Hence $v_1$, $\cdots$, $v_m$, $u_1$, $\cdots$, $u_n$ is linearly independent.

15. Solution: Note that $\varphi:V\to \mb F$ and $\varphi\ne 0$, there exists $v\in V$ such that $\varphi(v)\ne 0$. Then for any $a\in\mb F$, we have\[\vp\Big(\frac{av}{\varphi(v)}\Big)=\frac{a}{\varphi(v)}\vp(v)=a.\]Therefore $\mathrm{range}\,\vp=\mb F$.

By 3.91(d), we have $V/\mathrm{null}\,\vp\cong \mb F$. Hence $$\dim V/\mathrm{null}\,\vp=\dim \mb F=1.$$

16. Solution: Since $\dim V/U=1$, there exists a vector $v\notin U$ such that $v+U$ is a basis of $V/U$. Define $\theta: V/U\to \mb F$ by \[\theta(kv+U)=k,\]then $\theta\in\ca L(V/U,\mb F)$.

Let $\pi$ be the quotient map $\pi :V\to V/U$. Let $\vp=\theta\circ\pi$, then $\vp\in\ca L(V,\mb F)$. Now we check that $\mathrm{null}\,\vp=U$.

Let $w\in V$ such that $\vp(w)=0$, i.e. $\theta(\pi(w))=0$. By the definition of $\theta$, we have $$w+U=0v+U.$$ Hence $w\in U$, which implies that $\mathrm{null}\,\vp\subset U$.

On the other hand, let $u\in U$, then $\pi(u)=u+U=0+U$. Hence $\theta\circ \pi(u)=0$, which implies that $U\subset \mathrm{null}\,\vp$.

Therefore $\mathrm{null}\,\vp=U$. (Please note that such $\vp$ is not unique.)

17. Solution: Since $V/U$ is finite-dimensional, let $n=\dim (V/U)$, then there exist $w_1,\cdots,w_n\in V$ such that $w_1+U$, $\cdots$, $w_n+U$ form a basis of $V/U$. It is easy to see that $w_1,\cdots,w_n$ is linearly independent otherwise $w_1+U$, $\cdots$, $w_n+U$ would be linearly dependent. Consider the subspace $W$ of $V$ spanned by $w_1,\cdots,w_n$, then we have $\dim W=n=\dim(V/U)$.

Now we are going to show that $V=U\oplus W$. For any $v\in V$, since $w_1+U$, $\cdots$, $w_n+U$ is a basis of $V/U$, there exist $k_1,\cdots,k_n\in\mb F$ such that $$v+U=\sum_{i=1}^n k_i(w_i+U).$$This implies that $$v-\sum_{i=1}^n k_iw_i\in U.$$Therefore consider \[v=\Big(v-\sum_{i=1}^n k_iw_i\Big)+\sum_{i=1}^n k_iw_i,\]it tells us that $v\in U+W$. Since $v$ is chosen arbitrarily, we conclude that $V\subset U+W$. To show that $V=U\oplus W$, it suffices to show that $U\cap W=0$.

Suppose $v\in U\cap W$, then there exist $k_1,\cdots,k_n\in\mb F$ such that $$v=\sum_{i=1}^n k_iw_i.$$Hence we have $0=\pi(v)=\sum_{i=1}^n k_i(w_i+U)$, where $\pi$ is the canonical quotient map from $V$ onto $V/U$. However, note that $w_1+U$, $\cdots$, $w_n+U$ is a basis of $V/U$ we have $k_1=\cdots=k_n=0$. Thus$$v=\sum_{i=1}^n k_iw_i=0.$$This shows that $U\cap W=0$.

18. Solution: Suppose there exists $S\in\ca L(V/U,W)$ such that $T=S\circ\pi$. Since for any $u\in U$ we have $\pi u=0$, hence $$Tu=(S\circ \pi )u=S(\pi u)=S0=0.$$Hence $u\in \mathrm{null}\, T$. Note that this is true for any $u\in U$, we conclude that $U\subset \mathrm{null}\, T$.

Conversely, suppose $U\subset \mathrm{null}\, T$. Define a map $S: V/U\to W$ by $S(v+U)=Tv$. We have to check that this map is well-defined (since there may exist different $v_1,v_2\in V$ such that $v_1+U=v_2+U$ and we need to check that $Tv_1=Tv_2$).

For $v_1,v_2\in V$ such that $v_1+U=v_2+U$, we have $v_1-v_2\in U$. Note that $U\subset \mathrm{null}\, T$, it follows that $T(v_1-v_2)=0$, hence we have $Tv_1=Tv_2$. This implies that $S$ is well-defined.

Note that $T\in \ca L(V,W)$, it is easy to see that $S\in \ca L(V/U,W)$. By definition, we have $$S\circ \pi (v)=S(v+U)=Tv$$ for any $v\in V$. This implies that $T=S\circ \pi$.

19. Solution: For any given finite set $A$, denote by $|A|$ the number of elements in $A$. Let $S_i$, $i=1,\cdots,n$, be finite sets. The union $\bigcup_{i=1}^n S_i$ is a disjoint union if and only if $$|S_1|+\cdots+|S_n|=\Big|\bigcup_{i=1}^n S_i\Big|.$$

20. Solution: (a) To show that $\Gamma$ is linear, we need to show that for any $S_1,S_2\in\ca L(V/U,W)$ and $k_1,k_2\in\mb F$ we have $$\Gamma(k_1S_1+k_2S_2)=k_1\Gamma(S_1)+k_2\Gamma(S_2).$$This follows directly since \begin{align*}\Gamma(k_1S_1+k_2S_2)=&(k_1S_1+k_2S_2)\circ \pi\\=& k_1S_1\circ \pi+k_2S_2\circ \pi\\=& k_1\Gamma(S_1)+k_2\Gamma(S_2).\end{align*}(b) To show that $\Gamma$ is injective, one has to show that if $\Gamma(S)=0$ then $S=0$. If $\Gamma(S)=0$, it implies that $S\circ \pi=0$. Hence $(S\circ \pi) v=0$ for all $v\in V$. Thus $S(\pi(v))=S(v+U)=0$ for all $v\in V$. We conclude that $S=0$, which shows $\Gamma$ is injective.

(c) By Problem 18, we have\begin{align*}T\in \mathrm{range}\, \Gamma\iff & \text{there exists }S\in \ca L(V/U,W)\text{ s.t. }T=S\circ \pi\\ \iff & U\subset \mathrm{null}\, T\quad \text{ by Problem 18}\\ \iff &T\in\ca L(V,W): \, Tu=0 \text{ for every } u\in U.\end{align*}This completes the proof.

## Carl

21 Jan 2021One issue i consistently find when working through this book for self learning is that I will often run into a problem like #9 where I bang my head against a wall for 30 minutes, finally give up and look at the solution, only to find that the solution requires conclusions from an earlier problem. Since I am not doing every problem (mostly just odds), is it reasonable to expect, at the level of the text book, that I would intuitively see the path to solving the problem without having done the previous question?

To me that seems like a significantly more challenging premise than having been guided to a strong starting point from the previous question. Thanks :)

## HUYNH Huu Nhat

2 Jan 2021someone can check my solution for 11.c? Thank you so much

my solution:

Let W=span(v1, v2, ..., vm) then dim(W) -v belongs to W, take u belongs to U then u+v belongs to W, but -v belongs to W -> u belongs to W so that U is subset of W).

but U can't be equal to W because the restriction lamda_1+...lamda_n=1. So that dim(U)<=dim(W)-1 and we have dim(U)<=m-1.

## HUYNH Huu Nhat

2 Jan 2021Let W=span(v1, v2, ... vm) then dim(W) v+u belongs to W, and -v belongs to W-> u belongs to W -> U is subset of W. but U can't be equal to W because of the restriction lamda_1+lamda_2+...+lamda_m=1. So that dim(U) dim(U)<=m-1

## HongYi

23 Aug 2020Hi, I found another method for exercise 12. We can define a map between V and UxV/U as f(u+w)=(u,w+U), where u in U, w in W and U oplus W = V. Because each v inV can be uniquely decomposed as v=u+w, the definition of f makes sense, and this definition can handle the injectivity of f. However the definition of phi in your solution is somewhat weird, I think you need to state v_1,...v_n can’t represent any elements in U first.

## Kevin

8 Sep 2020This works if $V$ is finite-dimensional, so we can construct $W$. However, if $V$ is not finite-dimensional, showing that such a $W$ exists requires additional tools...

## HongYi

15 Sep 2020Yes, you’re right!

By the way, I have found the tool I need has been stated in exercise 17. ^_^

## HongYi

23 Aug 2020Hi, for exercise12, could I define the map as f(u+v)=(u,v+U) ? Then the condition of V/U is finite has been ignored. I wonder if it is ok for this problem

## HongYi

23 Aug 2020I think I have found the issue of that definition:u+v=u’+v’

## Marie

4 Aug 2020For Q13, since v + U = (a_1v_1+...+a_mv_m) + U this would mean v = a_1v_1+...+a_mv_m. Does this mean any v in V can be written uniquely as the linear combination of v_1,...,v_m ? However, we know that any v in V is actually written uniquely as a linear combination of the basis v_1,...,v_m,u_1,...,u_n. I'm confused by this and would appreciate help!

## George

8 Oct 2020v + U = (a_1v1+...+a_mv_m) + U is only true if v does not belong to U. Cause if it does, then v+ U is part of U, and can only be written as a liner sum of the basis of U. The way to think about it is that the additive identity in the affine subspace is U, it's not 0.

## Yuheng

22 Jul 2020I have a solution for 14(b). ((v)k means the kth entry of v)

We only need to show that for any integer m, there exists a linearly independent list of length m in the quotient space.

For i=1,2,...,m, k=1,2,...,infinity, define vi by (vi)k=1 if m divides k-i, (vi)k=0 otherwise.

Suppose a1(v1+U) + ... + am(vm+U) = 0, so a1v1+...+amvm = (a1,a2,...,am, a1,a2,...,am, a1,a2,...,am, ......) is in U, which means it should have finitely many nonzero entries. Because if any ai is not zero, then the vector has infinitely many nonzero entries, each ai must equal 0, and we are done.

## Tom C

14 May 2020Here's my try at 14b. Please correct me if I messed up. Suppose F^\infty / U is finite dimensional. Then it should have a basis. Let's say the dimension is n. Let v_i be a vector in F^\infty consisting of i zeros followed by all ones. Clearly v_1, ..., v_n are not in U because they have infinitely many nonzero elements. But they are also linearly independent. So v_1 + U, ..., v_n + U is a suitable basis. But then it is always possible to find another vector, namely v_{n + 1} that is also not in U, and not in span (v_1, ..., v_n). So v_1 + U, ..., v_{n + 1} + U is also a basis. But that is a contradiction since by assumption all bases must have length of n. Thus it must be infinite dimensional. That works right?

## Tom C

15 May 2020Wait.. never mind. This doesn't work. Sorry.

## Tom C

15 May 2020But maybe it can be fixed by redefining v_i as having i ones followed by (n + 1) - i zeros, repeating in an infinite pattern. That way, any linear combo of the v_i will still have an infinite repeating pattern with period n + 1 that contains at least one nonzero element, and thus is not in U. In this case v_{n +1} is just all ones.

## Linearity

17 May 2020You may try this list, $$v_i=(1,i,i^2,i^3,\dots).$$Show the list $\{v_i+U\}_{i\geqslant 1}$ is linearly independent over $\mathbb F^\infty / U$. Hint: Use Vandermonde determinant (may be not a good solution because det is used). See Dedekind's comment below. That is what I thought at the very beginning (taking ones at the $ip$-th component where $p$ is a prime number).

## samir

11 May 2020For #9, I think you should consider the case that there is no x and y (because the sets intersect at a single point x). Suppose A1 and A2 intersect at x only, then the affine subset = x + {0}, hence meets the definition.

## Linearity

11 May 2020Did I say $x\ne y$? 🤪

## samir

12 May 2020True - proof works even if x = y. Sorry for that.Still not thinking like a mathematician :-)

## Dedekind

1 Apr 2020Exercise 14 (b)

Let $a_{i, j} = 1$ if $j$ is a multiple of the $i$th prime number $\pi(i)$.

Let $a_{i, j} = 0$ if $j$ is not a multiple of the $i$th prime number $\pi(i)$.

Let $v_i = (a_{i, 1}, a_{i, 2}, \cdots)$.

Let $\alpha_1 (v_1 + U) + \cdots + \alpha_n (v_n + U) = 0 + U$.

Then, $\alpha_1 v_1 + \cdots + \alpha_n v_n \in U$.

Let $i$ be an arbitrary element of $\{1, 2, \cdots, n\}$.

Let $p$ be a prime number greater than $\pi(n)$.

Then, $1 = a_{i, m \pi(i)}$ for all $m \in \{1, 2, 3, \cdots\}$.

Let $j \in \{1, 2, 3, \cdots, n\}$ and $j \ne i$.

Then, $0 = a_{j, m \pi(i)}$ for all $m \in \{\pi(n+1), \pi(n+2), \pi(n+3), \cdots\}$.

$\therefore$

$1 = a_{i, m \pi(i)}$ for all $m \in \{\pi(n+1), \pi(n+2), \pi(n+3), \cdots\}$.

$0 = a_{j, m \pi(i)}$ for all $m \in \{\pi(n+1), \pi(n+2), \pi(n+3), \cdots\}$.

So, the $m \pi(i)$th slot of $\alpha_1 v_1 + \cdots + \alpha_n v_n$ is $\alpha_i$.

If $\alpha_i \ne 0$, then $\alpha_1 v_1 + \cdots + \alpha_n v_n \notin U$ by the definition of $U$.

So, $\alpha_i = 0$ for all $i \in \{1, 2, 3, \cdots, n\}$.

## Dedekind

1 Apr 2020I solved exercise 14.

## Henrik Young

6 Feb 2020No way, this section is so difficult to understand deeply, please tell me the rest of the book is easier.

## David

22 Sep 2019For number 18, it shows πu = 0. This "zero" is just meant to represent the additive identity of V/U, correct? That is, we could alternately say that πu = 0+U = U, and then Tu = S(πu) = S(U) = 0 for all u in U, because S being linear by assumption must map the additive identity of its domain to the additive identity of its codomain. Am I correct on this?

## Linearity

22 Sep 2019Yes.

## David

22 Sep 2019Thank you for these solutions, especially for this section. I've spent almost as much time on 3E by itself as I did on the rest of the book up to this point, and this site has helped tremendously.

## David

15 Sep 2019So for number 12, we can't say that V and U are finite if V/U is finite?

## Linearity

17 Sep 2019You can't say that.

## maowtm

11 Mar 2020Consider $V = F^\infty$ and $U = \{(0, x_2, x_3, \ldots)\ |\ x_n \in F\}$. $\mathrm{dim} V/U = 1$.

## David

15 Sep 2019Why does each section of the book cover finite-dimensional cases almost exclusively, then go on to have so many exercises that involve infinite-dimensional cases?

## Zerong Xi

30 Jul 2019A rudimental idea for exercise 14b:

If we simplify the problem to $x_j \in {0, 1}$, then using Cantor's diagonal argument, we can easily prove $U$ is a countable set while $F^\infty$ is an uncountable set. But still, there is no word about the operation of dimensions on infinite space so far.

## Zerong Xi

30 Jul 2019A rudimental idea for exercise 14b:

If we simplify the problem to $x_j \in \{0, 1\}$, then using Cantor's diagonal argument, we can easily prove $U$ is a countable set while $F^\infty$ is an uncountable set. But still, there is no word about the operation of dimensions on infinite-dimensional space so far.

## Rafael Deiga

28 May 2019In this link: https://pt.overleaf.com/read/kgjctnwvsvrz, I offer solutions to the problems 3, 5 and 6.

## David

15 Sep 2019Thank you.

## Zhu

11 Feb 2020Thx！This is very helpfulnl.

## Felix

3 Apr 2020Thanks a lot

## AliD

4 Sep 2018For question 15 (b),

Define vectors,

v1 = { 1 , 2 , 3 , 4 , 5 , 6 , ....}

v2 = { 1^2 , 2^2 , 3^2 , 4^2 , 5^2 , ....}

v3 = { 1^3 , 2^3 , 3^3 , 4^3 , 5^3 , ....}

v4 = { 1^4 , 2^4 , 3^4 , 4^4 , 5^4 , ….}

vn = { 1^n , 2^n , 3^n , 4^n , 5^n , ….}

These are all clearly vectors not in U.

Now, take some positive integer n. We just have to show that v1 + U, … , vn + U is a linearly independent list, since this will imply that F-inf / U is infinite dimensional.

Suppose a1*(v1 + U) + … + an*(vn + U) = U

then (a1*v1 + … + an*vn) + U = U

so a1*v1 + …. + an*vn must be in U.

Now, in order for this vector to be in U it can only be non-zero for finitely many indices. Therefore, there exists an integer N such that for all integers M > N, the M-th index is zero. I.E.

a1*M + a2*M^2 + … + an*M^n = 0

But, polynomials of degree n have at most n roots, and this poynomial has a infinite number of roots. Therefore our polynomial must be the zero polynomial, that is a1 = … = an. We have now showed out list is linearly independent so our proof is done.

Please let me know if you think there are any issues with this work or if I didn't explain something well. Thank you

## AliD

4 Sep 2018sorry this is problem 14 (b)

## Lucas Ferrari Soto

16 May 2018I'm guessing 10) is supposed to go by induction and thus holds for finite collections of affine subsets... is the statement true for infinite collections (the exercise doesn't specify)? how would I go about proving it?

Also, I think 8) can be easily modified to not require that there's a 2 in the underlying field.

## Mahdi Dibaiee

10 Feb 2018How do you use distributive property on function composition in 20? Function composition is not distributive, is it in this case? If so, what implies it?

## Mahdi Dibaiee

7 Feb 2018Why not just show that the two sets have equal dimensions, since two vector spaces are isomorphic to each other if and only if they have the same dimension in problem 4 and 5 and 6?

## Rafael Deiga

28 May 2019Because there the vector spaces can be infinite dimensional. What you said just holds for finite dimensional vector spaces.

## Kezer

19 Sep 201711) Whaaat the...

A simpler solution:

Suppose v + U contains v_1,v_2,...,v_m. Then there exists u_1,u_2,...,u_m € U such that v + u_i = v_i for i = 1,2,...,m.

Each element of A can be written as some sort of c_1v_1+...+c_mv_m with c_1+...+c_m = 1. Observe that

c_1v_1+...+c_mv_m = c_1(v+u_1)+...+c_m(v+u_m) = (c_1+...+c_m)v+(c_1u_1+...+c_mu_m) = v + c_1u_1+...+c_mu_m.

Obviously, the second term is an element of U. So that vector is contained in v + U.

Specifically, all possible vectors of A are contained in v + U and thus A is contained in v + U.

## Diego Ramos

6 Aug 2017Problem 16# may be solved in a lettle simpler way by stating that theta is simply any isomorfism from V/U onto F.

## Carson Rogers

20 Jul 2017The solution to problem 2 is wrong: the sum V_1+...+V_m is not defined unless all of the V_i are subspaces of a single larger space, which they are not assumed to be. Of course, each V_i is naturally isomorphic to a subspace of the product, but once you observe that, you have one sentence left to finish the solution.

## Mohammad Rashidi

27 Jul 2017Yes, you are correct. Thank you!

## Wu Jinyang

2 Aug 2017Hi. Can you post the solutions to 3, 4 and 6.

I have completely no idea on how to prove two vector space are isomorphic especially when the transformation is not given.

As for question 6, I found it totally incomprehensible, is it doable?

## Mohammad Rashidi

2 Aug 2017I uploaded a file for 4, you can try 6 similarly by constructing a reasonable linear map. Note that by using 4, 6 can be reduced to the case of n=1. If you have no idea what to do, then do the simplest case (n=1) first.

## Wu Jinyang

2 Aug 2017Also for question 11 (c). at the part "V can be written as", you miss a "v1" on the right side of the equation.

## Célio Passos

20 Jun 2017In exercise 8, what happens if A only contains one vector? The condition in the converse is clearly true but I don't think A is an affine subset.

## Célio Passos

20 Jun 2017I think I have a better proof for 7.

Because x is in v + U, it follows that there is some u in U such that x = v + u. If u is in U, because U is closed under addition, it follows that v + U = v + u + U. Hence v + U = x + U and x + U = x + W, which implies U = W (just subtract x from every element).

## David

15 Sep 2019Here's my proof for #7 (please reply with a reason if you think this proof isn't sufficient)

Since v + U = x + W for arbitrary v,x in V, then it must be true for v = x = 0, and this is only true if U = W, therefore U = W

## Linearity

17 Sep 2019It's wrong. The vectors $x$ and $v$ are fixed, not arbitrary.

## David

15 Sep 2019In #8, the result still holds for the case of one element, i.e. v = w.

## AJ

16 Feb 2017@Mohammad Rashidi

Did you need help with 1-6 and 14-20? Because I've gotten solutions for 1-6, 14, and 15 and when I get around to it, I'll do 16-20 as well.

## 李子鉴

17 Feb 2017Could you？

## AJ

17 Feb 2017I don't really want to type them all up here in the comments but if Mohammad hasn't added them because he doesn't have solutions for those then he can message me and I'll send images or something so he can add them.

## Mohammad Rashidi

17 Feb 2017I do have the solutions. The point is that I am lazy. I will update them today.

## Jin

17 Jul 2017Can you post #6 please?

## Christina DeMarzo

1 Mar 2017Anyway you could send me the solutions for 5 and 6? Those still haven't been posted.

## AJ

1 Mar 2017Sure ... for $5. :p

## Edgar Baquero

16 Mar 2017Do you have the solution for 5?

## AJ

16 Mar 2017Yes.

## Edgar Baquero

18 Mar 2017Can you give me the solution please??

## Elliott Mestas

3 Feb 2017On the 6th line of the correction of 11.c), you are missing v1 on the RHS

Also, for 11.b), I think I have a shorter proof:

Suppose there is an affine subset A' of V that contains v_1, ..., v_m. Then A' = v + U for some v in V and some U subspace of V.

v_1 = v + u_1, ..., v_m = v + u_m for some u_1, ..., u_m in U.

Consider any lambda_1, ..., lambda_m in F s.t. lambda_1 + ... + lambda_m = 1.

Then:

lambda_1 (v + u_1) + ... + lambda_m (v + u_m)

= v (lambda_1 + ... + lambda_m) + (u_1 lambda_1 + ... + u_m lambda_m)

= v + something in U

Therefore it is in A'.

## Mohammad Rashidi

3 Feb 2017Thank you. As for (b), are you using the result from (c)?

## Elliott Mestas

6 Feb 2017No, I am just using the definition of the affine subset and the definition of A

I have some answers for the other exercises in this section, would it be helpful for me to share them?

## Mohammad Rashidi

6 Feb 2017That is great. I really appreciate it, thanks.

## Marcel Ackermann

31 Dec 2016"is a basis of V" should be "is a basis of V/U"

## Mohammad Rashidi

15 Jan 2017For which problem? Thanks.

## Marcel Ackermann

15 Jan 20173E12, first sentence of solution

## Marcel Ackermann

23 Jun 20173E13

## Nuno Alvares Pereira

9 Sep 2016for 12), a much simpler solution uses just 3.76, 3.89, 3.73, and 3.59 . ;)

## Jinming Yu

16 Oct 2016but you dont know if V and U are finite dimensional

## Nuno Alvares Pereira

17 Oct 2016You're right... =D

## Ed Haarmann IV

14 Oct 2015Where are the solutions for 1,...,6 and 14,...,20?

## Jacqueline Garcia

23 Oct 2015Did you find the solutions?

## Edgar Baquero

16 Mar 2017I need the 5th