MaxisJaisi
commented
7 years ago @cpassos I was expecting a quick solve of Problem 22, Exercise 3.B by using the fundamental theorem of linear maps.
Since $S:U \to V$ and $T:U \to V$, we conclude that (by rearranging some inequalities) $\operatorname{dim \ null} S \geq \operatorname{dim} V - \operatorname{dim} W$, and $\operatorname{dim \ null} T \geq \operatorname{dim} U - \operatorname{dim} V$. Adding the inequalities yields $\operatorname{dim \ null} S + \operatorname{dim \ null} T \geq \operatorname{dim} U - \operatorname{dim} W$. Since the result is true, we must conclude that $\operatorname{dim} U - \operatorname{dim} W \geq \operatorname{dim \ null} ST$
But $ST : U \to W$ is also a linear map, so the fundamental theorem still applies, so we obtain (again after rearranging the inequality), $\operatorname{dim \ null} ST \geq \operatorname{dim} U - \operatorname{dim} W$.
I must've gone wrong somewhere. Where did I go wrong?
celiopassos
Chapter 3: Linear Maps (Part 1)
3.A
Exercise 3
We know that the list of $n$ n-tuples such that the $j$th n-tuple has a $1$ in the $j$th place and $0$ in the others form a basis for $\mathbf{F}^n$. Similarly, the list of $m$ m-tuples such that the $i$th m-tuple has a $1$ in the $j$th place and $0$ in the others forms a basis for $\mathbf{F}^m$. Then
$\begin{align} T(x_1, \dots, x_n) & = T[x_1(1,0,\dots,0)+x_2(0,1,\dots,0)+\dots+x_n(0,0,\dots,1)] \\ & = T(x_1(1,0,\dots,0))+T(x_2(1,0,\dots,0))+\dots+T(x_n(0,0,\dots,1)) \\ & = x_1 T(1,0,\dots,0)+x_2 T(0,1,\dots,0)+\dots+x_n T(0,0,\dots,1) \\ \end{align}$
Now each of $T(1,0,\dots,0),T(0,1,\dots,0),\dots,T(0,0,\dots,1)$ can in turn be expanded in terms of the standard basis of $\mathbf{F}^m$. Let us consider $T(1,0,\dots,0)$ for illustration purposes. In terms of the standard basis of $\mathbf{F}^m$, $T(1,0,\dots,0) = a{1,1} (1,0,\dots,0) + a{2,1} (0,1,\dots,0) + \dots + a_{m,1} (0,0,\dots,1)$. Doing this for each $T(1,0,\dots,0),T(0,1,\dots,0),\dots,T(0,0,\dots,1)$, and substituting in $x_1 T(1,0,\dots,0)+x_2 T(0,1,\dots,0)+\dots+x_m T(0,0,\dots,1)$, we obtain
$\begin{align} T(x_1,\dots,x_n) & = x1 [a{1,1} (1,0,\dots,0)+ a{2,1} (0,1,\dots,0) + \dots + a{m,1} (0,0,\dots,1) \\ & + x2 [a{1,2} (1,0,\dots,0)+ a{2,2} (0,1,\dots,0) + \dots + a{m,2} (0,0,\dots,1) \\ & + \dots \\ & + xn [a{m,1} (1,0,\dots,0)+ a{2,n} (0,1,\dots,0) + \dots + a{m,n} (0,0,\dots,1)) \\ \end{align}$.
Adding everything up, we obtain our desired result.
Exercise 4
Consider $a_1, \dots, a_m$ such that $a_1 v_1 + \dots + a_m v_m = \mathbf{0}$. Then $T(a_1 v_1 + \dots + a_m v_m) = a_1 T(v_1) + \dots + a_m T(v_m) = T(\mathbf{0}) = \mathbf{0'}$. By the linear independence of $T(v_1), \dots, T(v_m)$, we can conclude that $a_1 = \dots = a_m = 0$.
Exercise 7
Let $V$ be a $1$-dimensional vector space, and $T$ be a linear map from $V$ to itself. We are required to show that $T$ is of the form $T(v) = \lambda v$, for some $\lambda \in \mathbf{F}$.
Since $V$ is $1$-dimensional, it is generated by a vector, say $\mathbf{v} \in V$. Let $\mathbf{x}$ be an arbitrary vector in $V$. Then $T(\mathbf{x}) = T(a \mathbf{v})$, since $\mathbf{x}$ must be of the form $a \mathbf{v}$ for some $a \in \mathbf{F}$. Furthermore, we know that $T$ is a linear map, so $T(a \mathbf{v}) = a T(\mathbf{v})$. Note also that $T(\mathbf{v}) \in V$, so $T(\mathbf{v}) = \lambda \mathbf{v}$, for some $\lambda \in \mathbf{F}$. Since $a (\lambda \mathbf{v})= \lambda (a \mathbf{v}) = \lambda \mathbf{x}$, we have that $T(\mathbf{x}) = \lambda \mathbf{x}$. Thus $T$ is of the form $T(\mathbf{v}) = \lambda \mathbf{v}$, and we are done.
Exercise 8
Let $T$ be the "ruler" map, which means that it measures the "length" of any $\mathbf{v} \in \mathbf{R}^2$, which is given by $T(\mathbf{v}) = |\mathbf{v}|$. Take $\mathbf{v} = -\mathbf{x}$, and we immediately see that $|\mathbf{-x}| \neq -|\mathbf{x}|$.
Exercise 9
Let $T$ be the "protractor" map, which takes any complex number and returns its argument w.r.t the real line. $T$ satisfies $T(u+v) = T(u)+T(v)$ for any $u,v \in \mathbf{C}$, but stretching any complex number by a scalar will not change its argument, so $T(av) \neq aT(v)$.
Exercise 10
$S$ is a linear map on $U$, so it would be futile to use elements of $U$ to achieve a contradiction. Hence we should consider the sum of an element of $U$ and an element in $V - U$.
Let $u \in U$ such that $S(u) \neq 0$, $v \in V$ and $v \notin U$. Then $u+v$ cannot be in $U$, otherwise $(u+v)+(-u) = v \in U$, a contradiction.
Now suppose $T$ were a linear map. Then $T(u+v) = 0$, because by construction $T(x)=0$ for any $x \notin U$. But $T(u+v)=T(u)+T(v)=S(u)+T(v)=S(u)+0=S(u)=0$, contradicting the fact that $u$ wasn't a root of $S$. Hence $T$ wasn't a linear map after all.
Exercise 11
We construct the extension of $S$ to cover $V$, given by
$T(v)= \begin{cases} S(v), & \text{if $v \in U$} \\ v, & \text{if $v \notin U$} \end{cases} $
Since $S$ is already a linear map on $U$, we only need to check if $T$ is indeed linear for $v \in V-U$. Given $u,v \in V-U$, we clearly have $T(u+v) = u+v = T(u)+T(v)$, and given $\lambda u \in V-U$, we have $T(\lambda u) = \lambda u = \lambda T(u)$.
Addendum: The above "proof" is erroneous. The mistake lies in the fact that I was under no circumstances permitted to conclude that $u = T(u)$. It is indeed true that $u = S(u)$, but how do I know that $S(u) = T(u)$?
I have since learned that the proper way to go about this is to extend the basis of $U$ to a basis of $V$, then defining $T(u_j) = S(u_j)$ where each $u_j$ is a vector in the basis of $U$, and $T(v_k) = 0$ where each $v_k$ is a vector added to $U$'s basis to extend it to a basis of $V$.
Exercise 12
It suffices to exhibit a sequence of linear maps $T_1, T_2, T_3, \dots$ from $V$ to $W$ such that $T_1, T_2, \dots, T_j$ is linearly independent for every natural number $j$.
To do this, note that $V$ is finite dimensional. Let $n$ be its dimension. Then there exists a basis $B_1 = (v_1, v_2, \dots, v_n)$ for $V$. Meanwhile, $W$ being infinite dimensional implies that there exists a sequence $B_2 = w_1, w_2, \dots$ such any set consisting of a finite number of vectors from $B_2$ would be linearly independent.
I construct the sequence of maps $T_1, T_2, T_3,$ $\dots$ as follows:
$T{i} (v{j}) = w_{i+j-1}$ for every natural number $i$ (starting from $1$) and for $1 \leq j \leq n$.
For example, $T3$ is given by $T{3} (v_1) = w3$, $T{3} (v_2) = w4$, $T{3} (v_3) = w4$, $\dots$, $T{3} (vn) = w{n+2}$.
I claim that $T_1, T_2, \dots, T_n$ for every natural $n$ must be linearly independent. To see this, suppose we have $a_1 T_1 (\mathbf{v}) + a_2 T_2 (\mathbf{v}) + \dots + a_n T_n (\mathbf{v}) = 0$. This must be true for every $\mathbf{v} \in V$. In particular, it must also be true for $\mathbf{v} = v_1$. Thus we have $a_1 T_1 (v_1) + a_2 T_2 (v_1) + \dots + a_n T_n (v_1) = a_1 w_1 + a_2 w_2 + \dots + a_n w_n = 0$. But $w_1, w_2, \dots, w_n$ is linearly independent, hence $a_1 = \dots = a_n = 0$, and we're done.
Exercise 13
This problem stumped me, and I had to look online for clues. I don't want to dwell too long on a problem to the extent that I can't proceed further.
Anyway, the heart of this problem (I think) is the fact that attempting to define a linear transformation on a vector space by specifying where it sends vectors of a spanning set to another vector space is not enough, because the vectors in the spanning set might be linearly dependent. So we must have a basis in order to define a unique linear map taking a finite set of vectors to another finite set of vectors of the same length.
Suppose we had a linearly dependent list of vectors $v_1, \dots, v_m$ in $V$. That means there exist scalars $a_1, \dots, a_m$, not all zero, such that $a_1 v_1 + \dots + a_m v_m = 0$. If we try to guess a linear map $T$ to map $a_1 v_1 + \dots + a_m v_m $ to an element in $W$, we'll find $T(a_1 v_1 + \dots + a_m v_m) = a_1 T(v_1) + \dots + a_m T(v_m) = 0$. We are free to choose the image of $v_j$ under $T$ for each $1 \leq j \leq m$. Take any $a_j \neq 0$, and define $T(a_j) = 1$. Then we get a contradiction.
Exercise 14
We can take $V=\mathbf{R}^3$, and maps $S,T$ from $\mathbf{R}^3$ to itself such that $S,T$ are distinct rotations of the space about $(0,0)$.
3.B