p. 73, Ex. 4.

Yes, there is such a linear transformation: The two vectors $v _ 1 = (1,-1,1)$ and $v _ 2 = (1,1,1)$ are not scalar multiples of each other and so are linearly independent. Therefore they can be extended to a basis $v _ 1, v _ 2, v _ 3$ of $\mathbb{R}^3$. By Theorem 1, given any three vectors $w _ 1, w _ 2, w _ 3$ in a second vector space, which can be $\mathbb{R}^2$, there is a linear transformation taking $v _ i \mapsto w _ i$ for each $i$. We can choose $w _ 1 = (1,0)$, $w _ 2 = (0,1)$ and $w _ 3 =$ anything we want, say $(0,0)$.

In this problem we were not actually asked to find the transformation.

p. 73, ex. 5.

Notice that $\alpha _ 3 = - \alpha _ 1 - \alpha _ 2$. Since a linear transformation preserves linear relations, if such a $T$ exists then we should have $\beta _ 3 = - \beta _ 1 - \beta _ 2$ also. But instead, $\beta _ 3 = \beta _ 1 + \beta _ 2 \neq -\beta _ 1 - \beta _ 2$. Therefore such a linear transformation does not exist.

p. 73, Ex. 6.

By $\epsilon _ i$ the authors mean our standard basis vector $e _ i$. We know $T = \tau _ A$ for $A = \left[\begin{array}{cc}a&c\\ b&d\end{array}\right]$. To display the answer in a form like Example 1, calculate $\left[\begin{array}{cc}a&c\\ b&d\end{array}\right] \left[\begin{array}{c}x\\ y\end{array}\right] = \left[\begin{array}{cc}ax+cy\\ bx+dy\end{array}\right]$, which says $T(x,y) = (ax+cy, bx+dy)$.

p. 73, Ex. 8.

This is a transformation on $\mathbb{R}^3 \rightarrow \mathbb{R}^3$, so we can use $\tau _ A$ where $A$ is a $3 \times 3$ matrix. The range of $\tau _ A$ is the column space of $A$, so take $A$ to have columns $\left[\begin{array}{r}1\\ 0\\ -1\end{array}\right]$ and $\left[\begin{array}{c}1\\ 2\\ 2\end{array}\right]$. For the third column of $A$ we can take any vector already in the span of those two columns, say $\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$. So we can choose $A = \left[\begin{array}{rrr}1&1&0\\ 0&2&0\\ -1&2&0\end{array}\right]$.

For Problem O-1:

Advice: First write down some examples of functions $f:\{1,2,3\}\rightarrow \mathbb{R}$. You will see that you need a list of three numbers for each, so each function corresponds to a triple. Furthermore, functions add pointwise and triples coordinatewise, so that vector addition is compatible with the correspondence, and similarly for multiplication by scalars. That gives the following idea.

Given $f: \{1,\dots, n\}\rightarrow F$, define $\phi(f) = (f(1),\dots, f(n))$. Then $\phi:$Functions$(\{1,\dots, n\}\rightarrow F) \rightarrow F^n$ is one-to-one and onto. Also, $\phi$ is a linear transformation, because the pointwise operations on functions correspond to the coordinatewise operations on $n$-tuples.

The importance of this problem is the perspective that it gives: Our first examples of vector spaces included some consisting of $n$-tuples and some consisting of functions. Now it is becoming apparent that even the $n$-tuples were really made of functions in disguise. How about examples such as Mat$(F, m \times n)$? This could be viewed as Functions$(S,F)$, where $S$ is the set $\{(i,j)\vert 1 \leq i \leq m, 1 \leq j \leq n\}$ of pairs of indices, or more simply, $S = \{1,\dots, m\} \times \{1,\dots, n\}$.

For Problem O-2: (b) Just make sure that both of $T(e_1)$ and $T(e_2)$ are on the same line. One of them could be the zero vector.

A shear is a transformation that moves points parallel to some line. The line is the $x$-axis in the case of this program.

For Problem O-3:

(a) Yes: For closure under addition,, suppose $w _ 1, w _ 2 \in T(S)$. We must show that $w _ 1 + w _ 2 \in T(S)$. To say $w _ i \in T(S)$ is the same as saying that $w _ i = T(v _ i)$ for some $v _ i \in S$ ($i = 1,2$). Since $S$ is a subspace, $v _ 1 + v _ 2 \in S$, and since $T$ preserves addition, $T(S)$ contains $T(v _ 1 + v _ 2) = T(v _ 1) + T(v _ 2) = w _ 1 + w _ 2$. For closure under multiplication by scalars ...[continue with similar proof].

(b) Yes: For closure under addition, suppose $v _ 1, v _ 2 \in T^{-1}(U)$, which is the same as saying that $T(v _ i ) \in U$. We must show that $v _ 1 + v _ 2 \in T^{-1}(U)$, which is the same as saying that $T(v _ 1 + v _ 2) \in U$. Since $T$ is linear, $T(v _ 1 + v _ 2) = T(v _ 1) + T(v _ 2)$, which is in $U$ because $U$ is closed under addition. Therefore $v _ 1 + v _ 2 \in T^{-1}(U)$ as required. For closure under multiplication by scalars ...[continue with similar proof].

For Problem O-4:

(ii) is false, while (i), (iii), (iv) are true. (iv) is proved in the handout.

For (i):

$y \in f(A _ 1 \cup A _ 2) \stackrel{(1)}{\Leftrightarrow} y = f(x)$ for some $x \in A _ 1 \cup A _ 2$ $\stackrel{(2)}{\Leftrightarrow} y = f(x)$ for some $x$ with $x \in A _ 1$ or $x \in A _ 2$ $\stackrel{(3)}{\Leftrightarrow} y = f(x)$ for some $x \in A _ 1$ or some $x \in A _ 2 \stackrel{(4)}{\Leftrightarrow} y \in f(A _ 1)$ or $y \in f(A _ 2) \stackrel{(5)}{\Leftrightarrow} x \in f(A _ 1) \cup f(A _ 2)$, where the reasons for the double implications are (1) definition of $f($subset$)$, (2) definition of $\cup$, (3) the way ``for some'' (really $\exists$) interacts with ``or'', (4) definition of $f($subset$)$, (5) definition of $\cup$. Therefore $f(A _ 1 \cup A _ 2) = f(A _ 1) \cup f(A _ 2)$,

For (ii):

The clearest kind of example is one where $A _ 1 \cap A _ 2 = \emptyset$ but $f(A _ 1) \cap f(A _ 2)$ is not empty. For instance let $X = \{1,2\}$, let $Y = \{3\}$, define $f:X\rightarrow Y$ by $f(1) = f(2) = 3$, let $A _ 1 = \{1\}$, and let $A _ 2 = \{2\}$. Then $f(A _ 1 \cap A _ 2) = f(\emptyset) = \emptyset$ but $f(A _ 1) \cap f(A _ 2) = \{3\} \cap \{3\} = \{3\} \neq \emptyset$.

For (iii):

$x \in f^{-1}(B _ 1 \cup B _ 2) \stackrel{(1)}{\Leftrightarrow} f(x) \in B _ 1 \cup B _ 2 \stackrel{(2)}{\Leftrightarrow} f(x) \in B _ 1$    or $f(x) \in B _ 2 \stackrel{(3)}{\Leftrightarrow}$
$x \in f^{-1}(B _ 1)$    or $x \in f^{-1}(B _ 2) \stackrel{(4)}{\Leftrightarrow} x \in f^{-1}(B _ 1) \cup f^{-1}(B _ 2)$,

where the reasons for the double implications are (1) definition of $f^{-1}$, (2) definition of $\cup$, (3) definition of $f^{-1}$, (4) definition of $\cup$. Therefore
$f^{-1}(B _ 1 \cup B _ 2) = f^{-1}(B _ 1) \cup f^{-1}(B _ 2)$.