I need some help with exercise 10 in Chapter 2.2 of Beachy and Blair's Abstract Algebra with a Concrete Introduction.
The question is as follows:
Let $W$ be a subspace of a vector space $V$ over $\mathbb{R}$ (that is the scalars are assumed to be real numbers). We say two vectors $u,v \in V$ are congruent modulo $W$ if $u-v \in W$, written $u \equiv v \pmod{W}$.
- Show that $\equiv$ is an equivalence relation.
- Show that if $r,s$ are scalars and $u_1, u_2, v_1, v_2$ are vectors in $V$ such that $u_1 \equiv v_1 \pmod{W}$ and $u_2 \equiv v_2 \pmod{W}$, then $ru_1 + su_2 \equiv rv_1 + sv_2 \pmod{W}$.
- Let $[u]_W$ denote the equivalence class of the vector $u$. Set $U = \{[u]_W \mid u \in V\}$. Define $+$ and $\cdot$ on $U$ by $[u]_W + [v]_W = [u+v]_W$ and $r \cdot [u]_W = [ru]_W$ for all $u,v \in V$ and $r \in \mathbb{R}$. Show that $U$ is a vector space with respect to these operations.
- Let $V = \mathbb{R}^2$ and let $W = \{ (x,0) \mid x \in \mathbb{R} \}$. Describe the equivalence class $[x, y]_W$ geometrically. Show that $T : \mathbb{R} \to U$ defined by $T(y) = [0,y]_W$ is a linear transformation and is one-to-one and onto.
This is what I've got so far (verification on what I have would be great, but I need help on part 4):
- To prove that $\equiv$ is an equivalence class, we first show that for $u \in V$, we have $u \equiv u \pmod{W}$ because $u - u = \textbf{0}$ and $\textbf{0} \in W$ because $W$ is a vector space, thus showing $\equiv$ is reflexive. Now, for $u,v \in V$ such that $u \equiv v \pmod{W}$, we have $u - v \in W$, thus we must have the inverse of $u - v$ also in $W$, that is $v - u \in W$ so that $v \equiv u \pmod{W}$, showing that $\equiv$ is symmetric. Lastly, for $u,v,w \in V$ such that $u \equiv v \pmod{W}$ and $v \equiv w \pmod{W}$, we have $u - v \in W$ and $v - w \in W$, thus by the closure of addition in $W$, we have $(u - v) + (v - w) \in W$ or $u - w \in W$, that is $u \equiv w \pmod{W}$, showing that $\equiv$ is transitive.
- If $u_1 \equiv v_1 \pmod{W}$ and $u_2 \equiv v_2\pmod{W}$, then we have $u_1 - v_1 \in W$ and $u_2 - v_2 \in W$. This means that $ru_1 - rv_1 \in W$ and $su_2 - sv_2 \in W$ for $r,s \in \mathbb{R}$ by /scalar multiplication and the distributive property of scalar multiplication over addition. This means that $ru_1 - rv_1 + su_2 - sv_2 \in W$ by the closure property of addition, or $(ru_1 +su_2) - (rv_1 - sv_2) \in W$ by the axioms of vector spaces. By definition, $(ru_1 +su_2) - (rv_1 + sv_2) \in W$ is $ru_1 + su_2 \equiv rv_1 + sv_2 \pmod{W}$.
- We know that U must be closed under $+$ and $-$ because V is a vector space. We prove the commutativity of addition by seeing $[u]_W + [v]_W = [u + v]_W$ and $[v]_W + [u]_W = [v + u]_W$, but since vector addition is commutative in $V$, $[v + u]_W = [u + v]_w$, thus $[u]_W + [v]_W$. To show associativity of addition, we see for $u,v,w \in V$, $([u]_W + [v]_W) + [w]_W = [u + v]_W + [w]_W = [(u + v) + w]_W$ and $[u]_W + ([v]_W + [w]_W) = [u]_W + [v + w]_W = [u + (v + w)]_W$ and since $(u + v) + w = u + (v + w)$ by the associativity of addition in $V$, we have $([u]_W + [v]_W) + [w]_W = [u]_W + ([v]_W + [w]_W)$. Now we prove that $U$ has an additive identity, for $\textbf{0}_V, u \in V$, we have $[ \textbf{0}_V ]_W + [u]_W = [u]_W + [\textbf{0}_V]_W = [u + \textbf{0}_V]_W = [\textbf{0}_V + u]_W = [u]_W$, thus showwing that $[ \textbf{0}_V ]_W$ is the additive inverse in $U$. We continue by showing that $1 \in \mathbb{R}$ is the multiplicative identity in $U$, we see that for $u \in V$, $ 1 \cdot [u]_W = [1 \cdot u]_W = [u]_W$, thus $1$ is the multiplicative identity in $U$. Next, we show that there is an additive inverse in $U$, for $u \in V$, we have $-u \in V$, the additive inverse of $u$ in $V$, thus we see $[u]_W + [-u]_W = [-u]_W + [u]_W = [u - u]_W = [-u + u]_W = [\textbf{0}_V]_W$ which shows that $[-u]_W$ is the additive inverse of $[u]_W$ in $U$. Now we show the compatability of scalar multiplication in $U$, for $r,s \in \mathbb{R}$ and $u \in V$, we have $(rs) \cdot [u]_W = [rsu]_W$ and $r \cdot (s \cdot [u]_W) = r \cdot [su]_W = [rsu]_W$, thus $(rs) \cdot [u]_W = r \cdot (s \cdot [u]_W)$. To show the distributive property of scalar multiplication over addition, we see that for $r \in \mathbb{R}$ and $u,v \in V$, $r \cdot ( [u]_W + [v]_W ) = r \cdot [u + v]_W = [r (u + v)]_W = [ru + rv]_W = [ru]_W + [rv]_W = r \cdot [u]_W + r \cdot [v]_W$. Lastly, to show that field addition is distributive over scalar multiplication, we see that for $r,s \in \mathbb{R}$ and $u \in V$, $(r + s) \cdot [u]_W = [(r + s)u]_W = [ru + su]_W = [ru]_W + [su]_W = r \cdot [u]_W + s \cdot [u]_W$.
- The equivalence class $[ (x, \ y)]_W$ is the set of vectors in $\mathbb{R}^2$ that can be mapped via a shear transformation to the vector $(x, \ y) \in \mathbb{R}^2$ (that is, all vectors in $\mathbb{R}^2$ that have the same $y$ value). To prove that $T$ is a linear transformation, we see that for $y_1, y_2 \in \mathbb{R}$, $T( y_1 + y_2 ) = [ (0, \ y_1 + y_2) ]_W = [ (0, \ y_1) + (0, \ y_2) ]_W = [ (0, \ y_1) ]_W + [ (0, \ y_2) ]_W = T(y_1) + T(y_2)$ and for $c,y \in \mathbb{R}$, we have $T(cy) = [ (0, \ cy) ]_W = [ c(0, \ y) ]_W = c \cdot [ (0, \ y] = cT(y)$, thus showing that $T$ is a linear transformation. To show that $T$ is one-to-one, assume that $T(y_1) = T(y_2)$ for $y_1,y_2 \in \mathbb{R}$, thus we have $[ (0, \ y_1) ]_W = [ (0, \ y_2) ]_W$, but for these equivalence classes to be equal to each other, we must have $y_1 = y_2$.
I don't know how to show that $T$ is onto. (I hope I didn't make any typos)
Thanks!
Your work for 1.-3. is all correct.
For 4., observe that $(x,y)\equiv (0,y)\pmod W$ for any $(x,y)\in\Bbb R^2$, and thus, on one hand, the equivalence class of $(x,y)$ is the same as that of $(0,y)$, which is the horizontal line $\{(a,y):a\in\Bbb R\}$.
(Note that the congruence classes modulo any subspace $W$ are always the affine subspaces parallel to $W$, that is, the shifts $W+v$ of $W$ with all possible vectors $v$.)
On the other hand, it means $[(x,y)]_W=[(0,y)]_W=T(y)$, and since $(x,y)$ was arbitrary, it shows surjectivity of $T$.