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I'm having some trouble understanding the concept of quotient space of a vector space. The definition my teacher gave us is the following:

Let $V$ be a finite-dimensional vector space over a field $\mathbb K$ and $V' \leq V$.

We can define the following equivalence relation: for any $u,v \in V, u \simeq v$ iff $v -u\in V'$. Let, for any $u\in V$, $[u]=\{v \in V: u \simeq v\}$.

Then we can define: $$V/V':=\{[u],u\in V\}$$

In the first question, my teacher asked us to determine the vector space $\mathbb R^2 / span(1,2)$ and I have no clue how to proceed because I'm still confused about the definition. Is there any intuition on why we define it this way or the meaning of the quotient space? And, how can I determine the space $\mathbb R^2 / span(1,2)$?

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    $\begingroup$ Imagine that we consider elements of $V'$ to be "negligible" and we consider two elements of $V$ to be "the same" if their difference is negligible (that is, if their difference is in $V'$). The resulting group is $V/V'$. $\endgroup$
    – littleO
    Mar 23, 2021 at 21:51
  • $\begingroup$ That is a very nice way of imagining it, thanks. So, for example, what would $R2/span(1,2) $ be? @littleO $\endgroup$ Mar 23, 2021 at 21:53
  • $\begingroup$ What about the definition is confusing you? $\endgroup$ Mar 23, 2021 at 21:55
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    $\begingroup$ Another thing to consider is that a way to think about $V'$ being "negligible" means that it's in the kernel of some map. This is actual made formal in the first isomorphism theorem, so you can think of $V/V'$ as the image of some everywhere else (i.e. not $V'$) bijective map, but whose kernel is $V'$ $\endgroup$ Mar 23, 2021 at 22:10
  • $\begingroup$ Take some vector $v \in \mathbb{R}^2$. When is $u \sim v$? Obviously if $u - v \in span(1, 2)$. Then the equivalence classes are lines. $\endgroup$
    – ureui
    Mar 23, 2021 at 22:20

2 Answers 2

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The integers modulo a number are defined the same way. The integers $x$ and $y$ are congruent modulo $4$ if $x - y \in 4 \Bbb{Z}$, that is, if $x-y$ is an integral multiple of $4$.

This definition is really saying this:

  • let there be a large space, $V$ (the integers in modular arithmetic);
  • in $V$, let there be a small space $V'$ (the multiples of $4$ in the modular arithmetic example), all of whose elements we view as equivalent to zero in the quotient, "$V/V'$" (or $\Bbb{Z}/4\Bbb{Z}$);
  • we treat two elements of $V$ as equivalent in the quotient if they differ by an element of $V'$ (a multiple of $4$).

So, in the problem you are given, $V = \Bbb{R}^2$ and $V' = \mathrm{span}\left( (1,2) \right)$. We check that $\mathrm{span}\left( (1,2) \right)$ is a subspace of $V$, which it is. We want to treat everything in $V'$ is equivalent to $0$. (All real multiples of $(1,2)$ are equivalent to $0(1,2) = \vec{0}$.) One way to draw a sketch of what is going on is the line spanned by $(1,2)$ is being crushed onto the origin of $\Bbb{R}^2$ and every parallel line is being crushed the same way, leaving a line perpendicular to $(1,2)$.

Let's check that. What is $(2,3)$ equivalent to in the quotient? $(2,3)$ is equivalent to $y \in \Bbb{R}^2$ if $(2,3) - y \in \mathrm{span}\left( (1,2) \right)$. There are infinitely many solutions, but a very, very easy one to find is $(2,3) - (1,1) = (1,2) \in \mathrm{span}\left( (1,2) \right)$.

So $(2,3)$ and $(1,1)$ are equivalent in this quotient. How can we find everything that is equivalent to $(2,3)$ in this quotient? Notice that every element of $\mathrm{span}\left( (1,2) \right)$ is a $\Bbb{R}$-multiple of $(1,2)$, so $$ \mathrm{span}\left( (1,2) \right) = \{r(1,2) \mid r \in \Bbb{R} \} \text{.} $$ So, we ask for $y = (y_1,y_2)$ such that $$ (2,3) - (y_1, y_2) = r(1,2) $$ for some real $r$. But this is an easy vector equation. Taken in components, it is the system \begin{align*} 2 - y_1 &= r \\ 3 - y_2 &= 2r \text{.} \end{align*} Eliminating $r$ between these and simplifying a little, we obtain $$ 2y_1 - 1 = y_2 \text{.} $$ This means every vector $(y_1, 2y_1 - 1)$ is equivalent to $(2,3)$ in this quotient.

(As a quick check, $(2,3)$ should be a vector in the set of equivalences generated as $y_1$ ranges through the reals. This does happen -- when $y_1 = 2$, $2y_1 - 1 = 3$.)

Of course, the set $\{(y_1, 2y_1 - 1) \mid y_1 \in \Bbb{R}\}$ is a line, so an entire line is equivalent to $(2,3)$.

If you do more examples, you will find that each set of equivalent vectors lie along a line and all of the lines are parallel to the vector $(1,2)$. Now think of the subspace spanned by $(2,-1)$. This subspace is perpendicular to $(1,2)$ and no two vectors in $\mathrm{span}\left( (2,-1) \right)$ are equivalent in the quotient. How would we show that? We use the definition: two points in that span are equivalent if their difference is in $V'$, i.e., there are reals $r$ and $s$ such that there is a real $t$ where $$ r(2,-1) - s(2,-1) = t(1,2) \text{.} $$ As a system of equations among components, \begin{align*} 2(r-s) &= t \\ -1(r-s) &= 2t \text{.} \end{align*} Using the first to replace $t$ in the second, $$ s - r = 2(2(r-s)) \text{,} $$ which simplifies to $$ 5s = 5r \text{.} $$ So two points in the span of $(2,-1)$ are equivalent only when $r = s$, so when they are both the same point. Further, when $r = s$, $$ r(2,-1) - s(2,-1) = t(1,2) $$ reduces to $0 = t$, so the one point that we could be talking about is the origin. And, yeah, the origin is equivalent to itself in every quotient and the origin is in every subspace.

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Maybe a concrete example will help your intuition: imagine a $3$-dimensional scenario and a (scale $1{-}1$) $2$-dimensional map of that scenario. The map shows two points $(x, y, z)$ and $(x, y, z')$ as the same point, i.e., it identifies $(x, y, z)$ and $(x', y', z')$ if $(x, y, z) - (x', y', z') = (0, 0, z'')$ for some $z''$. So with $V= \Bbb{R}^3$ and $V' = \{ (0, 0, z'') \mid z'' \in \Bbb{R}\}$, you can think of the points on the map as the equivalence classes under the equivalence relation your teacher defined.

For $\Bbb{R}^2/\mathrm{span}\{(1, 2)\}$, you can describe the quotient space as the set of all lines equal to or parallel to the line $L$ comprising all points of the form $(t, 2t)$. You can identify any such line by its point of intersection with some chosen line $L'$ through $(0, 0)$ that is not parallel to $L$. $L'$ is a subspace of $\Bbb{R}^2$ and the vector space operations on $\Bbb{R}^2/\mathrm{span}\{(1, 2)\}$ are the same as those on $L'$ under this identification.

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