Trying to get a better understanding of the quotient V/U in linear algebra

[Karl Karlsson]

TL;DR Summary

I have read definitions that V/U = {v + U : v ∈ V} . U is a subspace of V. But v + U is also defined as the set {v + u : u ∈ U}. So V/U is a set of sets is this the correct understanding of this?

A linear map $\pi : V \rightarrow V/U$ defined by $\pi(v) = v + U$ . So here $\pi$ takes the vector v as input and returns the set {v + u : u ∈ U} ? How is even linear independence, null space and such defined in a situation where you have a vector as input and then a set as output ?

Hi! I want to check if i have understood concepts regarding the quotient U/V correctly or not.

I have read definitions that $V/U = \{v + U : v ∈ V\}$ . U is a subspace of V. But v + U is also defined as the set $\{v + u : u ∈ U\}$. So V/U is a set of sets is this the correct understanding of this?

A linear map $\pi : V \rightarrow V/U$ defined by $\pi(v) = v + U$ . So here $\pi$ takes the vector v as input and returns the set ${v + u : u ∈ U}$ ? How is even linear independence, null space and such defined in a situation where you have a vector as input and then a set as output ?

For example i also found the theorem that says: $dim(V/U) = dim(V) - dim(U)$ which they said holds because $dim(V) = dim(null(\pi)) + dim(range(\pi)) = dim(U) + dim(V/U)$
I assume this must mean that $dim(null(\pi)) = dim(U)$ , why is that?

Thanks in advance!

 

[fresh_42]

Yes, $V/U$ is a set of sets. However, it is better to think of it as a set of equivalence classes. We define a relation $\sim$ on $V\times V$ by
$$
v\sim w \Longleftrightarrow v-w\in U
$$
This splits $V$ into classes where we write $v= w+u$ and note all $v$ to be in the same class, if they have the same $w$. So $v=w+u$ belongs to the same class as $v'=w+u'$. So $V/U=\{w+U\}$. A homomorphism $V \longrightarrow V/U$ selects such a $w$ as representative of the entire class $w+U$.

You can visualize the principle if you take $V=\mathbb{R}^2$, $U=\{(x,0)\in \mathbb{R}^2\}.$
The calculation should give $V/U \cong \{(0,y)\in \mathbb{R}^2\}.$

 

[Karl Karlsson]

You can visualize the principle if you take $V=\mathbb{R}^2$, $U=\{(x,0)\in \mathbb{R}^2\}.$
The calculation should give $V/U \cong \{(0,y)\in \mathbb{R}^2\}.$

Hi!

Okay, so is V/U the set that contains all sets of equivalence classes? Shouldn't V/U in the calculations of the example you brought up: $V=\mathbb{R}^2$, $U=\{(x,0)\in \mathbb{R}^2\}.$
, $V/U \cong \{(0,y)\in \mathbb{R}^2\}$ instead be $V/U \cong \{(0,y)+U : y\in R\} = \{\{(x,y): x\in R\} : y\in R\}$.
By the way does $\cong$ mean equality between $V/U$ and $\{(0,y)\in \mathbb{R}^2\}$ ?

 

[fresh_42]

Hi!

Okay, so is V/U the set that contains all sets of equivalence classes?

The equivalence classes are the sets. Every such class is considered one element, and the set of these elements carries again a vector space structure.

Shouldn't V/U in the calculations of the example you brought up: $V=\mathbb{R}^2$, $U=\{(x,0)\in \mathbb{R}^2\}.$
, $V/U \cong \{(0,y)\in \mathbb{R}^2\}$ instead be $V/U \cong \{(0,y)+U : y\in R\} = \{\{(x,y): x\in R\} : y\in R\}$.

The example is a bit limited, since we can only draw two dimensional pictures. $V=\mathbb{R}^2$ is the two dimensional Euclidean plane. $U=\{(x,0)\}$ is the one dimensional subspace, the $x$-axis. This gives us an (injective) inclusion $\iota:U \rightarrowtail V$, and a (surjective) projection $\pi:V \twoheadrightarrow V/U$. The equivalence classes $w+U$ are represented by $w=\pi(v)$, $v$ modulo the kernel of $\pi$, which is $U$. This means we do not distinguish between deviations by elements of $U.$ $\pi(v)=\pi(v+u)$ for any $u\in U.$ Vector spaces have the property that those equivalence classes, which are a vector space, too, can be identified with a subspace of $V$. This is not generally true for structures like groups, but here it is. This goes as follows:
$$
\pi\, : \,V=\{(x,y)\}\ni (a,b) \longmapsto (a,b)+U=(a,b)+\{(x,0)\}= (0,b)+\{(x,0)\} \in V/U
$$
and $(0,b)$ is a representative of an equivalence class modulo $U$ which can be interpreted as vector of $\mathbb{R}^2=V$ again. I.e. $V/U$ is the plane where all first coordinates are irrelevant. Hence we can identify $V/U$ with all vectors which first coordinate is zero. This gives us the $y$-axis. That means $V/U \cong y$-axis. The symbol stands for an isomorphism, a bijective linear transformation. It is this identification, because we identified $(0,b)$ with the set $$(a,b) +U =(a,b)+\{(x,0)\}\stackrel{1:1}{\longleftrightarrow}(a,b)+\underbrace{(-a,0)}_{\in U}=(0,b)$$ The points on the $y$-axis aren't sets, which is why we cannot say equal. We identified those points with sets via a bijective function, written as $\cong$ for isomorphism.

 

[Karl Karlsson]

The equivalence classes are the sets. Every such class is considered one element, and the set of these elements carries again a vector space structure.
The example is a bit limited, since we can only draw two dimensional pictures. $V=\mathbb{R}^2$ is the two dimensional Euclidean plane. $U=\{(x,0)\}$ is the one dimensional subspace, the $x$-axis. This gives us an (injective) inclusion $\iota:U \rightarrowtail V$, and a (surjective) projection $\pi:V \twoheadrightarrow V/U$. The equivalence classes $w+U$ are represented by $w=\pi(v)$, $v$ modulo the kernel of $\pi$, which is $U$. This means we do not distinguish between deviations by elements of $U.$ $\pi(v)=\pi(v+u)$ for any $u\in U.$ Vector spaces have the property that those equivalence classes, which are a vector space, too, can be identified with a subspace of $V$. This is not generally true for structures like groups, but here it is. This goes as follows:
$$
\pi\, : \,V=\{(x,y)\}\ni (a,b) \longmapsto (a,b)+U=(a,b)+\{(x,0)\}= (0,b)+\{(x,0)\} \in V/U
$$
and $(0,b)$ is a representative of an equivalence class modulo $U$ which can be interpreted as vector of $\mathbb{R}^2=V$ again. I.e. $V/U$ is the plane where all first coordinates are irrelevant. Hence we can identify $V/U$ with all vectors which first coordinate is zero. This gives us the $y$-axis. That means $V/U \cong y$-axis. The symbol stands for an isomorphism, a bijective linear transformation. It is this identification, because we identified $(0,b)$ with the set $$(a,b) +U =(a,b)+\{(x,0)\}\stackrel{1:1}{\longleftrightarrow}(a,b)+\underbrace{(-a,0)}_{\in U}=(0,b)$$ The points on the $y$-axis aren't sets, which is why we cannot say equal. We identified those points with sets via a bijective function, written as $\cong$ for isomorphism.

So when we write $\pi:V \twoheadrightarrow V/U$ we actually don't mean that we are taking a vector from v as input and returning an equivalence class (a set that is an element). We actually mean that we are returning a vector that is in V but not in U because this represents the same as the set v+U (as you showed they are isomorphic) which is what the elements in V/U actually are?

Have I understood this correctly? Is this the reason why $\pi : V \rightarrow V/U$ where $\pi(v) = v + U, v \in V$ have the property $null(\pi) = U$ since the return value of $\pi$ is not interpreted as $v + U$ but as v modulo U?

 

[fresh_42]

So when we write $\pi:V \twoheadrightarrow V/U$ we actually don't mean that we are taking a vector from v as input and returning an equivalence class (a set that is an element).

No. We do mean the sets: $v+U$. It has a well defined vector space structure itself: $(v+U)+(w+U)=(v+w)+U .$ It would be a good exercise to check why this is well defined.

We actually mean that we are returning a vector that is in V but not in U because this represents the same as the set v+U (as you showed they are isomorphic) which is what the elements in V/U actually are?

No. This is a special property which vector spaces have. In general we start with a so called short exact sequence
$$
\{0\} \rightarrowtail U \stackrel{\iota}{\rightarrowtail} V \stackrel{\pi}{\twoheadrightarrow} V/U \twoheadrightarrow \{0\}
$$
Exact means that the image of $\iota$ equals the kernel of $\pi$, short means that the sequence has only three relevant objects aka vector spaces.

Now comes the special property if we deal with vector spaces, namely that there is a linear transformation back into $V.$
$$
\varphi\, : \,V/U \rightarrowtail V \text{ such that } \varphi\,\circ\, \pi = \operatorname{id}_V
$$
One says: the short exact sequence splits. It always splits in vector spaces, it does not (automatically) e.g. in groups or rings.

Have I understood this correctly? Is this the reason why $\pi : V \rightarrow V/U$ where $\pi(v) = v + U, v \in V$ have the property $null(\pi) = U$ since the return value of $\pi$ is not interpreted as $v + U$ but as v modulo U?

The example with the $x$-axis and the $y$-axis was a very simple one, which illustrated the principle without any further complications. One calls the projection $\pi$ canonical (= as usually done), and $\varphi$ could be called natural (= pops out without any further calculations) in this case.

$V/U$ has the structure of a one dimensional vector space. Of course it can be embedded in the higher dimensional space $V$ in many ways. We could even identify it with $U$ itself, since all one dimensional vector spaces are isomorphic. But this would cause confusion, and break down if we used a three dimensional $V$ and a two dimensional $U$. We could also identify it with any straight line $(a,b)+U \leftrightarrow \{(1,b)\}$ but this wouldn't be a linear function anymore, only a bijective mapping, which is unnecessarily inconvenient when dealing with vector spaces, hence the identification with the $y$-axis $\{(0,b)\}.$

The natural way in this case is the split to the $y$-axis. However, again: the split of the sequence is a special property of vector spaces. If we only consider $V \twoheadrightarrow V/U$, then we consider the elements of $V/U$ as equivalence classes. But as it is a vector space itself, there is no need to regard them as sets. You can distinguish the elements notationally e.g. by a bar: $v \longmapsto \bar{v}$ or just $\pi(v)$. If we deal with integers $z$, then $[z]$ is often used. The extra bar, however, is often omitted and people just write $v \in V/U.$ This should only be done if there is no risk of confusion. The word projection for $\pi: V\longrightarrow V/U$ can be taken literally. You can look at it as the projection of some higher dimensional onto some lower dimensional space, just as a projection of the real world onto a painting, or some photo dias.

 

[Karl Karlsson]

Doesn't V/U generally span V if V is a vectorspace since $V/U = {v+U:v\in V}$ and $v+U$ is defined as ${v+u:u\in U}$ if we but these together we get $V/U = {{v+u:u\in U}:v\in V}$ since v is any v V/U should always span V, right?

 

[fresh_42]

Doesn't V/U generally span V if V is a vectorspace since $V/U = {v+U:v\in V}$ and $v+U$ is defined as ${v+u:u\in U}$ if we but these together we get $V/U = {{v+u:u\in U}:v\in V}$ since v is any v V/U should always span V, right?

No. It doesn't span $U$.

 

[FactChecker]

Doesn't V/U generally span V if V is a vectorspace since $V/U = {v+U:v\in V}$ and $v+U$ is defined as ${v+u:u\in U}$ if we but these together we get $V/U = {{v+u:u\in U}:v\in V}$ since v is any v V/U should always span V, right?

Because elements of V/U are not in the V space, you can not say that they span V. V/U has a smaller dimension than V. Suppose you had collected data of heights and weights of people Data = set of (height, weight). For many applications, you would only care about height and would ignore weight. That means that you would consider people of the same height to be equivalent, regardless of any weight difference. That is, Height =~ Data/Weight. You should not say that Data/Weight spans Data, since elements of Data/Weight are the equivalence classes of a specific height regardless of weight, which are not in the Data space. In other words, elements of Data/Weight say nothing about Weight and ignore it completely, so it can not be said to span Data, which includes specific weights and distinguishes between different weights.

 

[WWGD]

Hi!

Okay, so is V/U the set that contains all sets of equivalence classes? Shouldn't V/U in the calculations of the example you brought up: $V=\mathbb{R}^2$, $U=\{(x,0)\in \mathbb{R}^2\}.$
, $V/U \cong \{(0,y)\in \mathbb{R}^2\}$ instead be $V/U \cong \{(0,y)+U : y\in R\} = \{\{(x,y): x\in R\} : y\in R\}$.
By the way does $\cong$ mean equality between $V/U$ and $\{(0,y)\in \mathbb{R}^2\}$ ?

Actually, V\U is a vector space over the same field as V. Define sum [a]+=[a+b] ; choose any two representatives. Similat for scaling. c[v]=[cv]. May be a good exercise to show these are well-defined, i.e... independent of the choice of class. See too, if you can define a basis for V/U.

 

[WWGD]

Like Factchecker wrote, V is a space of vectors while V/U is a space of equivalent classes. V by itself contains vectors and not equivalence classes. Besides, the dimension of V/U = DimV-DimU. So unless U is the 0-space, Dim V/U is less than Dim V.