Quotient Vector Spaces and $\dim (U+V+W)$

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This post is for those who know equivalence relation. With the concept of quotient spaces one can give another proof of \[\dim (U+V)=\dim U+\dim V-\dim (U\cap V)\] (example 18 of note 3) without the messy checking on the number of basis.

Quotient Vector Spaces

Let $V$ be a vector space and $U$ a vector subspace of $V$, it can be checked that for $x,y\in V$, the relation $x\sim y$ defined by \[

 x-y\in U

\] is indeed an equivalence relation on the set $V$. For $x\in V$, let $[x]$ denote the equivalence class of $x$, namely,\begin{align*}

[x]&=\{v\in V:v\sim x\}\\
&=\{v\in V: v-x\in U\}\\
&=\{v\in V:v\in x+U\} \\
&= x+U:=\{x+u:u\in U\}.

\end{align*} We denote $V/\text{$\sim$} = \{[v]:v\in V\}$ the collection of ALL equivalence class, read as the set $V$ modulo the relation $\sim$. This set is called a quotient space. Conventionally this is also denoted as \[

V/U:=V/\text{$\sim$} =\{[v]:v\in V\} = \{v+U:v\in V\},

\] read as "$V$ mod $U$". It is worth noting that quotient space is still a vector space with the naturally defined addition \[(x+U)+(y+U)=(x+y)+U\] and scalar multiplication \[\alpha (x+U)=\alpha x + U.\] It is easy to check that these definitions are well-defined. Namely, the definition of these operations are independent of the choices of representatives. Moreover, the zero element in $V/U$ is the class $0_{V/U} = [0]$. We shall denote this zero element in $V/U$ also by $0$. For $x+U\in V/U$, we observe that \[

x+U=0\iff [x]=[0] \iff x\sim 0\iff x-0\in U\iff x\in U.

\] Hence the representatives of zero element in $V/U$ are precisely every element in $U$. Let's look at the first result:

Theorem 1. Let $X$ be a finite dimensional vector space and $Y$ its subspace, then \[

\dim (X/Y)=\dim X-\dim Y.

\]

Proof.  Let $i:X\to X/Y$ be the natural "projection" map $x\mapsto x+Y$. Note that $i$ is onto. Also, \[

x\in \ker i\iff i(x)=0\iff x+Y=0\iff x\in Y,

\] hence $\ker i=Y$. By nullity-rank theorem applied to $i$, \[

\dim X= \dim Y+\dim (X/Y).

\] Thus the result follows.$\qed$

Recall that given two vector subspaces $X,Y$, the set $X+Y$ and $X\cap Y$ are still vector spaces. Now we have the following:

Theorem 2. Let $X$ and $Y$ be vector subspaces of some vector space, then \[

\frac{X+Y}{X\cap Y} = \frac{X}{X\cap Y}\oplus \frac{Y}{X\cap Y}.

\] Here $\frac{U}{V}$ is another notation for $U/V$.

Proof. Straightforward.$\qed$

Result. Let $U$ and $V$ be two finite dimensional subspaces of some vector space, then \begin{equation}\label{U+V}

\dim (U+V)=\dim U+\dim V-\dim (U\cap V).

\end{equation}

Proof. By theorem 2, \[\dim \frac{U+V}{U\cap V} = \dim \frac{U}{U\cap V}+\dim \frac{V}{U\cap V},\] by theorem 1, we have \[

\dim (U+V)-\dim (U\cap V)=\dim U-\dim (U\cap V)+\dim V - \dim (U\cap V).\qed

\]

Remark. Quotient is an important concept in almost everywhere. You will learn a much general concept in MATH3121 (Algebra I) or 3131 (Honors in Linear and Abstract Algebra II). For analytical aspect, one will learn this in MATH4063 (Functional Analysis) which is a good course for those who already have good background on real analysis (MATH3043) and general point-set topology (MATH4061 & MATH4225). Some of them (a set modulo some equivalence relation) even have smooth structure as a smooth manifold which we will learn in MATH4033 (Calculus on Manifold).

Another proof can be found here which uses the concepts that I tried to avoid here.

Do we have inclusion-exclusion principle for vector spaces?

For finite sets $A,B,C$, let $|A|,|B|,|C|$ denote their corresponding cardinalities, you have probably learnt from combinatorics the inclusion-exclusion principle:\begin{equation}\label{iep 1}

|A\cup B|=|A|+|B|-|A\cap B|

\end{equation} and  \begin{equation}\label{iep 2}

|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|C\cap A|+|A\cap B\cap C|.

\end{equation} Due to (\ref{U+V}) and (\ref{iep 1}), it is natural to study whether (\ref{iep 2}) has an analogue for finite dimensional subspaces. Namely, do we have given 3 vector subspaces:

\begin{equation}\label{is this true}

\begin{aligned}
\dim (U+V+W)\mathop{=}\limits^{\LARGE ?}&\dim U+\dim V+\dim W \\
&- \dim (U\cap V)-\dim (V\cap W)-\dim (W\cap U)\\
&+\dim (U\cap V\cap W).

\end{aligned}\end{equation}

The answer is NO! By applying (\ref{U+V}) to $U+V$ and $W$ and then also $U$ and $V$, one has \[
\dim ((U+V)+W) = \dim U +\dim V+\dim W-\dim U\cap V-\dim ((U+V)\cap W),
\] so (\ref{is this true}) is the same as \begin{equation}\label{equal?}

\dim ((U+V)\cap W)\mathop{=}\limits^{\LARGE ?}\underbrace{\dim (U\cap W)+\dim (V\cap W)-\dim (U\cap W\cap V)}_{\dim (U\cap W + V\cap W)}.

\end{equation} Finally, note that $U\cap W+V\cap W\subseteq (U+V)\cap W$, thus (\ref{equal?}) is the same as \[
U\cap W+V\cap W  \mathop{=}\limits^{\LARGE ?}  (U+V)\cap W.
\] Thus to disprove (\ref{is this true}), it is enough to raise example that \[

U\cap W+V\cap W \neq (U+V)\cap W.

\] Consider the following example,

let $V=\{(x,y,z): y=z\}$, $W=\{(x,y,z):y=2z\}$ and $U=\spann \{(0,1,-1)\}$. Since $U$ is a straight line pointing in the direction normal to the plane $V$, so $U+V=\R^3$ and \[

(U+V)\cap W = W.

\] However, $U\cap W$ is a point $\{(0,0,0)\}$ and $V\cap W$ is a straight line $\spann \{(1,0,0)\}$, so \[

U\cap W + V\cap W = \spann \{(1,0,0)\} \neq W.

\]

Or we may count both sides of (\ref{is this true}), since
\begin{align*}
\dim (U+V+W)&=3,\\
\dim U &= 1\\
\dim V&=2\\
\dim W&=2\\
\dim (U\cap V)&=0\\
\dim (V\cap W)&=1\\
\dim (W\cap U)&=0\\
\dim (U\cap V\cap W)&=0,
\end{align*}so (\ref{is this true}) becomes \[

3 \mathop{=}\limits^{\LARGE ?} 4,

\] which is of course wrong.