W1 + W3 = W2 + W3. Then W1 = W2

Question

Claim. Let $V$ be a vector space over $F$, and suppose that $W_1, W_2,$ and $W_3$ are subspaces of $V$ such that $W_1 + W_3 = W_2 + W_3.$ Then $W_1 = W_2.$

Proof

$W_1 + W_3 - W_3 = W_2 + W_3 - W_3$

$\therefore$ $W_1 = W_2$

My proof feels like cheating, is it even valid? Also is there anyway to 'say', prove or demonstrate the same thing in a more rigorous way i.e. set builder notation?

In regards to the rules of the site is my question to simple, if so what can I do to make it better?

Answer 1

Consider $V=\mathbb{R}^2 $ and $W_3= \{(x,0) : x\in \mathbb{R}\}$, $W_1= \{(0,y) : y\in \mathbb{R}\}$ and $W_2= \{(t,t) : t\in \mathbb{R}\}$

You should be able to verify that $W_1+W_3 = W_2 + W_3$

What does that tell you about your claim?

Answer 2

What does your claim say if $W_3$ is the whole space $V$?

You can't subtract subspaces.

Answer 3

There is a very simple counterexample. Since the statement is about arbitrary subspaces $W_1$, $W_2$ and $W_3$ only subject to the condition that $W_1+W_3=W_2+W_3$, it should in particular hold for $W_1=\{0\}$, $W_2=W_3=V$. Then from $$ \{0\}+V=V+V $$ (which is true), you'd conclude that $$ \{0\}=V $$ Now any non trivial vector space is a counterexample.

From a slightly higher point of view, the set $\mathscr{L}(V)$ of subspaces of $V$ is a commutative monoid under the $+$ operation, because it is associative and has the neutral element $\{0\}$, because $\{0\}+W=W$, for every $W\in\mathscr{L}(V)$.

However, this monoid cannot be cancellative (that is $W_1+W_3=W_2+W_3$ implies $W_1=W_2$) for several reasons, the most important one being that it has an absorbing element, namely $V$: $$ W+V=V $$ for every $W\in\mathscr{L}(V)$. An absorbing element cannot have a symmetric element, unless the monoid is trivial.

Also, if $U$ is a subspace of $V$ and $\mathscr{L}(V;\subseteq U)$ denotes the set of subspaces of $V$ contained in $U$, we have $\mathscr{L}(V;\subseteq U)=\mathscr{L}(U)$, which is thus a submonoid of $\mathscr{L}(V)$. Since every $U\in\mathscr{L}(V)$ is the absorbing element in a submonoid, no element can have a symmetric element, except for $\{0\}$.

Note. By “symmetric element” of an element $x$ in a monoid $(M,*,e)$ I mean an element $y$ such that $x*y=y*x=e$.