# Proof involving subspaces

I encountered this question in a document I found on a google search, it bugged me because my perception keeps telling me I'm wrong no matter what I do.

Let $U$, $W$ and $Z$ be subspaces of a vector space $V$ and the following conditions are satisfied:

• $U + Z = W +Z$
• $U \cap Z = W\cap Z$
• $U \subset W$

Prove that $U=W$

So I tried proving by contradiction that $U$ is not equal to $W$, by saying there exists some vector $w \in W \setminus U$ and if $w \in W \cap Z$ then $w \in U \cap Z$ which means $w \in U$. At this momment my mind is shouting "This isn't it", and I think I should be using a different approach.

• Hey just a little tip: you can use \  to get whitespace, instead of \space. For a big space there's also \quad. Good luck with your linalg! – GPerez Dec 21 '14 at 22:18
• Instead of $U \subset W$, do you mean $U \subseteq W$? Doesn't the former imply that $\exists u \in U \mathrel{\mathrm{s.t.}} u \notin W$, which in turn implies that $U \neq W$? – wchargin Dec 22 '14 at 3:57

No contradiction and no considerations about dimension. Let $w\in W$; then $w\in W+Z$, so $w=u+z$, for some $u\in U$ and $z\in Z$.

But now $z=w-u\in W$, since $U\subseteq W$. Therefore $w-u\in W\cap Z$, so $w-u\in U$, since $U\cap Z=W\cap Z$. Thus $$w=(w-u)+u\in U.$$

Another view at this. The lattice of subspaces of a vector space is modular; this means that, if $U\subseteq W$, then $$U+(W\cap Z)=(U+Z)\cap W$$ for every subspace $Z$ (see Wikipedia). Under our hypotheses the left hand side is $U+(U\cap Z)=U$, while the right hand side is $(W+Z)\cap W=W$. Thus $U=W$.

Proving the modular identity is a good exercise.

$$\frac{U}{U \cap Z} \cong \frac{U + Z}{ Z}$$ $$\frac{W}{W \cap Z} \cong \frac{W + Z}{ Z}$$ But $U + Z = W + Z$ so $$\frac{U}{U \cap Z} \cong \frac{W}{W \cap Z} = \frac{W}{U \cap Z}$$ This implies that $$\dim(U) = \dim (W)$$ but $U \subseteq W$ and so $$U = W$$

• oh wow to be honest I dont even know what these notations are :D im just a beginner in linear algebra, but I'll do a search. thanks. – Tal Dec 21 '14 at 20:53
• @Tal: search for isomorphism theorems – WLOG Dec 21 '14 at 21:01

You have your $w\in W\setminus U$. The vector $0\in Z$ so

$$w=w+0\in W+Z=U+Z\;,$$

and there must be $u\in U$ and $z\in Z$ such that $w=u+z$. $U\subseteq W$, so $w-u\in W$,and therefore

$$w-u=z\in W\cap Z=U\cap Z\;.$$

But then $z\in U$ and hence $w=u+z\in U$, contradicting the choice of $w$.

First, $U+Z=W+Z\implies \dim(U+Z)=\dim(W+Z)\implies$$\dim U+\dim Z-\dim (U\cap Z)=\dim W +\dim Z-\dim(W\cap Z)\implies$$ $$\dim U-\dim(U\cap Z)=\dim W-\dim(W\cap Z)\ \ \ \ (\star)$$ But$\dim(U\cap Z)=\dim(W\cap Z)$, by assumption. So putting this into$(\star)$, we get $$\dim U=\dim W$$ And then, since$U\subseteq W$, we know that$U=W$. First two conditions says that, dim$U =$dim$ W. $So using the fact that$U \subset W,$we have$U= W.$WARNING: This process can only be applied if all the subspaces involved (i.e.$U, W, Z$) are finite dimensional. But this is not a part of the assumption. Thanks to egreg for pointing it out. • Does the question say that the space in which the subspaces live is finite dimensional? – egreg Dec 21 '14 at 22:15 • @egreg: Sorry!! Somehow I missed it. This proof won't work if all the subspaces involved are NOT finite dimensional. But if all the subspaces involved are finite dimensional, it's fine. We don't need the whole space to be finite dimensional. – Krish Dec 22 '14 at 0:15 • The subspaces live in$U+W+Z$, which is finite dimensional if all three are. – egreg Dec 22 '14 at 0:29 • @egreg: By "whole space" I mean$V$, not$U + W + Z.\$ We are saying the same thing, just a little notational confusion. Anyway, I've edited my answer as pointed out by you. – Krish Dec 22 '14 at 0:36