0
$\begingroup$

Problem : If $ M $ and $N$ are two subspaces of the vector space $V$ such that $\forall v \in V $ , $ v \in M $ or $ v \in $ (or both) . Prove that at least one of the is equal to $ V $

My Approach Assume that $ M \neq V $ , we are required to prove $ N = V $ .

Assume contrary, let $\dim N < \dim V$.

We assume that $ M \cap N = \phi $ , in case the intersection is not null, we delete the elements from $ N $ . Now, $$\dim(M+N) + \dim (M\cap N) = \dim M + \dim N$$

Since $ \dim(M\cap N) = 0 $ and $M+N = V$. Let $\dim M =m , \dim N = n . \dim V = v$. Therefore we get $n+m = v$ We know there is an isomorphism between $V$ and $\mathbb F^v$.

Consider the vectors in $\mathbb V$ corresponding to $(1,0,0,\dots) , (0,1,0,0,\dots), \dots$ in $\mathbb F^v$.

Exactly $n$ of these belong to $N$ (which form a basis) and exactly m of these belong to $M$ (which again form a basis).WLOG we can assume that

$(1,0,0,\dots) , (0,1,0,\dots) , \dots , (0,0,\dots,1, 0,\dots) $ are the vectors present in $N $ ...$(*)$

Now consider the vector corresponding to $(1,1,1,\dots)$. It must belong to either $ N $ or $ M$ . WLOG assume it belongs to $N$ . Then we can get a set of $n+1$ linearly independent vectors in $N$. $n$ vectors as described at $(*)$ and one of the form $(0,0,0,\dots , 1,1,1,\dots)$ But since we know that number of elements in basis is greater or equal to the number of elements in an independent set, we arrive at a contradiction since $n+1 > n$

My proof inherently assumes that the vector space given is a finite dimensional vector space. And also I feel that my proof is really clumsy. Is there a better way to prove it ?

$\endgroup$
  • $\begingroup$ "in case the intersection is not null, we delete the elements from \mathbb{N}" : You can't delete elements from $\mathbb{N}$ and have $\mathbb{N}$ remain a subspace. $\endgroup$ – Morgan Rodgers Aug 24 '15 at 6:27
  • $\begingroup$ @Hirshy What about the vector (1,1) . It belongs to neither $W_1$ nor $W_2$ $\endgroup$ – Soham Aug 24 '15 at 6:29
  • $\begingroup$ @MorganRodgers and Lucyfer, of course, you're absolutely right. I shouldn't write something before having coffee. Just pretend that I said nothing... $\endgroup$ – Hirshy Aug 24 '15 at 6:30
  • $\begingroup$ @MorganRodgers I can delete elements and their span from N , then it would remain a subspace, right ? $\endgroup$ – Soham Aug 24 '15 at 6:31
  • $\begingroup$ @LucyferZedd No, for example think of taking a two-dimensional subspace and remove a one-dimensional subspace, you no longer have a subspace. $\endgroup$ – Morgan Rodgers Aug 24 '15 at 6:33
3
$\begingroup$

How about the following? It does not require the vector spaces to be finite dimensional.

If $M, N \neq V$, then pick $v_1 \in V \setminus M$ and $v_2 \in V \setminus N$. Since $v_1 \notin M$, $v_1 \in N$ by assumption. Similarly $v_2 \in M$. Let $v = v_1 + v_2$. Then $v \notin M$ for otherwise $v_1 = v + (-v_2) \in M$. Similarly $v \notin N$. Contradiction to the assumption.

Added: I would like to emphasize $V \setminus M$ is NOT a subspace. It may be easier if you look at a picture: removing the $x$-axis from the $x-y$ plane does not give you a subspace. In fact, $V \setminus M$ could never be a subspace since it does not contain the zero vector of $V$. In the context of vector spaces, disjoint union $M$ and $V \setminus M$ is not the way we are interested in decomposing a vector space $V$. Instead we are more interested in (direct) sums of vector spaces.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That was really slick :D thanks! $\endgroup$ – Soham Aug 24 '15 at 6:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.