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 ?

  • $\begingroup$ "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

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.

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

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