Let $V$ be a vector space over $\mathbb{R}$ of dimension $n$, and let $U$ be a subspace of
dimension $m$, where $m < n$. Show that if $m = n − 1$ then there are only two subspaces of $V$ that contain $U$ (namely $U$ and $V$), whereas if $m < n − 1$ then there are infinitely many distinct subspaces of $V$ that contain $U$.
I have two questions arising from this problem.
1) Now say $m=n-1$. Then since $U$ is a subspace of $V$, it is contained in $V$. It is also contained in itself, namely $U$. Other options are out of the game in this case, because say if we form a space of dimension $n-1$, it would have to be precisely $V$. This follows because if say $u_1,…,u_{n-1}$ is a basis for $U$, then a basis for another subspace of $V$ with $n-1$ elements, would have to span the space as does $u_1,…,u_{n-1}$ forming the same vector spac – $U$. Am I right here?
2) Now let $m<n-1$. Applying the same reasoning as above, we can see, that there are much more options now. What I don't understand, is that how do I prove, that we can form infinitely many subspaces of $V$, even though it is finitely dimensional. For example, let $\lambda _j$ be scalars from the field $\mathbb{R}$ and let $u_1,…,u_m$ with $m<n-1$ a basis for vector space $V$. While I think it follows without a proof (or do I need one?) than we can form infinitely many linearly independent combinations $\lambda_ju_j$ (for different set of bases), why would there be infinitely many spaces spanned by those combinations, forming infinitely many subspaces of $V$?
Hints would be appreciated! Thanks!
Best Answer
The subspaces $W$ with $U\subseteq W\subseteq V$ correspond 1-1 to the subspaces of $V/U$. If $V/U$ is onedimensional, the only subspaces are $0$ and $V/U$ itself. If $V/U$ is at least twodimensional there are infinitely many subspaces. For example, already $\mathbb R^2$ has - at least - the infinitely many (because $\mathbb R$ is infinite!) subspaces $\{(t,at)\in\mathbb R^2\mid t\in\mathbb R\}$, one for each value of $a\in \mathbb R$.