Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
How many lines (passing through zero) are there in a plane? – Vishal Gupta Jun 15 '13 at 13:17
That is, lines through the origin. – GEdgar Jun 15 '13 at 13:18
up vote 2 down vote accepted

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 $a\in \mathbb R$.

share|cite|improve this answer
Clear and concise! Thanks! – Sarunas Jun 15 '13 at 13:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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