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 finite dimensional vector space over a finite field. Suppose that $m \geq 2$ and $V_1$, ..., $V_m$ are non-zero subspaces of $V$ such that every non-zero vector belongs to one and only one of $V_i$'s. How to show that at least two of these subspaces must have equal dimension? Any suggestion would be helpful.

share|cite|improve this question
It will probably help you to know that $|V|$ and each $|V_i|$ are (nonzero) powers of some prime number $p$, and that $$|V| = |V_1| + |V_2| + \cdots + |V_m| - m + 1$$ I had a quick play around with this information but didn't get very far, but it might help you. – Clive Newstead Dec 29 '13 at 13:58
up vote 2 down vote accepted

Any vector space $W$ over $\Bbb F_q$ with $\dim W = n < \infty$ is isomorphic to $\Bbb F_q^n$, thus $|W| =q^{n}$.

So suppose $V_1,\dots ,V_m$ was a subspace partition of $V$ such that $m\geq 2$ and the $V_i$ have pairwise distinct dimension. Let $n=\dim V$. Then $\dim V_i<n$ and the $\dim V_i$ are distinct prime powers. Hence by counting vectors and excluding zeros (as Clive Newstead did in the comments) we get:$$p^n = |V| = 1+ \sum_{i=1}^m (|V_i|-1)\leq1-m+\sum_{i=1}^{n-1}p^i\leq p^n-1.$$ A contradiction.

share|cite|improve this answer

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.