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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.

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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

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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.

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