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While trying to understand the chosen answer to the question presented here, I got stuck wondering why the maximum number of non-zero elements for a proper $m$-dimensional subspace $U_i$ of a vector space $V$ over a field $\mathbb{F}$ with $|\mathbb{F}| = q < \infty$ is is $q^{m - 1} - 1$?

Evidently the $-1$ comes from eliminating the one zero element. Intuitively I figured that the power of $q$ is $m - 1$ instead of $m$ because if we assign anything other than a zero to some $i$th component, then we would inevitably get all elements of the field $F$, since $U_i$ is closed under addition.

However my knowledge of fields is not sufficient to conclude this thought, since the very next thing that came to my mind was that why in particular do we get (necessarily) all of the elements of $F$, if the $i$th component is anything other than $0$? Wouldn't this imply that all elements of the field $\mathbb{F}$ are generators for each other? If this is not the case, then could the number of elements for the proper subspace $U_i$ be $q^m - 1 - k > q^{m - 1} - 1$ for some $k \in \mathbb{Z}$?

Epsilon Away
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    The linked answer talks about a proper subspace of an $m$-dimensional vector space, not about a proper $m$-dimensional subspace. – cqfd May 06 '21 at 14:33
  • @ShiveringSoldier But doesn't the answer start with: "If $|F| = q < \infty$, and $V$ is $m$-dimensional ($m < \infty$), then any proper subspace $U_i$ has at most $q^{m - 1} - 1$ non-zero elements."? – Epsilon Away May 06 '21 at 14:46
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    Yes! So $U_i$ is a proper subspace of the $m$-dimensional space $V$, and hence the dimension of $U_i$ is $\leq m-1$. – cqfd May 06 '21 at 14:49
  • @ShiveringSoldier I see! So perhaps an alternative title to my question is is that why is the dimension of a proper subspace at most $m-1$? – Epsilon Away May 06 '21 at 15:05
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    It is a standard fact. For example, see https://proofwiki.org/wiki/Dimension_of_Proper_Subspace_is_Less_Than_its_Superspace – cqfd May 06 '21 at 15:17

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