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Let $V$ be a $n$-dimensional vector space and suppose we have a set $A$ of vectors which span $V$. Let $q$ be the smallest number such that for all $x \in V$, we can always write $x$ as a linear combination of $q$ vectors in $A$. What is the relationship between $|A|$ and $q$?

For example, here are the two examples:

  • By basic linear algebra, we can have $|A| = n$ when $q = n$.
  • Let $\mathbb{F} = \mathbb{Z}/p\mathbb{Z}$ be a finite field and $V = \mathbb{F}^n$. For any $q$, I can construct $A$ with size $|A| = p^{n-q} + q - 1$. Here it is: $$ A = \{(1, 0, \dots, 0, x_{q+1}, \dots, x_n) \mid x_{q+1}, \dots, x_n \in \mathbb{F}\} \cup \{e_2, \dots, e_q\},$$ where $e_2, \dots, e_q$ are standard basis vectors. This works because for any target $y = (y_1, \dots, y_n)$, we have $$ y = y_1(1, 0, \dots, 0, y_{q+1}y_1^{-1}, \dots, y_ny_1^{-1}) + y_2e_2 + \dots + y_qe_q.$$ I have no clue if a smaller set $A$ exists. (edit: may not work when $y_1 = 0$, thank you @jackson for pointing it out)

An easier question might be: Can the set $A$ be finite if we desire $q < n$, when $|\mathbb{F}| = \infty$?

A possibly harder question that I am also interested in is: What if we have $\mathbb{Z}$ instead of a field?

Glenn Sun
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    $A$ cannot be finite if the field is infinite; it's a classic exercise that in this case a vector space cannot be the union of finitely many proper subspaces (https://math.stackexchange.com/questions/60698/if-a-field-f-is-such-that-leftf-rightn-1-why-is-v-a-vector-space-over). Over $\mathbb{R}$ or $\mathbb{C}$ a vector space in fact cannot be the union of countably many proper subspaces, and one way to see this is that they have zero measure. – Qiaochu Yuan Sep 15 '23 at 19:29
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    I think your minimum $q$ is off by one: e.g. if $y_1 = 0$ and $y_{q+1} \ne 0$, then you need $y=y_{q+1}(1,0,\dots,0,1,y_{q+2}/y_{q+1},\dots,y_n/y_{q+1})-y_{q+1}e_1+y_2e_2+\cdots+y_qe_q$ – jackson Sep 15 '23 at 19:31

1 Answers1

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I think I can improve on this in the finite field case, but I don't have a general argument that this set is minimal. First consider $q = 1$: is it possible to construct $A \subset \mathbb F_p^n$ so that every element of $\mathbb F_p^n$ is a multiple of some $a \in A$? Yes, if (and only if) $A$ contains any representative set of $P(\mathbb F_p^n)$, so we have $|A| = p^{n-1} + \cdots + p + 1$.

More generally, for any $q$, we can take an approach inspired by yours and improve on it slightly. Use $q-1$ basis vectors as you do, and try to include all vectors in the complementary subspace to their span: $$ A' := \{e_1, \ldots, e_{q-1}\} \cup \{(0, \ldots, 0, x_q ,\ldots, x_n) \mid x_i \in \mathbb F_p\}.$$ But by including every vector in the subspace $V := 0^{q-1} \oplus \mathbb F_p^{n-q+1} \subset \mathbb F_p^n$, we introduce some redundancy. Instead, we can choose a representative set of $P(V)$, so $$|A| = |P(V)| + q - 1 = q - 1 + \sum_{i=0}^{n-q} p^i$$

I can take these ideas further, but still I don't know if I can prove this is minimal. If $\mathbb F_p^n$ is decomposed into two complementary subspaces $\mathbb F_p^n = V \oplus W$, then an arbitrary vector has a unique representation $v+w$. So for $q=2$ we could take $A' = V \cup W$. Or, instead, take a representative set of $P(V)$ and one of $P(W)$. To minimize $|A|$ under this scheme, let $\dim V = \lceil \frac n2 \rceil$ and $\dim W = \lfloor \frac n2\rfloor$, so $$|A| = \begin{cases}2\sum_{i=0}^{n/2} p^i \\ p^{\frac{n+1}2} + 2\sum_{i=0}^{(n-1)/2} p^i \end{cases},$$ depending on whether $n$ is odd or even.

For general $q$, we have $n = bq+r$ with $0\le r<q$. Then $$\mathbb F_p^n \cong \bigoplus_{j=1}^r \mathbb F_p^{b+1} \oplus \bigoplus_{j=r+1}^q \mathbb F_p^b,$$ and by letting $A$ be a union of representative sets of $P(V_j)$ for each of the $q$ subspaces $V_j$, we have $$|A| = rp^b + q\sum_{i=0}^{b-1} p^i$$

jackson
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