Try
In F₂³, each coordinate is 0 or 1 and addition is modulo 2. There are eight vectors. Try building an ordered basis: choose a first vector, then a second, then a third. The zero vector cannot be first. After choosing a nonzero vector v, its span contains exactly 0 and v, so six choices remain for the second vector. Two independent vectors span four vectors, leaving four choices for the third. There are therefore 7·6·4=168 ordered bases.
Write down the forbidden vectors after choosing e₁=(1,0,0) and e₂=(0,1,0). They are 0, e₁, e₂ and e₁+e₂. Predict whether changing to another independent pair changes the number forbidden. It does not: unique coordinates in any two-vector basis give exactly four combinations.
| Space | First independent vector | Second independent vector | Third independent vector | Ordered full bases |
|---|
| F₂³ | 8−1=7 | 8−2=6 | 8−4=4 | 168 |
| F₃² | 9−1=8 | 9−3=6 | Not applicable | 48 |
The entries are ordinary integer counts. F₃² has nine vectors, even though its scalar arithmetic has only three possible values. Counting 48 bases does not mean reducing 48 modulo 3.
Predict
Does 168 also count the three-dimensional subspaces of F₂³? No: there is only one, the whole space. An ordered basis describes a subspace, but many descriptions name the same object. Even removing order is insufficient: different sets of basis vectors may still span the same subspace.
Consider two-dimensional subspaces, which we will call planes through zero. Every plane has four vectors, including zero. There are 35 arbitrary four-element subsets containing zero, obtained by choosing three of the seven nonzero vectors. But ❴0,e₁,e₂,e₃❵ is not a plane: it omits e₁+e₂. Having the expected size is not a substitute for closure under vector addition and scalar multiplication.
Predict how many ordered pairs describe one genuine plane. It contains three nonzero vectors. Any of these can be first, and either remaining nonzero vector can be second, giving six bases. Thus the 7·6=42 independent ordered pairs in F₂³ describe only 42/6=7 planes.
Experiment
Build a paper enumeration rather than drawing a continuous plane. In F₂³ choose nonzero a and a different nonzero b. Record the set ❴0,a,b,a+b❵. Check its closure and list its six ordered bases. Repeat with the pair (a,a+b); the set stays the same. Mark that repetition in your record rather than counting another plane.
For example a=(1,0,0), b=(0,1,1) gives ❴000,100,011,111❵, where the three-digit strings are coordinate triples. The six ordered bases are (100,011), (011,100), (100,111), (111,100), (011,111), (111,011). Explain why two distinct nonzero vectors are automatically independent over F₂, and why this reasoning needs adjustment over F₃: there v and 2v are distinct but dependent.
Now conduct a changed calculation in F₃². There are eight first choices. Each first vector spans three vectors, so there are six second choices and 48 ordered bases. The number of one-dimensional subspaces is 8/2=4 because each line has two nonzero generators. These are finite sets, not Euclidean lines filled with infinitely many points.
Understand
Assume Fq is a field with q elements, with q a prime power. We use its vector-space properties; we are not constructing the field here. In particular, F₄ is not the ring of integers modulo 4: the latter has nonzero elements without multiplicative inverses. The arguments require a field.
A j-dimensional space has q^j vectors. Fix a basis; every vector has one and only one list of j scalar coordinates, each with q choices. Consequently, after j independent vectors in Fq^n, exactly q^j candidates are forbidden for the next choice. For 0≤k≤n, the number of independent ordered k-tuples is
(q^n−1)(q^n−q)⋯(q^n−q^(k−1)).
The multiplication principle applies because every valid prefix has the stated number of continuations. For k=0, there is one empty tuple and the product is 1. For k>n the count is zero. We do not extend the displayed counting argument into meaningless negative choice counts.
An n×n matrix is invertible exactly when its columns form a basis, and its columns are ordered. Therefore the same product with k=n is |GL_n(Fq)|, the size of the group of invertible matrices. For n=0 the zero-dimensional space has one linear map to itself: its identity. The empty matrix is invertible, consistently giving count 1.
To count k-dimensional subspaces, count pairs consisting of a subspace W and an ordered basis of W. Every independent k-tuple determines exactly one such pair through its span. Conversely, every k-dimensional W has exactly (q^k−1)(q^k−q)⋯(q^k−q^(k−1)) ordered bases, by the same coordinate argument within W. Dividing the tuple count by this constant number of descriptions gives the Gaussian binomial coefficient
G_q(n,k)=∏ from j=0 to k−1 of (q^n−q^j)/(q^k−q^j).
This is division of positive integer counts, not division inside Fq. The counting proof establishes that the resulting rational expression is an integer. For k=0 both products are empty and equal 1. For k=n numerator and denominator coincide: there is only the whole space. When n=0, the zero and whole space are the same single object. An unordered basis count instead divides the ordered basis count by k!, which removes ordering but still leaves many bases per subspace.
Apply
Change the ambient space to F₂⁴ and count its two-dimensional subspaces before checking the answer. There are 15·14 independent ordered pairs, and every plane has six ordered bases. Thus there are 210/6=35 planes. This coincidence with the earlier 35 arbitrary subsets in F₂³ is numerical only: the objects and counting arguments differ.
For a uniformly sampled n×n matrix whose entries are independent uniform Fq values, every one of the q^(n²) matrices has equal probability. Hence invertibility has probability ∏ from i=1 to n of (1−q^(−i)). In F₃² the value is 48/81=16/27. Predict whether the answer remains unchanged if both columns are forced to be equal.
Check the sampling change
Choose one uniform vector of F₃² and copy it into both columns. Every individual entry is still uniform, but the two columns are dependent and the matrix is always singular. The probability is zero. If instead each column is chosen independently and uniformly from the eight nonzero vectors, the probability is 6/8=3/4: after the first choice, six of the eight allowed second choices lie outside its span. A sampling rule is part of a probability calculation, not an optional description.
The practice includes a further derivation by splitting subspaces according to whether they lie in a fixed hyperplane. It requires proving the multiplicity for subspaces that leave the hyperplane, rather than guessing a familiar binomial recurrence. This is a new counting proof, not merely another evaluation of the product.
Continue
State what your objects are before using a formula: vectors, ordered independent tuples, unordered basis sets, subspaces, or invertible matrices. Record which choices are independent random draws and which are conditional exclusions in a counting argument. A denominator such as the number of bases per subspace belongs to integer counting even if its value becomes zero when reduced modulo the field characteristic.
These counts connect elimination to geometry over a finite field: rank determines how many independent columns can be selected, while equal-size fibers explain how multiple coordinate descriptions represent the same subspace. Finish with a changed case and a general proof, keeping the empty-dimensional boundaries visible rather than relying solely on familiar positive-dimensional examples.
