Recipe

Compute the dimension of an exterior power

Use when: you're given \dim V=n and an exponent k, and need \dim(\Lambda^kV)

Goal: the exact dimension of \Lambda^kV, correctly zero when k exceeds n and a binomial coefficient otherwise

Taught in Exterior powers Λᵏ V and the dimension count

1. 1

   Read off the ambient dimension n=\dim V and the exponent k you were asked about.

   Why: both numbers must be in hand before Step 2's fork can be decided.

2. 2

   Compare k against n directly.

   Example

   For n=3, k=4: no strictly increasing sequence of 4 distinct indices exists among only 3 basis vectors, so \dim\Lambda^4\mathbb R^3=0.

   • k>n — conclude \dim(\Lambda^kV)=0 immediately, and skip the binomial computation in Step 3 entirely.
   • k\le n — continue to Step 3 and compute the binomial coefficient.

   Why: no strictly increasing sequence of k distinct indices exists once k exceeds the number of available basis vectors, so the space collapses to \{0\} before any counting is needed.

3. 3

   Apply \binom{n}{k}=\dfrac{n!}{k!(n-k)!} directly to the values from Step 1.

   Example

   For n=4, k=2: \binom{4}{2}=\dfrac{4!}{2!\,2!}=6.

   ⚡ Shortcut

   Use the symmetry \binom{n}{k}=\binom{n}{n-k} to compute whichever of k or n-k is smaller. For \binom{10}{8}, compute \binom{10}{2}=45 instead of expanding 8! by hand.

   Why: this count is literally choosing which k of n basis vectors survive, since order and sign are already fixed by the alternating property.

4. 4

   Confirm k=0 gives dimension 1 (the scalars) and k=n gives dimension 1 (the top exterior power), both from \binom{n}{0}=\binom{n}{n}=1.

   Why: these two boundary values recur constantly in later sections, such as the determinant's own uniqueness proof, so catching a miscount here avoids compounding errors downstream.