What is finite dimension?

Example 19.5

$\dim K^n = n$ (canonical basis); $\dim K_n[X] = n + 1$ (monomials); $\dim_\R \C = 2$; the solution space of $y'' + ay' + by = 0$ has dimension $2$ (Theorem 5.10: the solutions are parametrized bijectively and linearly by $(\lambda, \mu) \in K^2$).

Example 19.6 (The same set, two dimensions)

The set $\C^2$ of pairs of complex numbers is a $\C$-vector space of dimension $2$ (canonical basis $e_1, e_2$) — and an $\R$-vector space of dimension $4$, with basis

every $(z, w) = (a + \iu b,\ c + \iu d)$ has real coordinates $(a, b, c, d)$, uniquely. Dimension is not a property of the set of vectors alone: it counts the degrees of freedom relative to the allowed scalars, and halving the scalar supply from $\C$ to $\R$ doubles the count. (The weekend problem exploits the extreme case of this sensitivity, with scalars shrunk all the way to $\Q$.)

Example 19.9 (Two-out-of-three, saving half the work)

Is $\bigl((1,1,0), (0,1,1), (1,0,1)\bigr)$ a basis of $\R^3$? Count: three vectors, dimension three — so freeness alone decides. A null combination gives $a + c = 0$, $a + b = 0$, $b + c = 0$; adding all three, $2(a + b + c) = 0$, and subtracting each original equation from $a + b + c = 0$ leaves $b = c = a = 0$: free, hence a basis, with the generating half of the verification supplied by the theorem for free. Compare Example 18.18, where the same double verification had to be done by hand — one chapter of theory converts into exactly that saving, on every basis check for the rest of the book.