3.4.9

Let $$ V = \mathbb{F}^n $$, let $$V_B $$ be the set of all bases of V, and let E be the standard basis $$(e_1,e_2,...,e_n) $$ for V. Given any basis B of V, where $$\textbf{B} = (b_1,b_2,...,b_n) $$, there is a invertible change of basis matrix P such that BP = E, by equation 3.4.19. Thus, we can define a mapping $$f: V_B \rightarrow GL_n(\mathbb{F}) $$ such that f(B) = P. This mapping is well-defined since given any basis B, the change of basis matrix P is unique (again from 3.4.19). We will show that f is a bijection.

Let P be any invertible matrix in $$GL_n(\mathbb{F}) $$, and construct a set of vectors $$ (b_1,b_2,...,b_n) $$ by letting $$b_j = \sum_{i=1}^n e_i(P^{-1})_{ij} $$, so that letting B be the ordered set of these n vectors, we have $$\textbf{B} = \textbf{E}P^{-1} $$. By 3.4.22, B forms a basis, and moreover satisfies BP = E, so that f(B) = P, showing that f is onto.

Now let A and B be bases of V such that f(A) = f(B) = P, where P is invertible. By the construction above, we then have $$\textbf{B} = \textbf{E}P^{-1} = \textbf{A} $$, showing that f is one-to-one. Thus f is a bijection.

= Alternative proof = Let $$B = (b_1, \dots, b_n)$$ be any basis for $$V$$, let $$f((b_1, \dots, b_n)) = [ b_1 \cdots b_n ] = A.$$ We will use exercise 3.4.6.: the column of $$A$$ are independent iff $$A$$ is invertible.

Clearly, $$f$$ is injective. Furthermore, $$f$$ is surjective by exercise 3.3.6.