首页笔记，数学

# Definition

To derive the determinant rigorously from the concept of nn -dimensional volume, we define a function

f:(Rn)n→Rf: (\mathbb{R}^n)^n \to \mathbb{R}

that maps nn vectors (v1,v2,…,vn)(\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n) to a scalar representing the signed volume of the parallelotope they span. For this function to behave like a measure of volume in Euclidean space, it must satisfy three fundamental axioms:

  1. Multilinearity: For any ii, ff is linear with respect to the ii-th vector:

    f(…,cvi+u,…)=cf(…,vi,…)+f(…,u,…)f(\dots, c\mathbf{v}_i + \mathbf{u}, \dots) = cf(\dots, \mathbf{v}_i, \dots) + f(\dots, \mathbf{u}, \dots)

  2. Alternating Property: If any two vectors are equal, the volume is zero (the shape collapses):

    f(v1,…,vi,…,vi,…,vn)=0f(\mathbf{v}_1, \dots, \mathbf{v}_i, \dots, \mathbf{v}_i, \dots, \mathbf{v}_n) = 0

  3. Normalization: The volume of the unit hypercube spanned by the standard basis {e1,…,en}\{\mathbf{e}_1, \dots, \mathbf{e}_n\} is unity:

    f(e1,e2,…,en)=1f(\mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n) = 1

From the alternating property, we can immediately derive that swapping any two vectors negates the result. Consider f(…,vi+vj,…,vi+vj,…)=0f(\dots, \mathbf{v}_i + \mathbf{v}_j, \dots, \mathbf{v}_i + \mathbf{v}_j, \dots) = 0. By expanding this using multilinearity and canceling the terms f(…,vi,…,vi,…)f(\dots, \mathbf{v}_i, \dots, \mathbf{v}_i, \dots) and f(…,vj,…,vj,…)f(\dots, \mathbf{v}_j, \dots, \mathbf{v}_j, \dots) which are zero, we arrive at:

f(…,vi,…,vj,…)=−f(…,vj,…,vi,…)f(\dots, \mathbf{v}_i, \dots, \mathbf{v}_j, \dots) = -f(\dots, \mathbf{v}_j, \dots, \mathbf{v}_i, \dots)

Now, let AA be an n×nn \times n matrix where the jj-th column is vj\mathbf{v}_j. We express each column vector in the standard basis: vj=∑i=1naijei\mathbf{v}_j = \sum_{i=1}^n a_{ij} \mathbf{e}_i. Substituting these into the volume function ff, we use multilinearity to expand the expression completely:

f(v1,…,vn)=f(∑i1=1nai11ei1,…,∑in=1nainnein)=∑i1,…,in=1nai11…ainnf(ei1,…,ein)f(\mathbf{v}_1, \dots, \mathbf{v}_n) = f\left(\sum_{i_1=1}^n a_{i_1 1} \mathbf{e}_{i_1}, \dots, \sum_{i_n=1}^n a_{i_n n} \mathbf{e}_{i_n}\right) = \sum_{i_1, \dots, i_n = 1}^n a_{i_1 1} \dots a_{i_n n} f(\mathbf{e}_{i_1}, \dots, \mathbf{e}_{i_n})

Due to the alternating property, any term in this summation where the indices (i1,…,in)(i_1, \dots, i_n) are not distinct vanishes. Therefore, we only sum over sequences that are permutations σ\sigma of the set {1,…,n}\{1, \dots, n\}. For any such permutation, the alternating property implies:

f(eσ(1),…,eσ(n))=sgn(σ)f(e1,…,en)f(\mathbf{e}_{\sigma(1)}, \dots, \mathbf{e}_{\sigma(n)}) = \text{sgn}(\sigma) f(\mathbf{e}_1, \dots, \mathbf{e}_n)

Applying the normalization axiom f(e1,…,en)=1f(\mathbf{e}_1, \dots, \mathbf{e}_n) = 1, the expression simplifies to the Leibniz formula, which serves as the formal definition of the determinant:

det⁡(A)=∑σ∈Snsgn(σ)∏j=1naσ(j),j\det(A) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{j=1}^n a_{\sigma(j), j}

This proof demonstrates that the determinant is the unique function that satisfies the geometric requirements of a signed volume scaling factor.