Minor (submatrix determinant) notation:

• For a matrix A, and index sets I = (i₁, i₂, ..., i_k) and J = (j₁, j₂, ..., j_k) Let A_I,J denote the k by k minor determinant of A in the rows indicated by I, columns indicated by J.
• Let Cⁿ_k denote the set of all length k increasing sequences from the size n index set {1, 2, ..., n}. The number of such sequences is the n,k binomial coefficient, i.e. coefficient of x^k in (x+1)ⁿ.
• Cauchy-Binet theorem: For I,J in Cⁿ_k , (AB)_I,J = Σ_{K in Cⁿk} A_I,KB_K,J.

Determinant facts:

• Definition det(A) = Σ_σ sign(σ) Π_i a_i,σ(i), where σ ranges over S_n, the set of all n! permutations of the n indices.
• Expansion on a row i: det(A) = Σ_j a_i,jA_I,J, where I is the set of all row indices excluding i and J is the set of all column indices excluding j.
• For characteristic polynomial C_A(x) = Σ_i=0 ⁿ c_ixⁱ, we have c_i = Σ_{I in Cⁿ n-i} (-1)^n-i A_I,I.

Determinant Algorithm:
Input: a matrix A over a finite field F of size q.
Output: the determinant of A, an element of F.

1. Choose random d₁, d₂, ..., d_n in F^*. Let D = diag(d₁, d₂, ..., d_n).
2. Compute m(x) = m_DA, the minimum polynomial of DA. // Wiedemann alg, probabilistic
3. If m₀= 0 or deg(m(x) = n, return m₀/(Πd_i). //success
   [m₀ = det(DA), Πd_i = det(D).]
4. return "fail" or go to 1.

Observe that if the return in step 3 is taken, the algorithm is correct. Either the matrix is singular or the minimal polynomial of DA, having degree n, is the characteristic polynomial of DA.

The question that remains is, "What is the probability of failure (or of having to repeat)?"

Theorem: Given nonsingular n by n matrix A, if δ₁, δ₂, ..., δ_n are distinct indeterminants, then for Δ = diag(δ₁, δ₂, ..., δ_n), the characteristic polynomial of ΔA is square free.

Corollary: If A is nonsingular, then c_ΔA(x) = m_ΔA(x) (charpoly = minpoly).

Call d₁, d₂, ..., d_n a good evaluation point if c_DA(x) = m_DA(x), for D = diag(d₁, d₂, ..., d_n).

Strategy:
(1) Construct nonzero polynomial whose roots are the bad evaluation points. Thus good evaluation points exist.
(2) Apply Schwartz lemma to show good evaluation points are ubiquitous.