Show that det(A) ≤ a_{11}a_{22}· · ·a_{nn}, where A is positive semidefinite and a_{11}, a_{22}, · · ·, a_{nn} are the diagonal entries.

Answer 1

Fix a positive number t. The matrix A + tI is positive definite with diagonal entries a₁₁ + t,...,a_nn + t. We first show that det(A + tI) ≤ (a₁₁ + t)•••(a_nn + t). Then, by continuity of the determinant, we obtain the desired result by taking the limit as t -> 0⁺.

Let D be the diagonal matrix whose i^th diagonal is (a_ii + t)^-1/2. Then P := D(A+tI)D is a positive definite matrix with diagonal entries all equal to 1. The determinant of P is the product of its (positive) eigenvalues, say λ₁,...,λ_n. The arithmetic-geometric mean inequality implies λ₁•••λ_n ≤ [(λ₁ + ••• + λ_n)/n]ⁿ, or set det(P) ≤ [trace(P)/n]ⁿ. Since P has all ones along its diagonal, trace P = n, and thus det(P) ≤ 1. On the other hand,

det(P) = det(D) det(A + tI) det(D) = det(A + tI)/[(a₁₁+t)•••(a_nn+t)]

Hence, det(A + tI) ≤ (a₁₁ + t) ••• (a_nn + t). In the limit as t -> 0⁺ the result follows.