Let \(A\), \(B\) be positive semi-definite matrices satisfying \(A\le B\). Then \(\det A \le \det B\).

Proof

We assume \(0(0x6787b04b) .

Then, \(A\) is invertible and \(B^{-1/2} A B^{-1/2}\) is symmetric and

\begin{equation*} 0 where \(I\) denotes the identity matrix. This implies that all eigenvalues of \(B^{-1/2}AB^{-1/2}\) lie between 0 and 1 and therefore \(\det B^{-1/2} A B^{-1/2} \le 1\). Using determinant rules finishes the proof.