# On the difference of two positive semi-definite matrices

I am relatively new to linear algebra, and have been struggling with a problem for a few days now. Say we have two positive semi-definite matrices $A$ and $B$, and further assume that $A$ and $B$ are such that $A - B$ is also positive semi-definite. Can it be shown that $det(A) \ge det(B)$? In my own attempts, I can see that $Tr(A) \ge Tr(B)$, but I do not think this is enough to prove the desired result. Perhaps there is something to be said about the relative magnitudes of the eigenvalues of $A$ and $B$, but I can't see it. In any case, I would appreciate any help. Thanks a lot.

Consider that the determinant is the product of the (non-negative) eigenvalues. The $k$th largest eigenvalue $\lambda_k$ of an Hermitian operator $P$ can be expressed as $$\lambda_k(P) = \min_{\dim S = k} \max_{x\in S,x\neq 0}\frac {\langle Px, x \rangle} {\langle x,x\rangle}$$ by the Courant minimax principle. Thus $\lambda_k(A) \geq \lambda_k (B)$ follows from the definition of $A-B$ positive semi-definite, since $$\langle (A-B)x, x\rangle \geq 0 \Rightarrow \langle Ax, x\rangle \geq \langle Bx, x\rangle,$$
and the inequality continues to hold once we divide by $\langle x,x\rangle$, and take max and min. So the inequality holds for the product of the eigenvalues.
• No problem. A good rule is: whenever you're working with eigenvalues and positivity (or matrix inequalities), keep in mind the "Rayleigh quotient" $$\frac {\langle Px,x \rangle} {\langle x,x \rangle},$$ because this is what relates $\langle Px,x \rangle$ to the spectrum of $P$. – Eric Auld Aug 12 '13 at 21:34