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.


Yes, it can be shown as follows:

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.

  • $\begingroup$ Thank you very much for your quick response. I must admit that I did not know about the minimax principle. You have given me a new avenue to learn from. $\endgroup$ – drkula Aug 12 '13 at 21:31
  • $\begingroup$ 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 $P$. $\endgroup$ – Eric Auld Aug 12 '13 at 21:34
  • $\begingroup$ Hello Eric, I have been trying to understand the minimax theorem, but I still can't quite understand your explanation. Would it be possible to elaborate a bit more? Thanks. $\endgroup$ – drkula Aug 14 '13 at 21:11
  • $\begingroup$ @drkula Sure, I would be happy to elaborate. Is it the minimax principle which you're asking about, or how I have applied it, or both? $\endgroup$ – Eric Auld Aug 19 '13 at 4:51
  • $\begingroup$ Hello Eric, I actually struggled through first understanding the principle itself for quite a while. As I had mentioned earlier, I am new to Linear Algebra. Once I had a basic understanding of the theorem and its proof, I was trying to see how you applied it to my problem. After a lot of headache, I think I saw a way yesterday! That insight helped me better understand the min-max principle also, I think. I would be happy to share my thoughts with you, if you have the time. Thanks. $\endgroup$ – drkula Aug 20 '13 at 20:06

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