5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.