I have real positive semidefinite matrices (symmetric) $A$ and $B$, both are $n \times n$.
I am looking for upper bounds and lower bounds on the $m$th largest eigenvalue of $AB$, in terms of the eigenvalues of $A$ and $B$.
I know, for example, that the $\det(AB) = \det(A)\det(B)$ and that the determinant of a square matrix equals the product of its eigenvalues, but that wouldn't give me what I am looking for, because I want to focus only on the top $m$ largest eigenvalues of $AB$.
Best Answer
Let the eigenvalues of $A$ be $a_1\ge a_2\ge \dots \ge a_n$, and the eigenvalues of $B$ be $b_1\ge b_2\ge \dots \ge b_n$. According to the discussion in comments, we are looking for bounds on the singular values of $AB$, denoted $s_1\ge s_2\ge \dots \ge s_n$. The min-max principle is the natural tool to use. Actually, the Wikipedia article does not state the form I'd like to use here: the $k$th largest singular value of $M$ is $\inf\{\|M-T\|:\mathrm{rank}\, T\le k-1\}$. This formula appears in another article with attribution to Allakhverdiev (and no reference). I think the result of Allakhverdiev was the extension to compact operators in Hilbert spaces, but don't know for sure. In general, Wikipedia articles on singular values are a toxic mess.
Claim 1: $s_k\le \min\{ a_i b_{j} : i+j=k+1\}$.
Proof: Let $T$ be an operator of rank $\le i-1$ such that $a_i=\|A-T\|$. Similarly, let $S$ be an operator of rank $\le j-1$ such that $b_{j}=\|B-S\|$. Since $\|(A-T)(B-S)\|\le a_i b_{j}$, it remains to estimate the rank of $(A-T)(B-S)-AB$. Writing $(A-T)(B-S)-AB=-T(B-S)-AS$, we see that the rank is at most $j-1+i-1=k-1$, as desired.
Claim 2: $s_k\ge \max\{ a_i b_{j} : i+j=k+n\}$.
If $A$ and $B$ are invertible, Claim 2 follows by applying Claim 1 to $(AB)^{-1}$. Indeed, the $k$th largest singular value of $AB$ is the reciprocal of the $(n+1-k)$th largest singular value of $(AB)^{-1}$. The $(n+1-k)$th largest singular value of $(AB)^{-1}$ does not exceed $\min \{a_{n+1-i}^{-1}b_{n+1-j}^{-1} : i+j=n+2-k\}$. Decoding these sums of indices, we get Claim 2. The general case follows by considering $A+\epsilon I$ and $B+\epsilon I$, and letting $\epsilon\to 0$.