A bound on the eigenvalues of a product

Question

Let $A,B\in \mathbb{S}^n$ be such that $A\succ0,B\succ0$.

(a) Show that all eigenvalues of $BA$ are real and positive

(b) Let $B^{-1}=diag\{||{a_1}^T||_1,...,||{a_n}^T||_1\}$, where ${a_i}^T$ , $i=1,...,n$, are the rows of $A$.Prove that $$0<\lambda_i(BA)\le1 \ \ \forall i=1,...,n,$$ (c) Prove that $$\rho(I-\alpha BA)<1, \ \forall \alpha\in (0,2).$$

Answer 1

Good title, for $(a)$, we have that the spectre $\{\sigma\} $ of $XY$ is the same as $XY$ for all square matrices. So $\{\sigma(AB)\}=\{\sigma (B^{\frac{1}{2}}AB^{\frac{1}{2}})\}$ and that is $\in \mathbb{R}^+$. For $(b)$ resp $(c)$ it is sufficient to prove that $\lambda_{m}(BA)\le 1$ resp $|\lambda_m(I-\alpha BA)|\le 1$ ($\lambda_m$ is the maximal eigenvalue). It is well known using your notations that $|\lambda_m( A)|\le \text{max}_i \|a_i^T\|_1$. Applying that to $BA$ of row $R_i$, $ \|R_i\|_1\le 1$