Let A be a symmetric matrix and B be a diagonal matrix with positive entries. Is there any way to prove that
$\lambda_k(AB) \leq \lambda_k(A)\lambda_k(B)$,
where $\lambda_k(A)$ denotes the $k^{th}$ eigenvalue of A.
I've checked this answer but they bound the eigenvalues according to the weak majorization order. Also, this answer shows a bound $\lambda_i(AB)\leq \lambda_i(A)\lambda_n(B)$, where $\lambda_i(A)$ denotes the ith eigenvalue of A, but it is only for positive definite matrices.
