Eigenvalues of product matrix

TravisJ already answered the question for the upper bound on the eigenvalues of the product of two symmetric positive definite matrices, $$\lambda_\text{max}(AB) \le \lambda_\text{max}(A)\lambda_\text{max}(B).$$ To get a lower bound, apply the upper bound inequality to the inverse: $$\lambda_\text{max}\left((AB)^{-1}\right) \le \lambda_\text{max}(A^{-1})\lambda_\text{max}(B^{-1}).$$ Since the eigenvalues of the inverse are the inverse of the eigenvalues, this is the same as, $$\frac{1}{\lambda_\text{min}(AB)} \le \frac{1}{\lambda_\text{min}(A)}\frac{1}{\lambda_\text{min}(B)}.$$ Multiplying through to clear the denominators yields the nice minimum eigenvalue bound, $$\lambda_\text{min}(A)\lambda_\text{min}(B) \le \lambda_\text{min}(AB).$$

The largest eigenvalue of such a matrix (symmetric) is equal to the matrix norm. Say your two matrices are $A$ and $B$.

$$\Vert AB\Vert \leq \Vert A\Vert \Vert B\Vert = \lambda_{1, A}\lambda_{1, B}$$

where $\lambda_{1,A}$ is the largest eigenvalue of $A$ and $\lambda_{1,B}$ is the largest eigenvalue of $B$. So the largest eigenvalue of the product is upper-bounded by the product of the largest eigenvalues of the two matrices. For a proof of what I just asserted, see: Norm of a symmetric matrix equals spectral radius

In terms of the smallest, it looks like the product of the smallest two eigenvalues also gives you a lower bound on the smallest eigenvalue of the product. For a complete reference on how the eigenvalues are related, see: https://mathoverflow.net/questions/106191/eigenvalues-of-product-of-two-symmetric-matrices