I have two matrices, both positive definite, real symmetric and one is diagonal. What can I say about lower and upper bound of the eigenvalues of the product matrix in terms the of lower and upper bounds on eigenvalues of those two matrices.
$\endgroup$3 Answers
$\begingroup$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 inverses exist because the matrices are assumed to be positive definite in the question. The result extends to the case where the matrices are positive semi-definite as well. In this case, either $A$ or $B$, are not invertible, and the above proof would divide by zero. However, in this case either $A$, or $B$ has zero as the minimum eigenvalue, and $AB$ has zero as the minimum eigenvalue, so the inequality becomes $0 \le 0$.
$\endgroup$ 1 $\begingroup$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:
$\endgroup$ 4 $\begingroup$I think there is something wrong with the identity. For example consider L as lower triangular matrix of 1, i.e. Lower triangular entries are all 1 including diagonal and rest are zero. The clearly the eigenvalues of L are all 1. Similarly eigenvalues of transpose of L are all 1. However eigenvalues of product of $L^TL$ are dispersed widely, as the sum of diagonal of $L^TL$ is n(n+1)/2, where n is the dimension of matrix L.
$\endgroup$ 1
