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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.

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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 $0 \le 0$.

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  • $\begingroup$ @tranisstor Thanks for bringing up the issue of invertibility in your edit. The pseudoinverse is not required though, as the result holds directly as 0=0. I've done another edit to address this. $\endgroup$
    – Nick Alger
    Jan 28, 2022 at 16:31
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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

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  • $\begingroup$ I am sure this does not give any lower bound, on the contrary of what you said $\prec$ is not a component-wise comparison. $\endgroup$
    – user5644
    Mar 17, 2015 at 20:10
  • $\begingroup$ @timhortons, after more clearly reading "weak majorizing" it appears that you are correct... at least for the lower bound. The majorizing means each sum of the first $k$ elements from the vector satisfy the inequality. The upper-bound on the solution is certainly correct. I will remove my erroneous comment. $\endgroup$
    – TravisJ
    Mar 17, 2015 at 20:27
  • $\begingroup$ If we have lower and upper bounds for the eigenvalues, then we have lower and upper bounds for the determinant of the product matrix. This combined with the upper bound on the largest eigenvalue of the product provides a lower bound on the smallest eigenvalue of the product matrix. DOne! $\endgroup$
    – user5644
    Mar 17, 2015 at 23:04
  • $\begingroup$ Shouldn't it be that $\lambda_{1,A}$ and $\lambda_{1,B}$ are the absolutely largest eigen values of $A$ and $B$ respectively? $\endgroup$ Feb 7, 2017 at 15:03
  • $\begingroup$ @pikachuchameleon (late comment, left mostly for those seeing this now). A positive definite matrix (as in OP) only has positive eigenvalues. So the largest and absolutely largest eigenvalues are the same thing(s). $\endgroup$
    – TravisJ
    Oct 7, 2022 at 8:34
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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.

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    $\begingroup$ The statement is about symmetric matrices. $\endgroup$
    – user5644
    Nov 1, 2016 at 19:53

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