# Bound of the eigenvalues of a matrix product of two diagonal and one symmetric PSD matrices

Let B be a positive diagonal matrix, and C a PSD symmetric matrix with eigenvalues $\Lambda$, with $\lambda_i = {0,1}$. I'm trying to find a reference for the following bounds or similar

$B^{1\over 2}DB^{1\over 2} \preceq \Lambda B$

$DB \preceq \Lambda B$.

I've checked this answer but they bound the eigenvalues according to the weak majorization order. Also, this answer shows a bound $\lambda_i(AB)\leq \lambda_i(A)\lambda_i(B)$, where $\lambda_i(A)$ denotes the ith eigenvalue of A, but it is only for positive definite matrices.

• The eigenvalues of $B^{1/2}DB^{1/2}$ are the same as the eigenvalues of $DB$, and you can treat positive semidefinite matrices using a limit argument. So you should be able to use the result on $\lambda_i(AB)$. – Federico Poloni Jun 20 '17 at 20:37
• Also, it's not clear to me what you mean with $\preceq \Lambda B$, since $\Lambda B$ may not be symmetric. – Federico Poloni Jun 20 '17 at 20:39
• I mean that the eigenvalues of $DB$ are smaller than the eigenvalues of $D$, ie $\Lambda$, multiplied by $B$. – Pedro G. Jun 20 '17 at 21:45
• OK, thanks for the clarification. I normally associate that notation to the Loewner ordering, which is a stronger property. – Federico Poloni Jun 20 '17 at 21:49
• I do not understand what you mean with by using a limit argument. Can you give me a reference? – Pedro G. Jun 21 '17 at 5:46