# Type-I seesaw: order of magnitude of the eigenvalues of effective $M_\nu$

Consider a square matrix $C$ constructed out of two other square matrices $A$ and $B$ as $$C=-A^TB^{-1}A.$$ Suppose all the elements of $B$ are very large compared to those of $A$. In such a case, is it guaranteed that all the matrix elements (or all the eigenvalues) of $C$ will also be small?

The preamble above looks like a math question but there is some physics context to it. In the type-I seesaw, the expression for the effective light neutrino mass matrix is given by $$M_\nu=-m_D^TM_R^{-1}m_D$$ where $m_D$ is the Dirac mass and $M_R$ is Majorana mass for the right-handed electroweak singlet $N_R$.

It is often stated that since "the matrix elements of $m_D$ $\ll$ the matrix elements of $M_R$", the eigenvalues of $M_\nu$ (representing light neutrino masses) will also be small? Or is it one of those hand-waving argument by physicists?

Take A to be an orthogonal rotation, so it does not alter the eigenvalues of 1/B. Choose then, for enormous M and minuscule ε, $$B= \begin{pmatrix} M&M-\epsilon\\ M-\epsilon &M \end{pmatrix} \qquad \Longrightarrow\qquad B^{-1}=\frac{1}{2\epsilon M-\epsilon^2} \begin{pmatrix} M&\epsilon-M\\ \epsilon -M&M \end{pmatrix}.$$