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?


It is "one of those hand-waving arguments", aimed at well-meaning, alert consumers. While it plausibly works for most reasonable mass matrices in semi-realistic models, of course you can concoct freak counterexamples that violate your expectation. Here is one.

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, $$ B= \begin{pmatrix} M&M-\epsilon\\ M-\epsilon &M \end{pmatrix} \qquad \Longrightarrow\qquad B^{-1}=\frac{1}{2\epsilon M-\epsilon^2} \begin{pmatrix} M&\epsilon-M\\ \epsilon -M&M \end{pmatrix}. $$

The (1,1) eigenvector of B has eigenvalue ~ 2 M so a minuscule 1/(2 M ) such for the inverse B-1. However, (1,-1) has a small one, ε, for B, and thus a huge one, 1/ ε , for the inverse B-1.

So C could, in principle, have huge eigenvalues. It is up to your model to avoid such pathologies.

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