Let $A$ and $D$ be symmetric positive definite matrices and consider the symmetric block matrix
$$ M := \begin{pmatrix} A & \alpha B \\ \alpha B^\top & D \end{pmatrix} $$
where $\alpha \in \mathbb{R}$ is a scalar parameter. Is it possible to say something about the positive definiteness of $M$ as a function of $\alpha$?
Because of the facts that (i) the eigenvalues are continuous w.r.t. the matrix parameters, and (ii) for $\alpha = 0$, $M$ is positive definite and its eigenvalues are those of $A$ and $D$ (since block diagonal), it seems that if $\alpha$ is "small enough", the matrix $M$ will be positive definite. Hence, it seems possible to relate the positive definiteness of $M$ to conditions on the smallest eigenvalue $\lambda_\min$ of $A$ and $D$ and some metric of $\alpha B$ (maybe some norm). Does someone have an idea?
I was thinking on the Schur complement and looking at the matrix
$$A – \alpha^2 B D^{-1} B^\top$$
but I have difficulties showing the positive definiteness of that as a function of $\alpha$. However, again it is clear that if $\alpha=0$, everything works out. Does someone have some idea?
Best Answer
Let me write $M_{\alpha}$ for $M$ to stress the $\alpha$-dependence. Denote the smallest eigenvalue of $M_{\alpha}$ by $\lambda_{\alpha}$ and note that $\lambda_{\alpha} = \inf\limits_{\|x\| = 1} \|M_{\alpha}x\|$. From this it is easy to see that $$|\lambda_{\alpha} - \lambda_0| \leq \|M_{\alpha}-M_0\| = \|\alpha B\|.$$ Hence, if $\|\alpha B\| < \lambda_0$, then $\lambda_{\alpha} > 0$.