# How to show that a continuous family of symmetric matrices is uniformly positive?

My problem : I have a family of $$4 \times 4$$ symmetric matrices. More precisely consider an interger $$d$$, a real $$\lambda> 0$$ and define the family $$S_{\lambda}$$:

$$\{A(\lambda,x_1,x_2) ; (x_1,x_2) \in (\mathbb{R}^d)^2 \} \subset S_4(\mathbb{R})$$ where $$S_4(\mathbb{R})$$ is the set of $$4 \times 4$$ symmetric real matrices.

In addition $$A(\lambda,x_1,x_2)$$ has the form (blocks)

$$A(\lambda, x_1,x_2)= \left[ {\begin{array}{cc} B(\lambda,x_1,x_1) & B(\lambda,x_1,x_2) \\ B(\lambda,x_1,x_2)^T & B(\lambda,x_2,x_2) \\ \end{array} } \right]$$

where for each $$(x,y) \in (\mathbb{R}^d)^2$$, $$B(\lambda,x,y)$$ is a $$2 \times 2$$ matrix. Furthemore, I can show that if $$\lambda$$ is big enough then for every $$x \in \mathbb{R}^d$$, $$B(\lambda,x,x)$$ is a positive semi-definite and symmetric matrix. Note that the coefficients of the matrices $$A(\lambda,x_1,x_2)$$ are complicated (sums and integrals of functions of the different parameters $$x_1$$ and $$x_2$$) which makes any computation really tedious.

My question : I want -if it's possible- to find a big enough $$\lambda$$ such that the family $$S_{\lambda}$$ is uniformly positive semi-definite (e.g. such that each of its elements is a positive semi-definite matrix). It doesn't have necessarily to be for all coefficients in $$\mathbb{R}^{2d}$$ : a subdomain would be enough.

What I've tried so far: I first thought about Sylvester's criterion (positivity of the leading principal minors) but the coefficients are too complicated to check the positivity of the determinants. Then I tried to use gershgorin's theorem (and also its block version) to estimate the lowest eigenvalue of the matrices but there's no diagonal dominance.

Do you any idea or any reference on a method used to deal with this kind of issue? Many thanks.

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