Show that the smallest eigenvalue is strictly positive. I have a $2\times 2$ real, symmetric and positive definite matrix $B_{x,n}$ which depends on a point $x\in[0,1]$ and $n\in\mathbb{R}$. I want to show that for sufficiently large $n$, the smallest eigenvalue of $B_{x,n}$ is strictly positive uniformly on $x$.
To do so, I managed to show that $B_{x,n}\to B$ as $n\to\infty$ with $B$ positive definite (and so, has only strict positive eigenvalues).
It is known that the smallest eigenvalue equals the solution $\inf_{\lVert z \rVert=1}z'B_{x,n}z$. Then for each nonzero $z\in\mathbb{R}^2$, $z'B_{x,n}z\to z'Bz>0$. However, I believe that the desired result can be shown only if $z'B_{x,n}z\to z'Bz>0$ uniformly on $\{z:\lVert z\rVert=1\}$. Is it right?
Can you give me suggestions?
This is Lemma 1.5 of Tsybakov's Introduction to nonparametric estimation, by the way.

Since $B$ is positive definite then $\det(B)>0$ and $\operatorname{tr}(B)>0$.
Since the determinant and the trace are continuous functions and $\displaystyle\lim_{n\rightarrow \infty}B_{x,n}=B$ then $\displaystyle\lim_{n\rightarrow \infty}\det(B_{x,n})=\det(B)$ and $\displaystyle\lim_{n\rightarrow \infty}\operatorname{tr}(B_{x,n})=\operatorname{tr}(B)$.
So there is $N$ such that if $n>N$ then $\det(B_{x,n})>\frac{1}{2}\det(B)>0$ and $\operatorname{tr}(B_{x,n})>\frac{1}{2}\operatorname{tr}(B)>0$.
Therefore the product and the sum of the two eigenvaues of $B_{x,n}$ are positive for $n>N$. So the eigenvalues are positive for $n>N$. 
