Evaluate the mininal eigenvalue of positive definite matrix

Question

Let $\{A_n\}$ be a sequence of positive definite matrices. Denote the minimal eigenvalue of matrix A by $\lambda_{min}(A)$. If $tr(A_n) \to +\infty$ when $n \to +\infty$, then what can we say about $\lambda_{min}(A_n)$ ?

Answer 1

Suppose $$ A_n= \left( \begin{array}{cc} n & 0 \\ 0 & a_n \end{array} \right) $$ and that $n>a_n>0$. Then $A_n$ is positive definite and $tr(A_n)\to \infty$ as $n \to \infty$.

However we can say nothing interesting about the behavior of $\lambda_{min}(A_n)=a_n$ as $n \to \infty$. It could tend to $0$ or $\infty$, anything in between, or $\lim_{n \to \infty}\lambda_{min}(A_n)$ might not exist.

Edit:

If we also know that $A_n=b_1b_1^T+\cdots+b_nb_n^T$ (where $b_k$ are vectors in $\mathbb R^K$), then we can say a little more. As in user251257's comment, we can use the variational characterization of the eigenvalues of a hermitian matrix to say that $\lambda_{min}(A_n)$ is a (non-strictly) increasing function.

As can be seen at this link, \begin{align*} \lambda_{min} (A_n) &= \min_{x \neq 0} \frac{(A_nx,x)}{(x,x)}=\min_{x \neq 0} \frac{((A_{n-1}+b_nb_n^T)x,x)}{(x,x)}\\ &= \min_{x \neq 0} \frac{(A_{n-1}x,x)}{(x,x)}+\frac{((b_nb_n^T)x,x)}{(x,x)}\\ & \geq \min_{x \neq 0} \frac{(A_{n-1}x,x)}{(x,x)}\\&=\lambda_{min}(A_{n-1}). \end{align*}