Limit of largest eigenvalue [closed]

Question

For positive definite matrix, if we increase the dimension to the infinity, is it true that the largest eigenvalue stays bounded from above? In other words does the following limit exists:

$$\lim_{n\rightarrow \infty }\frac{\lambda_{max}}{n}=C<\infty,$$

where $\lambda_{max}$ is the largest eigenvalue of $n\times n$ symmetric positive definite matrix.

For every $n$ the eigenvalues are bounded from above, but is it true in the limit? It is not exactly infinite matrix per se, but finite matrices increasing in its size.

This question is related to my previous question, but is of independent interest in general and no information from my other question is necessary.

UPDATE

We have a sequence of matrices, but the mechanism of the sequence formation is not known. Only known is that it has to be kept positive definite and symmetric. Therefore we have a sequence of matrices such that each new $n^{th}$ matrix contains the previous matrix plus new added row and column. Hence we have a sequence of positive definite matrices, increasing in size with previous matrix being a sub-matrix. In addition, all values of the matrix has to be bounded from above (and below), no matter how large the matrix gets.

Probably this is way too general a problem to be able to infer anything usefull.

Answer 1

A partial answer can be given by the Gershgorin circle theorem. If all the values of the matrix are bounded by $K$, then the largest eigenvalue of an $n \times n$ matrix is bounded by $nK$. Therefore at least the limsup is finite: $\limsup \limits_{n \to \infty} \frac{\lambda_\text{max}}{n} \le K$.

I'm trying to construct a counterxample for the claim by a sequence of sparse matrices, but I had no success so far. My idea is that you extend your matrix in most cases only with zeroes, but sometimes (relatively rare) add columns and rows whose entries are large. With these it should be possible to get eigenvalues near the bound nK. When you add more zeroes, the eigenvalues stay the same, aside from the multiplicity of the eigenvalue 0. Therefore $\frac{\lambda_\text{max}}{n} $ decreases. Maybe someone else can complete this idea.