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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.

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closed as unclear what you're asking by Alex Degtyarev, Fedor Petrov, Marco Golla, Stefan Waldmann, Suvrit Aug 12 '15 at 13:10

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ What do you mean by "increasing the dimension" of a matrix? Aren't you supposed to speak about a sequence of matrices? Isn't the answer supposed to depend on the sequence? Voting to close. $\endgroup$ – Alex Degtyarev Aug 12 '15 at 8:22
  • $\begingroup$ @AlexDegtyarev I added an update. Hopefully this makes it more clear. $\endgroup$ – Tomas Aug 12 '15 at 8:45
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    $\begingroup$ This setting is way too general, just take a diagonal matrix that has the value $i^2$ at the position $(i, i)$. $\endgroup$ – Dominik Aug 12 '15 at 8:47
  • $\begingroup$ @Dominik Thanks for the suggestion, but all values of the matrix has to be bounded from above for a matrics, no matter how large it gets. Added this information to my question. $\endgroup$ – Tomas Aug 12 '15 at 8:49
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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.

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