# Random matrix is positive

This is a follow up question on my previous question here that was on solved in the deterministic setting by Denis Serre, when the perturbation can be separated. Therefore, I decided to split the deterministic case from the random question initially posed in the very same question. So let me state the random case:

We consider a symmetric matrix $$A \in \mathbb R^{n \times n}$$ defined by

$$A_{ij}= i \delta_{ij} - X_{ij}\varepsilon \,.$$

Here $$\delta_{ij}$$ is the Kronecker delta and $$X_{ij}$$ are iid Bernoulli (but of course $$X_{ij}=X_{ji}$$ to make it hermitian).

We first note that this matrix is not diagonally dominant if the matrix dimension $$n$$ is large enough, independent of what $$\varepsilon$$ is.

This is because $$\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \vert A_{i,1}\vert=\infty >\vert A_{1,1} \vert.$$

Numerical experiments with matrix size $$n=50$$ or $$n=200$$ show that the lowest eigenvalue stays far above $$0$$, if $$\varepsilon>0$$ is sufficiently small.

Question: How can one show that $$A$$ is positive definite independent of the dimension if $$\varepsilon$$ is sufficiently small but fixed ?

To elaborate on the numerical experiments. We find that the lowest eigenvalue of $$A$$ in some interval $$[0,96,1.08]$$ if we choose $$\varepsilon=0.1$$ and $$n=50$$(upper plot) and $$n=200$$(lower plot), as I illustrate in the following plot where I sampled the lowest eigenvalue for 100 realizations:

Please let me know if you have any questions!

Let $$u\in \mathbb{R}^n$$ as $$u_i=i^{-1}$$. Then $$\mathbb{E}\big[\langle u Au\rangle \big]=\sum_{i=1}^n i^{-1}-\frac{\epsilon}{2}\sum_{i,j\leq n}\frac{1}{ij}\approx \log n -\frac{\epsilon}{2}(\log n)²$$ Which is stricly negative for $$\log n > 2\epsilon^{-1}$$. (Remark : In order to see it numerically one should choose $$n$$ exponentially large with $$\epsilon^{-1}$$.)
You can find other counter-examples along the vein of $$v\in \mathbb{R}^n$$ $$v_i=i^a$$ for certain $$a$$, making sure that the sum of the negative terms outweighs that of the positive components of the diagonal elements. Also, a matrix may have all positive eigenvalues yet not be positive definite; see <https://math.stackexchange.com/questions/4336/if-eigenvalues-are-positive-is-the-matrix-positive-definite>.