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Playing around with Matlab I noticed something very peculiar:

Take the symmetric matrix $A \in \mathbb R^{n \times n}$ defined by

$$A_{ij}= i \delta_{ij} - \frac{\varepsilon}{\sqrt{i}\sqrt{j}}\,.$$

Here $\delta_{ij}$ is the Kronecker delta.

We first note that this matrix is not diagonally dominant if $n$ is large enough.

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

It is obvious that we require $\varepsilon<1$ in order for $A$ to be positive definite, since otherwise $A_{1,1}\le 0.$

However, I noticed that for let's say $\varepsilon=0.1$ one can make the dimension as large as one wants and the matrix remains positive definite.

To my surprise this seems to be even true if one takes the random matrix

$$A_{ij}= i \delta_{ij} - X_{ij}\varepsilon$$ where $X_{ij}$ are iid Bernoulli and we enforce no decay of the perturbation for large entries.

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

Playing around with Matlab I noticed something very peculiar:

Take the symmetric matrix $A \in \mathbb R^{n \times n}$ defined by

$$A_{ij}= i \delta_{ij} - \frac{\varepsilon}{\sqrt{i}\sqrt{j}}\,.$$

Here $\delta_{ij}$ is the Kronecker delta.

We first note that this matrix is not diagonally dominant if $n$ is large enough.

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

It is obvious that we require $\varepsilon<1$ in order for $A$ to be positive definite, since otherwise $A_{1,1}\le 0.$

However, I noticed that for let's say $\varepsilon=0.1$ one can make the dimension as large as one wants and the matrix remains positive definite.

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

Playing around with Matlab I noticed something very peculiar:

Take the matrix $A \in \mathbb R^{n \times n}$ defined by

$$A_{ij}= i \delta_{ij} - \frac{\varepsilon}{\sqrt{i}\sqrt{j}}.$$

Here $\delta_{ij}$ is the Kronecker delta.

We first note that this matrix is not diagonally dominant if $n$ is large enough.

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

It is obvious that we require $\varepsilon<1$ in order for $A$ to be positive definite, since otherwise $A_{1,1}\le 0.$

However, I noticed that for let's say $\varepsilon=0.1$ one can make the dimension as large as one wants and the matrix remains positive definite.

To my surprise this seems to be even true if one takes the random matrix

$$A_{ij}= i \delta_{ij} - X_{ij}\varepsilon$$ where $X_{ij}$ are iid Bernoulli and we enforce no decay of the perturbation for large entries.

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

Playing around with Matlab I noticed something very peculiar:

Take the symmetric matrix $A \in \mathbb R^{n \times n}$ defined by

$$A_{ij}= i \delta_{ij} - \frac{\varepsilon}{\sqrt{i}\sqrt{j}}\,.$$

Here $\delta_{ij}$ is the Kronecker delta.

We first note that this matrix is not diagonally dominant if $n$ is large enough.

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

It is obvious that we require $\varepsilon<1$ in order for $A$ to be positive definite, since otherwise $A_{1,1}\le 0.$

However, I noticed that for let's say $\varepsilon=0.1$ one can make the dimension as large as one wants and the matrix remains positive definite.

To my surprise this seems to be even true if one takes the random matrix

$$A_{ij}= i \delta_{ij} - X_{ij}\varepsilon$$ where $X_{ij}$ are iid Bernoulli and we enforce no decay of the perturbation for large entries.

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