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Let $A=(a_{ij})$ be an arbitrary $n \times n$ real symmetric matrix and $n\geq 2$. Let $\| \cdot \|$ denote the operator $2$-norm or equivalently the maximum absolute value of eigenvalues for symmetric matrices. Define

$$d(A) := \min_{D \; \text{is diagonal}} \|A-D\|,$$

as the minimum distance of $A$ from diagonal matrices. What is the best constant $C$ that we can put in the following inequality?

$$d(A) \leq C \max_{i \neq j} |a_{ij}|.$$

The constant $C=n-1$ obviously works, by considering the distance of $A$ from its own diagonal, but it can be shown that it's not the smallest constant for $n>2$.

Let $A=(a_{ij})$ be an arbitrary $n \times n$ real symmetric matrix and $n\geq 2$. Let $\| \cdot \|$ denote the operator $2$-norm or equivalently the maximum absolute value of eigenvalues for symmetric matrices. Define

$$d(A) := \min_{D \text{ is diagonal}} \|A-D\|,$$

as the minimum distance of $A$ from diagonal matrices. What is the best constant $C$ that we can put in the following inequality?

$$d(A) \leq C \max_{i \neq j} |a_{ij}|.$$

The constant $C=n-1$ obviously works, by considering the distance of $A$ from its own diagonal, but it can be shown that it's not the smallest constant for $n>2$.

Let $A=(a_{ij})$ be an arbitrary $n \times n$ real symmetric matrix. Let $\| \cdot \|$ denote the operator $2$-norm or equivalently the maximum absolute value of eigenvalues for symmetric matrices. Define

$$d(A) := \min_{D \; \text{is diagonal}} \|A-D\|,$$

as the minimum distance of $A$ from diagonal matrices. What is the best constant $C$ that we can put in the following inequality?

$$d(A) \leq C \max_{i \neq j} |a_{ij}|.$$

The constant $C=n-1$ obviously works, by considering the distance of $A$ from its own diagonal, but it can be shown that it's not the smallest constant.

Let $A=(a_{ij})$ be an arbitrary $n \times n$ real symmetric matrix and $n\geq 2$. Let $\| \cdot \|$ denote the operator $2$-norm or equivalently the maximum absolute value of eigenvalues for symmetric matrices. Define

$$d(A) := \min_{D \; \text{is diagonal}} \|A-D\|,$$

as the minimum distance of $A$ from diagonal matrices. What is the best constant $C$ that we can put in the following inequality?

$$d(A) \leq C \max_{i \neq j} |a_{ij}|.$$

The constant $C=n-1$ obviously works, by considering the distance of $A$ from its own diagonal, but it can be shown that it's not the smallest constant for $n>2$.

Let $A=(a_{ij})$ be an arbitrary $n\times n$ real symmetric matrix. Let $||.||$ denote the operator 2-norm or equivalently the maximum absolute value of eigenvalues for symmetric matrices. Define $$d(A) = \min_{D \; \mathrm{diagonal}} ||A-D||,$$ as the minimum distance of $A$ from diagonal matrices. What is the best constant $C$ that we can put in the following inequality? $$d(A) \leq C\max_{i\neq j} |a_{ij}|.$$ The constant $C=n-1$ obviously works, by considering the distance of $A$ from it's own diagonal. But it can be shown that it's not the smallest constant.

Let $A=(a_{ij})$ be an arbitrary $n \times n$ real symmetric matrix. Let $\| \cdot \|$ denote the operator $2$-norm or equivalently the maximum absolute value of eigenvalues for symmetric matrices. Define

$$d(A) := \min_{D \; \text{is diagonal}} \|A-D\|,$$

as the minimum distance of $A$ from diagonal matrices. What is the best constant $C$ that we can put in the following inequality?

$$d(A) \leq C \max_{i \neq j} |a_{ij}|.$$

The constant $C=n-1$ obviously works, by considering the distance of $A$ from its own diagonal, but it can be shown that it's not the smallest constant.