I am an engineer working in radar research. I came accross a problem on which I cannot seem to find literature. I can ask it in two different ways. Perhaps depending on the reader, the alternative question is easier to answer.

**First way**

Assume I have a real symmetric matrix $\mathbf{C} \in \mathbb{R}^{M \times M}$.

I know its eigenvalues that are non-negative: $\lambda_1, \ldots, \lambda_M$. And The trace of the matrix, i.e., the sum of all eigenvalues is $\lambda_1+\cdots+\lambda_M = M$.

The diagonal matrix of eigenvalues is $\mathbf{\Lambda}$ and the matrix with eigenvectors in its colums is $\mathbf{V}$. The eigendecomposition is then $\mathbf{C} = \mathbf{V} \mathbf{\Lambda} \mathbf{V}^T$.

Also the diagonal of the matrix is all ones, i.e., $\operatorname{diag}(\mathbf{C}) = [1,\ldots,1]$.

Define $$c_\max = \max\limits_{i\neq j}|c_{ij}|$$ where $c_{ij}$ is the element of $\mathbf{C}$ in the $i$-th row and $j$-th column. Given that I can choose $\mathbf{V}$ freely, i.e., any matrix with those eigenvalues, what is the minimum of the maximum of all off-diagonal elements that I can attain (in absolute value)? In other words what is the minimum of $c_\max$?

**Second way**

Given that you have $M$ vectors $\{\mathbf{v}_1, \ldots, \mathbf{v}_M\}$.

They are orthonormal $\mathbf{v}_i^T \mathbf{v}_j = \delta(i-j)$ by standard dot product definition.

They have norm one $\| \mathbf{v}_i \|=1$ by standard dot product definition.

Define the weighted inner product as $\mathbf{v}_i^T \mathbf{\Lambda} \mathbf{v}_j$, where $\mathbf{\Lambda} = \operatorname{Diag}(\lambda_1,\ldots,\lambda_M)$ and $\operatorname{trace}(\mathbf{\Lambda}) = M$.

$\{\mathbf{v}_1,\ldots,\mathbf{v}_M\}$ also have norm one $\|\mathbf{v}_i\|_w=1$ by this new weighted inner product definition.

What is then the minimum value for the maximum inner product (in absolute value) among all vectors $\{\mathbf{v}_1,\ldots,\mathbf{v}_M\}$ given they can be chosen freely as far as they satisfy the conditions?

$$\min\limits_{\mathbf{v}_1,\ldots,\mathbf{v}_M} \left(\max\limits_{i\neq j}(\mathbf{v}_i^T\mathbf{\Lambda}\mathbf{v}_j)\right)$$

identity matrixsatisfies your hypothesis. Also, since the matrix that you have is a correlation matrix (positive semidefinite, ones on diagonal), we know that the off-diagonals must be bounded in absolute value by 1. $\endgroup$ – Suvrit Apr 25 '12 at 17:27