Hello,

I am en engineer working in radar research. I came accross a problem I cannot seem to find math literature on it.

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 which 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 eivenvalues 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 $c_{ij}$ are the elements 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 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 contions?

$\min\limits_{\mathbf{v}_1,\ldots,\mathbf{v}_M}$

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

Thank you

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. – Suvrit Apr 25 '12 at 17:27