Let $A$ be a real symmetric $n\times n$ matrix normalised such that $Tr[A]=1$. Define 'fischer basis' as the basis in which all diagonal elements are equal to $\frac{1}{n}$. The motivation for this terminology is that the fischer information(entropy of diagonal elements) is maximised in this basis.

The problem is to find constraints on two matrices $A,B$ so that they have a common fischer basis, for $n=4$. We need to examine the existance of a fischer basis for a single matrix before we proceed.

For complex spaces, there is an obvious fischer basis for every hermitian matrix, given by the fourier transform of the eigen basis. If ${a_i}$ is the eigen basis, and $\lambda_i$ are the corresponding eigen values, then, $A=\sum_i\lambda_i a_i^Ta_i$

A fischer basis, ${b_k}$ is given by $b_k=\frac{1}{\sqrt{n}}\sum_{j}e^{i\frac{2\pi}{n}kj}a_j$

It is easy to check that this is an orthonormal basis, and the diagonal elements $b_j^TAb_j$ are given by $b_j^TAb_j=\frac{\lambda_1+\lambda_2+\cdots \lambda_n}{n}$

However, the solution is more complex in real spaces. For $n=2$, there is a simple solution. If $a_1,a_2$ form the eigen basis, the unique fischer basis is given by

$b_1=\frac{a_1+a_2}{\sqrt{2}}, b_2=\frac{a_1-a_2}{\sqrt{2}}$

In this case, it is clear that, given two matrices $A,B$ have a common fischer basis iff they have a common eigen basis, i.e, $[A,B]=O$

The problem is now to generalise it to $n=4$. The condition $[A,B]=O$, in this case is sufficient, but not necessary. The problem is to find a necessary and sufficient condition, and a method of finding a common fischer basis, whenever it exists.

Has this problem been addressed before? What are the references available on it?