Let $A$ be a real symmetric $n\times n$ matrix, normalized such that $Tr[A]=1$. Define a '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 eigenbasis. If $\{a_i\}$ is an eigenbasis, and $\{\lambda_i\}$ are the corresponding eigenvalues, then $A=\sum_i\lambda_i a_i^Ta_i$, and a Fischer basis $\{b_k\}$ is given by $b_k=\frac{1}{\sqrt{n}}\sum\limits_{j}e^{ \frac{2\pi i}{n}kj}a_j$.

It is easy to check that this is an orthonormal basis and that 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 eigenbasis, 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 two given matrices $A,B$ have a common Fischer basis iff they have a common eigenbasis, i.e. $[A,B]=O$.

The problem is now to generalize that to $n=4$. The condition $[A,B]=O$ is in this case 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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.