MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.