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

I want to determine some structures of matrices that can be transformed into a symmetric matrices using similarity transformation, i.e.,


where $T$ is the similarity transformation matrix. Here, $A$ is a non-negative matrix and the diagonal elements of $A$ is supposed to be the same $(diag(A)=[a, a, \ldots , a])$.

Any ideas?

share|cite|improve this question
The diagonal matrix in question belongs to the center of $GL(n,R)$ and the question itself does not belong to MO. Voting to close. – Misha Jun 4 '13 at 18:21
The matrix $A$ is not necessarily a diagonal matrix. The question may not be central, but still is reasonable. – Dietrich Burde Jun 4 '13 at 19:40
Oh, sorry, I misread the question. – Misha Jun 4 '13 at 21:46
up vote 2 down vote accepted

We can say something on such matrices $B$ by characterizing its eigenvalues, which coincide with the eigenvalues of $A$. Since $A$ is a real symmetric matrix, it has real eigenvalues. Hence a necessary condition on $B$ is that it has real eigenvalues. But one can say more. There are necessary and sufficient conditions known for real numbers $\lambda_1,\ldots ,\lambda_r$ to be the eigenvalues of a non-negative symmetric matrix. See for example the paper of M. Fiedler, Eigenvalues of nonnegative symmetric matrices here:

Update: We still have the additional condition that the diagonal of $A$ is $(a,a,\ldots ,a)$, but this is also discussed in the paper of Fiedler. Given $2n$ real numbers $\lambda_i$, one can say whether there is a nonnegative symmetric matrix with eigenvalues $\lambda_1,\ldots ,\lambda_n$ and diagonal $\lambda_{n+1},\ldots ,\lambda_{2n}$.

share|cite|improve this answer

An addition to Dietrich Burde's answer: Since symmetric matrices are diagonizable, they are semisimple (each invariant subspace has an invariant complement). Thus $B$ has to be semisimple ($\iff$ the minimal polynomial is square free).

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.