Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a quick way to determine if a $2\times 2$ matrix, $M\in M_2(\mathbb R)$, is congruent to $I_2$ over $\mathbb R, \mathbb C, \mathbb Q$? Without explicitly finding the matrices $P\in M_2$ s.t. $I_2=P^TMP$?

Brainstorm: Perhaps usig ranks and signatures? But if I use that how would I determine over what field is the congruence?

share|cite|improve this question
The matrix is symmetric to begin with and since the congruence transformation does not change the signature, it must be moreover positive definite. Hence you can perform a Cholesky decomsposition (not sure about $\mathbb{Q}$) Therefore, with some assumptions all positive definite $M$ are congruent to $I$. – user13838 Nov 29 '11 at 21:28
@percusse: Thanks! If the matrix has a different signature i.e. not 2, does that mean that it is not congruent to The identity matrix over ANY field? e.g. $\mathbb C$? i am asking this because Sylvester's law only talks about reals or at least the version I was told of... – alec Nov 30 '11 at 0:08
Since the matrix $M$ is symmetric or Hermitian its eigenvalues are real and Cholesky is possible at each case. And the law of inertia is valid for Hermitian matrices too. I don't know for over the rationals. – user13838 Nov 30 '11 at 0:26

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.