1
$\begingroup$

I'm trying to assess and prove whether a matrix containing only non-negative values (generated from the standard normal distribution) and diagonal values which are always positive (obtained as a product of the identity matrix and constants: $\alpha$ (size of the matrix)) and $\beta$ (a real number) is positive definite.

The matrix might be composed of two other matrices that easily create these results. For example: A multiplied by its transpose summed with the identity matrix.

$ AA' + \alpha\beta I $

This example would show the matrix to be symmetric, so we also need to show the positive definiteness

Is there some proof relating to the properties that allows a positive definite proof to be created without proving all eigenvalues to be positive for example?

EDIT: Originally forgot to specify that values of the matrix are from the standard normal distribution and there is a multiplier on the diagonal using the product of the length of the matrix and another real parameter

$\endgroup$
0
$\begingroup$

No, this won't necessarily happen, for fairly basic reasons: for example, $$ \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} $$ has negative determinant, so it can't be positive-definite.

$\endgroup$
  • $\begingroup$ Thanks @Chappers, I realised from this basic example I might need to update my question as the identity matrix is multiplied by a constant relating to the size of the matrix and the rest of the values are generated from the standard normal distribution! $\endgroup$ – jfive Jan 7 '16 at 21:11
  • $\begingroup$ Have you met Sylvester's Criterion? That's one simple way of determining if a Hermitian matrix is positive-definite without going all the way to the eigenvalues. $\endgroup$ – Chappers Jan 7 '16 at 21:16
  • $\begingroup$ I have updated the question with the additional information, perhaps now I can use this property I found elsewhere somehow: Namely, AA is positive-definite if all the diagonal entries are positive, and each diagonal entry is greater than the sum of the absolute values of all other entries in the corresponding row/column. See ece.uwaterloo.ca/~dwharder/NumericalAnalysis/04LinearAlgebra/… $\endgroup$ – jfive Jan 7 '16 at 21:21

Your Answer

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

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