One thing that is true is this: if A and B are hermitian (or real
symmetric) with all their eigenvalues in [0, a] and [0, b]
respectively, then A B has all its eigenvalues in [0, a b].
A matrix is positive definite if all of its eigenvalues are positive. In other words, when the matrix is multiplied by any non-zero vector, the result will always be a positive number.
To determine if a matrix is positive definite, you can use the Cholesky decomposition method. This involves factoring the matrix into a lower triangular matrix and its transpose. If the resulting lower triangular matrix has positive diagonal elements, then the original matrix is positive definite.
Yes, it is possible for AB-A^2 to be positive definite. This would occur when the eigenvalues of AB-A^2 are all positive.
To prove that AB-A^2 is positive definite, you can use the Sylvester's criterion. This involves checking that all of the leading principal minors of the matrix are positive. If this is true, then the matrix is positive definite.
No, it is not possible for AB-A^2 to be negative definite. This is because the eigenvalues of AB-A^2 would need to be negative, but the matrix cannot have negative eigenvalues if it is positive definite.