For HW, let det(A - \lamda I)=0. Simplify to \lambda^2 - 5\lambda + 3 = 0. Solve for \lambda, we get \lambda = (1/2)(5 \pm \sqrt(13)). Eveidently, both \lambda values are real and >0.
This criterion for positive definiteness in terms of leading submatrices shows up in the Second Derivative Test, at least the way it's usually taught in the US. At a critical point the gradient is 0 so the function is approximated by a quadratic form. The critical point is a local minimum if the quadratic form is positive-definite.
@@brightsideofmathsYeah and I'm trying to understand the motivation for it... If you define it the standard way you can have non symmetric matrices with postive definite quadratic form. So why add selfadjointness?
