I am trying to prove that eigenvalues of $A$ are positive iff $\det(A_k)> 0$ for all $k = 1, ..., n$ for a real symmetric matrix $A$ where $A_k$ is the $k \times k$ matrix obtained by deleting the $(k + 1)$ through $n$ row and columns from $A$ for $n = 2$.
Proving this for $k = 1$ is easy, since $\det(A_1) = a > 0$ for $A_1 = a$. $\det(A_1 - \lambda I_1) = a - \lambda = 0$ therefore $\lambda = a > 0$
I am having some difficulty with $k = 2$
I know that $\det(A_2) = ad - bc > 0$ for $A_2 = \begin{pmatrix} a && b \\ c && d \\ \end{pmatrix}$
and that $\det(A_2 - \lambda I_2) = \lambda(a - d - \lambda) + ad - bc = 0$ but I am not sure where to go from there.