Prove that if $A$ is a symmetric matrix with all eigenvalues greater than $0$, then it is positive definite.
If $A$ is symmetric then there exists an orthogonal matrix $S$, such that $S^TAS$ is a diagonal matrix. I'm not sure about this, but if a matrix is diagonalizable, then the eigenvalues of that matrix are equal to the corresponding entries in the diagonal of its diagonalized form, correct? So if $S^TAS$ is diagonal, then the entries in the diagonal equal to the eigenvalues?
Either way, even if they were and assuming they were positive, I could only show that $x^TAx>0$ if $x$ is an eigenvector of $A$, but how would one prove this for all $x \neq 0$?