Let $A\in \mathbb R^{n \times n},\ A^T=A$ and the eigenvalues $\lambda_i>0$. Then $v^TAv>0$ for every nonzero vector $v$.
I know how to prove the above statement by using the fact that if $A$ is real symmetric then there exists an orthonormal basis $\mathcal B=\{v_1,v_2,....,v_n \}$ that consists of eigenvectors of $A$.
Is there a proof where one doesn't need $\mathcal B$ in order to prove that $A$ is positive definite?