A real symmetric matrix $A$ positive definite if all its eigenvalues are positive 
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?
 A: Let $A$ be a $n \times n$ real symmetric matrix with eigenvalues $\lambda_1,\lambda_2,\dots ,\lambda_n > 0$. Then $A$ is orthogonally diagonalizable, that is, $A = QDQ^T$ for some orthogonal matrix $Q$ and diagonal matrix $D$. Now if we consider $f(x) = v^TAv$, then \begin{align*}
f(v) & = v^TAv \\
& = v^TQDQ^Tv \\
& = (Q^Tv)^TD(Q^Tv) \\ 
& = w^TDw \\
& = \lambda_1 {w_1}^2 + \dots \lambda_n {w_n}^2 \\
& > 0.
\end{align*}
A: This was answered in the MO thread "real symmetric matrix has real eigenvalues - elementary proof". See in Alexandre Eremenko's answer and its subsequent comments. It suffices to show that the minimum value of the Rayleigh quotient $v^TAv/v^Tv$ on $\mathbb R^n\setminus0$ is an eigenvalue of $A$, hence positive by assumption.
The main idea of the proof is that if $\lambda$ is the minimum and $v$ is a minimiser, we must have
$$
(v+u)^TA(v+u)\ge\lambda(v+u)^T(v+u)\tag{1}
$$
for every vector $u$, meaning that
$$
2u^T(Av-\lambda v)\ge\lambda u^Tu-u^TAu\tag{2}
$$
because $A$ is symmetric. Since the LHS is linear in $u$ (so that you can control its sign if it is nonzero) but the RHS of $(2)$ is $o(\|u\|)$ when $\|u\|$ is small, $Av-\lambda v$ must be zero. Hence $\lambda$ is an eigenvalue.
