Symmetric matrices and eigenvalues 
If the eigenvalues of a symmetric matrix $A$ are greater than 0, show that $v^{\top}Av > 0$ for every $v \ne 0$

I am trying to prove this as follows:
If $v$ is an eigenvector of $A$, then $Av = \lambda v$ and we can rewrite the expression as follows:
\begin{align*}
v^\top A v = v^\top \lambda v= \lambda \left(v^\top v\right)
\end{align*}
As $\lambda > 0$ and $\left(v^\top v\right) = \vert\vert v\vert \vert_2 > 0$ for all $v \ne 0$. Therefore, $v A v > 0$ has to be true.
What if $v$ is not an eigenvector of $A$? is it possible? I know this should be a proof for symmetric matrices to be positive-definite, but how can I strengthen this proof?
 A: By Spectral Theorem, $A$ is orthogonally similar to a diagonal matrix, i.e
$$
P^{T}AP=\pmatrix{\mu_1 \\ & \ddots \\ && \mu_n}
$$
where $\mu_i>0$ is eigenvalue of $A$, and $\space P^{T}P=P^{-1}P=I$.
For any $v$, let $v=Pu$. Then
$$
v^TAv=u^TP^TAPu=\sum_{k=1}^n\mu_ku_k^2>0
$$
A: If $A$ is a real symmetric matrix, then you can form an orthonormal basis for $\mathbb{R}^n$ from its eigenvectors. What happens when you expand $v$ in this basis?
A: @hermes already provided the way forward.  I thought that it might be instructive to see the use of tensor notation herein.  So, here we go ...
Let $U$ be the orthogonal matrix that diagonalizes $A$, such that $UU^T=U^TU=I$ and $(U^TAU)_{ij}=\lambda_i\delta_{ij}$, where $\lambda_i$ is the $i$'th eigenvalue of $A$ and $\delta_{ij}$ is the Kronecker Delta.
Using tensor notation, with summation implicit over repeated indices, we have
$$\begin{align}
(v^TAv)_{in}&=v_iU_{ij}U^T_{jk}A_{k\ell}U_{\ell m}U^T_{mn}v_n\\\\
&=v_iU_{ij}\lambda_j\delta_{jm}U^T_{mn}v_n\\\\
&=v_iU_{im}\lambda_mU^T_{mn}v_n\\\\
&=\lambda_m\left(v_iU_{im}\right)\left(v_nU_{nm}\right)\\\\
&=\sum_{m}\lambda_m\left(v_iU_{im}\right)^2\\\\
&>0
\end{align}$$
as expected!
