why $||e_i||^2$ implies $\lambda_i\gt 0$ $\forall i$?

Question

I have some questions on the following proof:

Let $S$ be an $n \times n$ symmetric matrix. Then:

$S$ is p.d (positive definite) if and only if every eigenvalue of $S$ is strictly positive $(\gt0)$

Proof:

Suppose $S$ is p.d. Let $\lambda_1,...,\lambda_n$ be the eigenvalues of $S$. Let $e_1,...,e_n$ be the corresponding eigenvectors. Hence,

$$0\lt e^T_iSe_i=e^T_i\lambda_ie_i=\lambda_ie^T_ie_i=\lambda||e_i||^2$$

for every $i$ and since $||e_i||^2\gt 0$ $\Rightarrow$ $\lambda_i\gt0, \forall i.$

Conversely... (and the proof continues)

What I don't understand is why $||e_i||^2$ implies $\lambda_i\gt 0$ $\forall i$?